Elementary Algebra - MathHands

CONTENTS

4 Polynomials 4.1 Adding & Subtracting Polynomials 4.2 Multiply Monomials & Binomials . 4.3 Multiplying Famous Polynomials . 4.4 Dividing by a Monomial . . . . . . 4.5 Dividing by Polynomials ii . . . . . 4.6 Factor Polynomials by DL . . . . . 4.7 Factor By Grouping . . . . . . . . 4.8 Factor by Splitting . . . . . . . . . 4.9 Factor Famous Polynomials . . . . 4.10 Chapter Review . . . . . . . . . . .

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Bibliography

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Index

85

1

2

CONTENTS

CHAPTER 4

POLYNOMIALS We concluded chapter 3 with a modest discussion of solving linear equations, a very important milestone. In this chapter, we expand on that or more accurately we begin sharpening our tools so that we can expand on that, and solve not just linear equations, but many other types of equations as well. Roughly speaking, this involves being comfortable when working with ’polynomials’. Recall, chapter 2 was essentially a chapter on integers, that is positive and negative numbers, and how to add, subtract, and multiply them. In chapter 3 we did the similar type of work but for fractions. A good way to think about this chapter is that polynomials are to this chapter, what rationals were to chapter 3, or what integers were to chapter 2. Once we get comfortable with polynomials, we will begin to solve much more interesting equations, than the [linear] ones we solved at the end of chapter 3.

1

2

CHAPTER 4. POLYNOMIALS

4.1

Adding & Subtracting Polynomials

Let us begin by defining a ’monomial’. Definition A monomial in x with real coefficient is defined as an expression that can be written as Axn Where A represents some constant real number, x is a variable, and n is a whole number. While x is called the variable, n is called the degree of the term, and A is called the coefficient. Another word for a monomial is a term. We could very easily make variations of this definition. For example: Definition A monomial in x with integer coefficient is defined as an expression that can be written as Axn Where A represents some integer number, x is a variable, and n is a whole number. The sum or difference of two monomials is called a binomial , for example: 3x2 + 5x The sum of difference of three monomials is called a trinomial. Generally, we define, Definition A polynomial is defined as the sum or difference of a collection of one or more monomials. For example, x5 + 2x3 − 5

is a polynomial. Often, it is convenient to name the polynomials. In the name we may often include some information about the polynomial, such as what is the variable. For example we could name the x5 + 2x3 − 5 as simply p(x), meaning its name is p and the variable is identified as x. Thus we could write p(x) = x5 + 2x3 − 5 Subsequently, we could just refer to x5 + 2x3 − 5 as simply p(x), under the agreement that these are interchangeable. Meanwhile we could have a different polynomial with a different name, such as h(x) = x5 + x + 2

4.1. ADDING & SUBTRACTING POLYNOMIALS

3

Then when we write p(x) everyone should concur we are referring to x5 + 2x3 − 5 while if we mention h(x) everyone should think of x5 + x + 2 If we restrict the possible coefficients to only integers, and the possible variables to only x’s, then we can call then entire set of such polynomials Z[x]. Analogously, the set Q[t] represents all polynomials with only rational coefficients and only t variables, while R[x] represents the set of all polynomials in x with √ real coefficients. Moreover, we often abbreviate, instead of ”3x2 + 5x + 3 is a polynomial with real coefficients” we write: 3x2 + 5x +

√ 3 ∈ R[x]

where the symbol ∈ is read as ’is an element of..’. Thus. 3x2 + 5x + can be read as ”3x2 + 5x +

√

√ 3 ∈ R[x]

3 is an element of the set of real polynomials”

Definition If q(x) is a polynomial, we define the degree of polynomial q(x) as the highest of the degrees of any of its terms. If q(x) is a constant we say that its degree is zero. Examples 4x5 + 82x + 9

(degree for this polynomial is 5)

4x9 + 27x + 9

(degree for this polynomial is 9)

5x3 + 9 + 27

(degree for this polynomial is 3)

12

(degree for this polynomial is 0)

12 + x2 + 5x

(degree for this polynomial is 2)

Now that we have some of the basic vocabulary tackled, we are ready to start adding/subtracting polynomials. Example Add (3x2 + 4x + 1) + (x2 − 4x + 5)

4

CHAPTER 4. POLYNOMIALS solution: (3x2 + 4x + 1) + (x2 − 4x + 5)

= (3x2 + 1x2 ) + (4x + −4x) + (1 + 5) 2

= (3 + 1)x + (4 + −4)x + (1 + 5)

(ALA, CoLA) (DL)

2

(BI)

2

(0MT)

2

(AId)

= 4x + 0x + 6 = 4x + 0 + 6 = 4x + 6

It should be noted that based on the above example, it should be clear that the underlying principle in adding polynomials is to use ALA and CoLA to group associate like terms together, and like terms for us will mean same degree. That is, we collect degree two terms together, degree one terms together and so on. Then we can use DL to factor out the variable in effect just adding the coefficients. This should be reminiscent of adding natural numbers as you did in grade-school where you collect units, collect tens, and collect 100’s more or less. Such as

+

4825 5307 10132

Based on this idea and the the above comments, we can alternatively add polynomials the ’Kindergarden’ [KG] way such as:

3x2 + 4x + 1 + x2 − 4x + 5 − − − − − − − − −− 4x2 + 6

Example Subtract (3x2 + 4x + 1) − (x2 − 4x + 5)

(KG)

5

4.1. ADDING & SUBTRACTING POLYNOMIALS solution: (3x2 + 4x + 1) − (x2 − 4x + 5)

= (3x2 + 4x + 1) + −1 · (x2 + −4x + 5) 2

(def a-b)

2

= (3x + 4x + 1) + −1x + −1 · −4x + −1 · 5 2

(DL)

2

= 3x + 4x + 1 + −1x + 4x + −5 2

2

= (3x + −1x ) + (4x + 4x) + (1 + −5)

(DL) (ALA, CoLA)

2

= (3 + −1)x + (4 + 4)x + (1 + −5) 2

= 2x + 8x + −4

(DL) (BI)

OR we can try the KG method. . 3x2 + 4x + 1 −

x2 − 4x + 5

− − − − − − − − −− 2x2 + 8x − 4

4.1.1

Exercises

1. Compute and simplify:

1 − r4 + 7r6 + 2r5 + 6r4 + 5r3 + 3r2 + 2r + 4 2. Compute and simplify: 2q 3 + 4q + 2 + 2q 4 − 4q 3 + q 2 + 7q

3. Compute and simplify: 4w6 + 6w5 + 7w4 + w2 − 5w + 3
− −w6 + 5w5 + 5w3 + 5w − 4

(KG)

6

CHAPTER 4. POLYNOMIALS

4. Compute and simplify: −

−3

(−5y − 5)

5. Compute and simplify:

−3w2 − 2w − 1 + 3w4 + 3w3 − 2w2 + 6w + 3 6. Compute and simplify:

4r3 + 2r2 + 4 + 3r3 − 4r2 − 2 7. Compute and simplify:

−

− 3r6 + 7r3 − 5r2 + 7r
−4r5 − 5r4 + r

8. Compute and simplify:

2t3 + 7t + 4 + 6t3 − 2t2 + 3t 9. Compute and simplify:

−

− y 4 + y 3 + 2y 2 − 4y + 5
3y 5 − y 4 + 7y 3 + 7

10. Compute and simplify: 7r6 − 5r5 + 5r4 − r2 + 5r − 4
− r6 + 6r3 − 5r 11. Compute and simplify: 3r5 + 7r4 + 5r2 + 2 +

− 3r5 + 3r4 + r3 + r2 + 7r + 5

4.1. ADDING & SUBTRACTING POLYNOMIALS 12. Compute and simplify: (3q − 5) + (6q − 2) 13. Compute and simplify:

−

− 4v 4 − 3v 3 − 3v 2 − 5v
−3v 4 − v 3 + 5v

14. Compute and simplify:

−

4q 6 − 5q 3 + 4q − 5

−2q 6 − q 5 + 7q 3 + 2




15. Compute and simplify: (0) + (3t) 16. Compute and simplify:

−4z 2 − 5z + −2z 5 − z 4 − 4z 3 + 5z 2 − z + 7 17. Compute and simplify: 5v 2 − 3

+ 4v 4 + 6v 3 − 2v 2 − 3v + 7 18. Compute and simplify: 2y 2 + 3y + 2 + 6y 3 + 6y 2 − y 19. Compute and simplify: 1 −

(w + 2)

7

8

CHAPTER 4. POLYNOMIALS

20. Compute and simplify: 7v 2 + 7v +

6v 2

9

4.2. MULTIPLY MONOMIALS & BINOMIALS

4.2

Multiply Monomials & Binomials

At the heart of multiplication of polynomials is the distributive law. We begin with the easiest case, namely a multiplication of a monomial times a polynomial. Example Multiply (3x3 )(5x2 + 4x − 1) solution: (3x3 )(5x2 + 4x − 1)

= (3x3 )(5x2 + 4x + −1) 3

2

3

(def a-b) 3

= (3x )5x + (3x )4x + (3x )(−1) 3 2

3

(DL) 3

= (3 · 5)x x + (3 · 4)x x + (3 · −1)x 5

4

3

= 15x + 12x + −3x

(ALM, CoLM) (BI)

Example Multiply (−2x)(5x2 + 4x − 1) solution: (−2x)(5x2 + 4x − 1)

= (−2x)(5x2 + 4x + −1) 2

= (−2x)5x + (−2x)4x + (−2x)(−1) 2

= (−2 · 5)x · x + (−2 · 4)x · x + (−2 · −1)x 3

2

= −10x + −8x + 2x

Example Multiply (−2x + 1)(5x2 + 4x − 1)

(def a-b) (DL) (ALM, CoLM) (BI)

10

CHAPTER 4. POLYNOMIALS

solution: (−2x + 1)(5x2 + 4x − 1)

= (−2x + 1)(5x2 + 4x + −1)

(def a-b)

2

= (−2x + 1)5x + (−2x + 1)4x + (−2x + 1)(−1) 2

(DL)

2

= (−2x)5x + (1)5x + (−2x)4x + (1)4x + (−2x)(−1) + (1)(−1) 3

2

= −10x + 5x + −8x + 4x + 2x + −1 3

(DL)

2

2

(DL)

2

= −10x + (5x + −8x ) + (4x + 2x) + −1 3

(ALA)

2

= −10x + (5 + −8)x + (4 + 2)x + −1 3

(DL)

2

= −10x + −3x + 6x + −1

(BI)

Alternative solution: Multiplying using the KinderGarden Method: (−2x + 1)(5x2 + 4x − 1) =

+

5x2 + 4x + −1 ×(−2x + 1) 5x2 + 4x + −1

− 10x3 − 8x2 + 2x

−10x3 + −3x2 + 6x + −1

4.2.1

Exercises

1. Compute and simplify using KG: u4 + 6u2 + 3u − 5

×

7u2 − 5u + 5

2. Compute and simplify using KG: −5w5 − 4w3 + 2w ×

6w6

(KG M)

11

4.2. MULTIPLY MONOMIALS & BINOMIALS 3. Compute and simplify use DL: −2z 7 + 3z 6 + 7z




−2z 4




4. Compute and simplify using KG: y 4 + 2y 3 + 7y 2 + 5y 5y 2 + 4y − 2

×

5. Compute and simplify using KG: w6 − 5w4 + 7w3 ×

7w6

6. Compute and simplify using KG: 4z 3 − 4z 2 − 5z + 7

×

3z 2 − 2z − 4

7. Compute and simplify using KG: −w3 + 5w2 − 3w − 4 × 4 − 2w

8. Compute and simplify using KG: −4v 5 + 6v 3 + 3v 2 ×

3v 5

12

CHAPTER 4. POLYNOMIALS

9. Compute and simplify using KG: 3u4 + 4u2 − 2u − 1

u2 − 5u − 2

×

10. Compute and simplify use DL: t2 + 5t − 5




t5 − 2t2




11. Compute and simplify using KG: −4t3 − 5t2 − t − 2 ×

6t2 + 3t

12. Compute and simplify using KG: −3r3 + 7r2 + 1 7r3 − 5r2

×

13. Compute and simplify using KG: −2z 2 + z + 4

×

2z 3 − 2z

14. Compute and simplify using KG: 4r5 + r3 − 5r × −r

13

4.2. MULTIPLY MONOMIALS & BINOMIALS 15. Compute and simplify using KG: 5w4 + 7w3 + 6w + 3 ×

− 4w2 + 4w + 3

16. Compute and simplify use DL:
−y 5 − 5y 3 − 3y 2 (5y) 17. Compute and simplify use DL: 6z 5 + 4z 3 − 1




5z 7




18. Compute and simplify using KG: 6z 3 − 4z 2 − 5z 6z 3 − 2z

×

19. Compute and simplify use DL: 3y 2 + y − 4




−3y 3 − 2y




20. Compute and simplify using KG: 6y 4 + y + 4 ×

− 4y 2

14

4.3

CHAPTER 4. POLYNOMIALS

Multiplying Famous Polynomials

In the previous section we introduced the kindergarten method for multiplying polynomials. In this section, we introduce a list of very famous polynomials. In particular, we want to prove, understand, and own every one of these famous products of polynomials. We will see these again, and it will be essential that we recognize them when seen forwards and backwards. Moreover, we introduce a very special kind of multiplication, namely the type when we multiply a binomial times another binomial. For this type of multiplication we introduce, as an option, the FOIL method. As we shall soon see, the FOIL method is simply an abbreviated version of a the distributive law applied a couple times. Formally, and without further ado: Theorem 4.3.1. FOIL Suppose a, b, c, d are numbers or variables for which our axioms apply, then: (a + b)(c + d) = ac + ad + bc + bd

[FOIL]

The name is intended to help one remember the four terms on the right. First we have the first terms from each of the binomials on the left, the a and c are each the First terms thus the F , the the Outter, the Inner, and the Last terms, thus the FOIL acronym. Outer First

(a + b)(c +

d)

Inner Last F

O

I

L

z}|{ z}|{ z}|{ z}|{ = ac + ad + bc + bd Proof. (a + b)(c + d) = a(c + d) + b(c + d) = a·c+a·d+b·c+b·d

(DL) (DL)

15

4.3. MULTIPLYING FAMOUS POLYNOMIALS Example (3x + 2)(4x + 1) = (3x)(4x) + (3x)(1) + (2)(4x) + (2)(1)

(FOIL)

2

(BI)

2

(BI)

= 12x + 3x + 8x + 2 = 12x + 11x + 2 One more time... Example

(3x + 2)(5x2 + 7x) = (3x)(5x2 ) + (3x)(7x) + (2)(5x2 ) + (2)(7x) (FOIL) = 15x3 + 21x2 + 10x2 + 14x 3

(BI)

2

= 15x + 31x + 14x

(BI)

It should be noted that the FOIL method is an option for the case when we are multiplying binomial times a binomial. It becomes less relevant when we are multiplying trinomials or monomials. Meanwhile, the methods from the last section, DL and KG are good for all multiplication of all polynomials from out class. Next we turn our attention to some very famous polynomial products. We will include here some of the proofs, while some of these and leave the others as important exercises. For all of these, we assume our variables are ones for which our axioms apply. Theorem 4.3.2. Difference of Squares [DS] (a − b)(a + b) = a2 − b2 Theorem 4.3.3. Difference of two Cubes [DC] (a − b)(a2 + ab + b2 ) = a3 − b3 Proof. (a − b)(a2 + ab + b2 ) =

+

a2 + ab + b2

(KG M)

×(a + −b)

−a2 b + −ab2 + −b3

a3 + a2 b a3 +

= a3 − b 3

0+

+ ab2 0+

−b3

(BI)

16

CHAPTER 4. POLYNOMIALS

Theorem 4.3.4. Difference of two Cubes [SC] (a + b)(a2 − ab + b2 ) = a3 + b3 To prove SC and each of the other polynomials here one may proceed by using DL or KG. While we do a couple of these here, it is most important and beneficial for these modest challenges to be taken on by the reader. Most of these appear as important exercises. Theorem 4.3.5. Pascal Polynomials • (a + b)2 = a2 + 2ab + b2 3

3

2

[PP#2] 2

• (a + b) = a + 3a b + 3ab + b

3

[PP#3]

• (a + b)4 = a4 + 4a3 b + 6a2 b3 + 4ab4 + b4

[PP#4]

• (a + b)5 = a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5

[PP#5]

Theorem 4.3.6. Geometric Series Polynomials • (x − 1)(x + 1) = x2 − 1

[GS#2]

• (x − 1)(x2 + x + 1) = x3 − 1 3

2

[GS#3] 4

• (x − 1)(x + x + x + 1) = x − 1

[GS#4]

• (x − 1)(x4 + x3 + x2 + x + 1) = x5 − 1

[GS#5]

As an example we can prove GS#5, and we can prove it at least a couple ways, by KG method or by using DL a couple times. Here is the proof using KG. Proof. x4 + x3 + x2 + x + 1 ×

x−1

−x4 − x3 − x2 − x − 1

x5 + x4 + x3 + x2 + x

x5 − 1

4.3. MULTIPLYING FAMOUS POLYNOMIALS

17

Theorem 4.3.7. General Geometric Series Polynomials • (x − y)(x + y) = x2 − y 2

[GGS#2]

• (x − y)(x2 + xy + y 2 ) = x3 − y 2

[GGS#3]

• (x − y)(x3 + x2 y + xy 2 + y 3 ) = x4 − y 4

[GGS#4]

• (x − y)(x4 + x3 y + x2 y 2 + xy 3 + y 4 ) = x5 − y 5

[GGS#5]

4.3.1

Exercises

1. State and prove the famous polynomial product, [GGS2] 2. Compute and simplify the following product, using FOIL: (7w + 5) (5 − 4w) 3. Compute and simplify the following product, using FOIL: (3t + 2) (4t − 5) 4. Use [PP2] to computing and simplify the following: 2

(v + 1)

5. prove [PP3] by computing and simplifying the following: 3

(w + y)

6. Compute and simplify the following [famous] product: (β + 1) (β − 1) 7. Compute and simplify the following product, using FOIL: (3q − 1) (3 − q) 8. Use [PP2] to computing and simplify the following: 2r3 + 3


2

18

CHAPTER 4. POLYNOMIALS

9. Compute and simplify the following product, using FOIL: (5z + 1) (7z + 4) 10. Compute and simplify the following [famous] product: 5u2 + u




5u2 − u




11. prove [PP3] by computing and simplifying the following: (r + z)3 12. Use [PP2] to computing and simplify the following: 4r3 + 7r


2

13. Prove [PP2] by computing and simplify the following: (θ + φ)

2

14. Prove [PP2] by computing and simplify the following: (β + φ)

2

15. Compute and simplify the following product, using FOIL: 6q 4 − q 3




7q 6 − 5q 7




16. Use [PP3] to computing and simplify the following: 3

(4 − y)

17. Use [PP3] to computing and simplify the following: 3

(4 − z)

18. Use [PP2] to computing and simplify the following: 7v 3 + 2


2

4.3. MULTIPLYING FAMOUS POLYNOMIALS 19. Use [PP2] to computing and simplify the following:
2 3 − 4t3 20. State and prove the famous polynomial product, [GS2] 21. Use [PP2] to computing and simplify the following: 2

(3y + 1)

22. Compute and simplify the following product, using FOIL: (2v + 5) (4 − 5v) 23. Compute and simplify the following [famous] product: 2q 6 + 4q 2




2q 6 − 4q 2




24. Compute and simplify the following [famous] product: 7z 5 − z 4




7z 5 + z 4




25. Compute and simplify the following [famous] product: 11z 5 + 4z 2




11z 5 − 4z 2




26. Compute and simplify the following product, using FOIL: (−4u − 3) (−5u − 3) 27. State and prove the famous polynomial product, [GS2] 28. Compute and simplify the following [famous] product: (11v + 5) (11v − 5) 29. Prove [PP2] by computing and simplify the following: 2

(β + ω)

30. Compute and simplify the following product, using FOIL: 5y 5 − y 7




7y 6 − y 7




19

20

CHAPTER 4. POLYNOMIALS

31. Compute and simplify the following [famous] product: (µ + 3) (µ − 3) 32. Use [PP2] to computing and simplify the following: 6q − 2q 3


2

33. Compute and simplify the following [famous] product: (µ + 1) (µ − 1) 34. Compute and simplify the following product, using FOIL: (3 − 4r) (1 − r) 35. Prove [DC] by computing and simplify the following: (µ + φ) (φ − µ)

4.4. DIVIDING BY A MONOMIAL

4.4

21

Dividing by a Monomial

We will now learn a basic methods used to divide polynomials. The task can be divided into two categories, dividing by a monomial, and dividing my a non-monomial. We will practice dividing by a monomial in this section and leave the dividing by a general polynomial for the next section. We do not need to learn any new axioms or definition, as all the necessary axioms and definition which apply to the real numbers apply to polynomials. We have harnessed all the necessary ideas already to the point that all left to do is to take a look a couple examples. Once you become good at looking at the example you will be ready to practice doing some of these yourself in the provided exercises.

Example divide 6x4 ÷ 2x

6x4 2x 3 · 2x3 x1 = 2x 3x3 · 2x = 1 · 2x 3x3 2x = 1 2x 3x3 = 1 = 3x3

6x4 ÷ 2x =

(def ÷) (JAE,TT) (CoLM, N-Expo, Mid) (MAT) (JOT,Mid) (OUT)

Notice, this process is nearly identical to the work we did in simplifying rational numbers, where we prime factorized each the numerator & denominator and we eliminated the gcd, ie the common prime pieces. Let’s try it one more time.

Example divide 16x4 ÷ 2x3

22

CHAPTER 4. POLYNOMIALS

16x4 2x3 8 · 2x1 x3 = 1 · 2x3 8x · 2x3 = 1 · 2x3 8x 2x3 = 1 2x3 8x ·1 = 1 = 8x

16x4 ÷ 2x3 =

(Def ÷) (JAE,TT) (CoLM, +Expo) (MAT) (JOT) (OUT,Mid)

Next, we turn our attention to diving a general polynomial by a monomial.

Example divide (4x3 + 10x2 + 8x + 1) ÷ 2x2

(4x3 + 10x2 + 8x + 1) ÷ 2x2

4x3 + 10x2 + 8x + 1 2x2 3 4x 10x2 8x 1 = 2+ + 2+ 2 2x 2x2 2x 2x 4 1 = 2x + 5 + + 2 x 2x

=

Example divide (6x3 + 4x2 + 1) ÷ 2x

(def ÷) (ATT) (BI)

23

4.4. DIVIDING BY A MONOMIAL

6x3 + 4x2 + 1 2x 4x2 1 6x3 + + = 2x 2x 2x 2x · 3x2 2x · 2x 1 = + + 2x · 1 1 · 2x 2x 2x 2x 1 2x 3x2 + + = 2x 1 1 2x 2x 2x 1 3x2 + ·1+ =1· 1 1 2x 1 = 3x2 + 2x + 2x

(6x3 + 4x2 + 1) ÷ 2x =

(Def a/b, ÷) (ATT) (JAE,CoLM,BI) (MAT) (JOT) (Mid, OUT)

1 . Alternatively, we say We can conclude (6x3 +4x2 +1)÷2x = 3x2 +2x+ 2x 3 2 2 (6x + 4x + 1) ÷ 2x = 3x + 2x with reminder 1. This method used the idea that every division problem can be expressed as a fraction, while making use of our fraction skills. While it has its virtues, it also has drawbacks. This method may not be very useful when we try to divide polynomial by polynomial. Ultimately, we will have to resort to long division, which is king in the world of Q[x] under division. Here is a second look at the same problem above, solved using long division.

Example divide (6x3 + 4x2 + 1) ÷ 2x First we set it up as, we do when we divide integers using long division,
2x

6x3 + 4x2 + 1

then the idea is to see howmany times 2x goes into the leading term 6x3 . 3 In other words we calculate 6x 2x . We can do this by inspection [BI] or as 3 2 example 1, above. In either case, we conclude 6x 2x = 3x . This becomes the first part of the quotient. 3x2
2x

6x3 + 4x2 + 1

It is customary to try to keep columns ordered by degree. Observe the 3x2 was placed on the degree 2 column. The next step, as with integers, is to multiply the 3x2 by the divisor 2x and subtract it from the leading term 6x3 . The result is shown,

24

CHAPTER 4. POLYNOMIALS 3x2
2x

6x3 + 4x2 + 1 − 6x3

We subtract and bring down the next term 4x2 to obtain: 3x2
2x

6x3 + 4x2 + 1 − 6x3 4x2

We now ask how many times will 2x go into 4x2 ? That is, we find 2x. This becomes the second part of the quotient. We obtain

4x2 2x

=

3x2 + 2x
2x

6x3 + 4x2 + 1 − 6x3 4x2

The next step is to multiply the 2x in the quotient by the divisor, 2x to obtain 4x2 which we write down below and subtract from 4x2 to obtain 3x2 + 2x
2x

6x3 + 4x2 + 1 − 6x3 4x2 − 4x2

1 Finally, we bring down the 1. Since 2x does not divide 1 evenly ( 2x does not reduce) it is left as a remainder. We conclude with the expected, 1 (6x3 + 4x2 + 1) ÷ 2x = 3x2 + 2x + 2x OR (6x3 + 4x2 + 1) ÷ 2x = 3x2 + 2x with reminder 1.

A couple more examples are in order Example divide (3x3 + 15x2 − 6x − 12) ÷ (3x)

25

4.4. DIVIDING BY A MONOMIAL


3x

x2 + 5x − 2

3x3 + 15x2 − 6x − 12 − 3x3 15x2 − 15x2

− 6x 6x

Finally we bring down the -12 and find the remainder is -12, thus (3x3 + − 12 15x2 − 6x − 12) ÷ (3x) = x2 + 5x − 2 + 3x Example (3x3 + 5x2 − 16x − 2) ÷ (3x)
3x

x2 + 35 x −

16 3

3x3 + 5x2 − 16x − 2 − 3x3 5x2 − 5x2 − 16x 16x

Finally we bring down the (remainder) -2 and conclude, by Long Division − − [LD] (3x3 + 5x2 − 16x − 2) ÷ (3x) = x2 + 35 x + 316 + 3x2

4.4.1

Exercises

1. Divide using Long Division:
36y 4 − 48y 3 − 30y 2 − 18y + 27 ÷ (3y) 2. Divide using Long Division:
−24t4 + 42t3 + 30t2 − 36t + 9 ÷ (−t) 3. Divide using [ATT] and [def ÷]:

−2q 9 + 8q 8 − 20q 2 ÷ 3q 2

26

CHAPTER 4. POLYNOMIALS

4. Divide using Long Division:
6u4 + 51u3 + 36u2 + 24u + 54 ÷ (2u) 5. Divide using Long Division:
51y 4 + 18y 3 + 36y 2 − 48y − 12 ÷ (−y) 6. Divide using [ATT] and [def ÷]:

23w8 − 2w6 + 11w5 ÷ −2w3 7. Divide using Long Division:
3u4 − 39u3 + 33u2 − 30u − 9 ÷ (2u) 8. Divide using [ATT] and [def ÷]:

−17y 6 − 22y 4 − 12y 3 ÷ −2y 3 9. Divide using Long Division:
−6y 4 + 48y 3 + 60y 2 + 39y − 33 ÷ (y) 10. Divide using [ATT] and [def ÷]:

−26v 5 + 11v 4 + 14v 2 ÷ −2v 2 11. Divide using Long Division:
45y 4 − 12y 3 − 24y 2 + 51y − 36 ÷ (−3y) 12. Divide using [ATT] and [def ÷]:

−20v 7 + 15v 5 − 18v 3 ÷ −3v 2 13. Divide using Long Division:
12x4 − 57x3 + 21x2 − 30x + 6 ÷ (−2x)

4.4. DIVIDING BY A MONOMIAL 14. Divide using [ATT] and [def ÷]:
2z 8 − 6z 7 + 26z 6 ÷ (3z) 15. Divide using Long Division:
−33x4 + 24x3 − 30x2 − 57x + 42 ÷ (3x) 16. Divide using Long Division:
−3t4 + 48t3 + 27t2 − 18t + 54 ÷ (−3t) 17. Divide using Long Division:
−48y 4 − 57y 3 + 12y 2 − 15y − 27 ÷ (3y) 18. Divide using [ATT] and [def ÷]:

−22r10 + 15r9 − 12r6 ÷ −2r3 19. Divide using [ATT] and [def ÷]:
26t7 + 13t5 + 23t3 ÷ (−3t) 20. Divide using Long Division:
−15u4 − 33u3 + 42u2 + 36u + 60 ÷ (−u) 21. Divide using [ATT] and [def ÷]:

−v 9 + 19v 7 − 14v 2 ÷ −v 3 22. Divide using Long Division:
33u4 − 27u3 − 30u2 − 60u − 54 ÷ (−u) 23. Divide using Long Division:
−27y 4 − 42y 3 + 18y 2 + 30y + 21 ÷ (2y)

27

28

CHAPTER 4. POLYNOMIALS

24. Divide using [ATT] and [def ÷]:

−11y 8 + 3y 5 − 22y 2 ÷ 3y 3 25. Divide using Long Division:
15x4 + 21x3 − 39x2 − 9x − 3 ÷ (−2x)

4.5. DIVIDING BY POLYNOMIALS II

4.5

29

Dividing by Polynomials ii

Last section we divided polynomials by monomials exclusively. We now tackle the general problem of dividing by a binomial or any polynomial. Consider dividing using long division 2 ÷ 5. If you think about it for a second you will realize the quotient is 0 with a remainder of 2. This will usually happen when trying to divide an small integer by a larger one. The same holds for polynomials. You will not get very far if you are trying to divide polynomial of small degree by a polynomial of large degree. The quotient will be 0 with remainder equal to the remainder being the original polynomial. For example, (5x + 1) ÷ x3 = 0 with remainder 5x + 1 OR 5x+1 x3 . Therefore, we will practice our diving skill mostly with cases where we divide a polynomial of large degree by one of smaller degree. The Long Division method will be the primary tool. Example Divide (3x3 + 5x2 − 6x − 2) ÷ (x − 1) Most steps are identical. One small difference is that we always concentrate on the leading terms, momentarily ignoring the other term. For example, we just concentrate on the leading term 3x3 inside and x, the lead3 ing term inside. We calculate by inspection 3xx = 3x2 to obtain our first term of the quotient. 3x2 x−1




3x3 + 5x2 − 6x − 2

Now, we multiply the 3x2 by the entire divisor (x − 1), thus we have to use DL, to get 3x2 (x − 1) = 3x3 − 3x2 . This is the quantity we subtract (i.e. change the sign). We obtain... 3x2 x−1




3

3x + 5x2 − 6x − 2 − 3x3 + 3x2

Since we have changed the signs already (to subtract), we can simply add the terms and bring down the next term to obtain, 3x2 x−1




3x3 + 5x2 − 6x − 2 − 3x3 + 3x2 8x2 − 6x

30

CHAPTER 4. POLYNOMIALS

We now calculate the quotient for leading terms term of the quotient...

8x2 x

= 8x to get the next

3x2 + 8x x−1




3x3 + 5x2 − 6x − 2 − 3x3 + 3x2 8x2 − 6x

We now multiply 8x(x − 1) = 8x2 − 8x and subtract, OR change sign and add to obtain.... 3x2 + 8x x−1

Finally, we calculate

2x x




3

3x + 5x2 − 6x − 2 − 3x3 + 3x2 8x2 − 6x − 8x2 + 8x 2x − 2

= 2 for the last part of the quotient... 3x2 + 8x + 2

x−1




3x3 + 5x2 − 6x − 2 − 3x3 + 3x2 8x2 − 6x − 8x2 + 8x

2x − 2

Then we have 3x2 + 8x + 2 x−1




3

3x + 5x2 − 6x − 2 − 3x3 + 3x2 8x2 − 6x − 8x2 + 8x

2x − 2 − 2x + 2 0

Therefore, by [LD] we have (3x3 + 5x2 − 6x − 2) ÷ (x − 1) = 3x2 + 8x + 2 Example Divide (10x4 + 13x3 + 8x2 + 8x + 3) ÷ (2x + 1)

31

4.5. DIVIDING BY POLYNOMIALS II 5x3 + 4x2 + 2x + 3 2x + 1




10x4 + 13x3 + 8x2 + 8x + 3 − 10x4 − 5x3 8x3 + 8x2 − 8x3 − 4x2

4x2 + 8x − 4x2 − 2x

6x + 3 − 6x − 3 0

Example Divide (4x3 +− 4x2 +− 23x + 22) ÷ (2x − 5) 2x − 5




2x2 + 3x − 4

4x3 − 4x2 − 23x + 22 − 4x3 + 10x2 6x2 − 23x − 6x2 + 15x − 8x + 22 8x − 20 2

thus by [LD] (4x3 +− 4x2 +− 23x + 22) ÷ (2x − 5) = 2x2 + 3x − 4 + Example Divide (4x4 + 8x3 + −3x2 − x − 4) ÷ (2x2 + 3x − 4) 2x2 + x + 1




2x2 + 3x − 4

4x4 + 8x3 − 3x2 − x − 4 − 4x4 − 2x3 − 2x2 6x3 − 5x2 − x − 6x3 − 3x2 − 3x

− 8x2 − 4x − 4 8x2 + 4x + 4 0

thus by [LD] (4x3 +− 4x2 +− 23x + 22) ÷ (2x − 5) = 2x2 + 3x − 4 Example Divide (4x4 + 8x3 + −3x2 − x − 4) ÷ (3x + 2)

2 2x−5

32

CHAPTER 4. POLYNOMIALS

3x + 2




4 3 3x 3

16 2 9 x 2

4x4 + 8x − 3x − 4x4 − 38 x3 −

4.5.1

+

16 3 3 x 16 3 3 x

−

− 3x2 2 − 32 9 x −

59 2 9 x 59 2 9 x

59 27 x

91 81

−4

−x

−x

118 27 x 91 27 x 91 − 27 x

+

+

−

−

−4 182 81 506 81

Exercises

1. Divide using Long Division:
−96t2 + 40t − 1 ÷ (8t − 2) 2. Divide using Long Division:

−84x6 + 68x5 + 92x4 − 59x3 + 26x2 − x − 50 ÷ −7x2 + x + 6 3. Divide using Long Division:

30y 6 + 76y 5 + 81y 4 + 40y 3 − 20y 2 + 19y − 9 ÷ −6y 2 − 8y + 3 4. Divide using Long Division:
90y 3 + 48y 2 + 33y + 12 ÷ (−9y − 3) 5. Divide using Long Division:
−12y 4 + 70y 3 − 88y 2 − 42y + 71 ÷ (6y − 11) 6. Divide using Long Division:
−20x5 − 105x3 − 95x2 + 60x + 31 ÷ (10x + 5)

4.5. DIVIDING BY POLYNOMIALS II 7. Divide using Long Division:
132u5 + 88u4 + 12u3 − 16u2 + 128u + 98 ÷ (−12u − 8) 8. Divide using Long Division:
40y 4 − 72y 3 − 78y 2 + 36y − 34 ÷ (4y − 10) 9. Divide using Long Division:
−9u2 + 6u + 23 ÷ (−3u − 4) 10. Divide using Long Division:
5y 5 − 10y 4 + 6y 3 − 24y 2 + 15y + 25 ÷ (2 − y) 11. Divide using Long Division:
99y 5 + 76y 4 + 84y 3 − 11y 2 − 87y − 25 ÷ (−9y − 2) 12. Divide using Long Division:
−28y 5 + 63y 4 − 20y 3 + 13y 2 + 56y + 28 ÷ (9 − 4y) 13. Divide using Long Division:
−24u4 − 36u3 − 48u2 − 96u + 88 ÷ (−6u − 12) 14. Divide using Long Division:

99y 3 − 63y 2 − 33y + 17 ÷ 3 − 9y 2 15. Divide using Long Division:
−55u5 + 96u4 + 90u3 − 62u2 − 53u + 45 ÷ (5u − 11)

33

34

4.6

CHAPTER 4. POLYNOMIALS

Factor Polynomials by DL

Recall ’To factor’ means to break up into multiples. The main tool here will be the distributive law. To factor completely means to factor, factor until you can factor no more. Another name for factoring completely is prime factorization. Here we may not learn to all methods to factor polynomials completely. However, we will learn one method of factoring polynomials. This method called the DL METHOD involves looking at each of the terms, finding the gcd for the collection of terms, and factoring this gcd using DL among other axioms and theorems. If the gcd is 1, then this method does not provide any non-trivial factorization of the polynomial. Example factor 6x + 4 note the gcd(6x, 4) = 2, thus using the DL method we can factor 2 from each of the terms. 6x + 4 = 2 · 3x + 2 · 2 = 2(3x + 2)

(T.T) (D.L.)

Example factor 16x2 + 40x + 24. note the gcd(16x2 , 40x, 24) = 8, thus using the DL method we can factor 8 from each of the terms. 16x2 + 40x + 24 = 8 · 2x2 + 8 · 5x + 8 · 3 2

= 8(2x + 5x + 3)

(T.T) (D.L.)

Example factor 16x5 + 40x4 + 24x3 note the gcd(16x5 , 40x4 , 24x3 ) = 8x3 , thus using the DL method we can factor 8x3 from each of the terms. 16x5 + 40x4 + 24x3 = 8x3 · 2x2 + 8x3 · 5x + 8x3 · 3 3

2

= 8x (2x + 5x + 3)

(B.I.) (D.L.)

Example factor (blah)x3 + (blah)2x (blah)x3 + (blah)2x = (blah)(x3 + 2x) 2

= (blah)(x + 2)x

(D.L.) (D.L., BI)

4.6. FACTOR POLYNOMIALS BY DL

35

Example factor (x + 5)x3 + (x + 5)2x (x + 5)x3 + (x + 5)2x = (x + 5)(x3 + 2x)

(given) (D.L.)

= (x + 5)(x2 · x2 + 2x)

(J.A.E.)

= (x + 5)(x2 · x + 2x)

(N.Expo)

= (x + 5)(x2 + 2)x

(D.L., BI)

It should be noted that this method for factoring is not always successful. Specially if there is nothing common amongst the terms. Example factor (x + 4)x3 + (x + 5)2x (x + 4)x3 + (x + 5)2 3

=(x + 4)x + (x + 5)2

(given) (nothing we can factor)

This expression is as factored as it will get, at least using the DL method from this section. As always, the homework is an essential part of the learning process. At this point the only thing left to do is to get to it. The exercises should help you practice spotting gcd’s and factoring it from each of the terms using DL.

4.6.1

Exercises

1. Factor using DL: 8t11 − 11t6 − 7t3 2. Factor using DL: 3u5 − 5u4

36

CHAPTER 4. POLYNOMIALS

3. Factor using DL: 9t12 + 7t7 + 8t4 4. Factor using DL: −8t6 + 11t3 + 10t 5. Factor using DL: 12y 7 − 7y 4 − 8y 2 6. Factor using DL:

−5 t3 − 9 t3 u 7. Factor using DL:


11u5 u3 − 12u2 u3 + 7 u3 8. Factor using DL: 9t8 + 5t5 − 4t3 9. Factor using DL: −6y 12 + 4y 7 + 7y 4 10. Factor using DL: 12t5 − 2t2 + 7t 11. Factor using DL:


12 x3 + 8 + 8 x3 + 8 u5 − x3 + 8 u2 12. Factor using DL: t12 + 12t7 + 8t4 13. Factor using DL: 3 (5x + 12) u − 8 (5x + 12)

4.6. FACTOR POLYNOMIALS BY DL 14. Factor using DL: 7t8 + 3t5 − 8t3 15. Factor using DL: 5u2 ω 2 + uω 2 − 8ω 2 16. Factor using DL: y 4 − 12y 5 17. Factor using DL: 7t2 − 9t 18. Factor using DL: 10µ4 + 8µ4 x3 − 12µ4 x2 + 3µ4 x 19. Factor using DL: 11t8 (11u + 2) − 7t3 (11u + 2) − (11u + 2) 20. Factor using DL:


7t5 x3 − 2t2 x3 − 11 x3 21. Factor using DL:


2 x3 − 4 x3 t4 + x3 t 22. Factor using DL: −7λ4 − 8λ4 x 23. Factor using DL: −11θ2 − 2θ2 x 24. Factor using DL: −7 (7y + 11) + 4 (7y + 11) t8 + 6 (7y + 11) t3

37

38

CHAPTER 4. POLYNOMIALS

25. Factor using DL: −4µ2 − 6µ2 t2 − 3µ2 t 26. Factor using DL: 11x2 + 4x 27. Factor using DL: −10u7 + 9u4 + 11u2 28. Factor using DL: 11u4 + 6u3 29. Factor using DL:


6u4 x6 − 9u x6 − 10 x6 30. Factor using DL:


12 y 2 + 11 y 2 x5 + 9 y 2 x2

39

4.7. FACTOR BY GROUPING

4.7

Factor By Grouping

In the our section, we learned to factor using DL, once we identified the gcd for the terms of a polynomial. It turns out this method often fails. Particularly, if there is nothing in common within the terms, that is, it fails if the gcd is 1. Such as: 3 + 5y + 6y 3 + 10y 4 In this section, we discuss one possible remedy for this, It is called factor by grouping. While it works with many polynomials, we will pay special attention to the case when we have a quad-nominal, that is a polynomial with four terms such as the one state above. The basic idea is to group the terms, and see where that leads. Example 3 + 5y + 6y 3 + 10y 4

(given)

3

= (3 + 5y) + 6y + 10y

4




(ALA)

We have grouped but nothing has happened... but wait.. 3 + 5y + 6y 3 + 10y 4 3

= (3 + 5y) + 6y + 10y

(given) 4




3

=1 · (3 + 5y) + 2y (3 + 5y)

(ALA) (BI, DL)

But now we see, both of these groups have something in common, indeed, we proceed, 3 + 5y + 6y 3 + 10y 4 = (3 + 5y) + 6y 3 + 10y 3

(given)
4

=1 · (3 + 5y) + 2y (3 + 5y)
= 1 + 2y 3 (3 + 5y)

(ALA) (BI, DL) (DL)

... and the factor by grouping method has succeeded! It does not always work, and it is not always clear when it will work. You must try it and practice. If you can’t factor a polynomial using this method, relax. Later we will learn a few more ideas to factor.

40

CHAPTER 4. POLYNOMIALS

Example factor 1 + 3x + x2 + 3x3 1 + 3x + x2 + 3x3 = (1 + 3x) + (1 · x2 + 3x · x2 ) 2

= (1 + 3x) · 1 + (1 + 3x)x 2

= (1 + 3x)(1 + x )

(ALA and M.Id) (MId, D.L.) (D.L.)

Example factor 1 + 3x − x2 − 3x3 1 + 3x − x2 − 3x3

= (1 + 3x) + (−x2 + −3x3 )

(ALA)

2

2

= (1 + 3x) + (1 · −x + 3x · −x ) 2

= (1 + 3x) + (1 + 3x)(−x )

2

= (1 + 3x) · 1 + (1 + 3x)(−x ) 2

= (1 + 3x)(1 + (−x )) 2

= (1 + 3x)(1 + −x )

(MId, N-Expo, CLM) (D.L.) (M.Id) (D.L.) (D.L.(to be continued...))

At last, the most important thing to do is for you to practice the factor by grouping method.

4.7.1

Exercises

1. Factor by grouping: 2 + −4x2 + −4x3 + 8x5 2. Factor by grouping: 12 + 4ρ2 + 3ρ3 + ρ5 3. Factor by grouping: 1 + −t3 + −t2 + t5

4.7. FACTOR BY GROUPING 4. Factor by grouping: 4 + 3Q2 + 4Q3 + 3Q5 5. Factor by grouping: 20 + 8w + 5w2 + 2w3 6. Factor by grouping: 2 + 10x + 2x2 + 10x3 7. Factor by grouping: 6 + 10v 2 + 6v 3 + 10v 5 8. Factor by grouping: 6 + 3v + 10v 2 + 5v 3 9. Factor by grouping: 16 + 12θ2 + 16θ3 + 12θ5 10. Factor by grouping: −2 + 5T 2 + −8T 3 + 20T 5 11. Factor by grouping: 10 + 20z + 10z 2 + 20z 3 12. Factor by grouping: 15 + 3T 2 + 15T 3 + 3T 5 13. Factor by grouping: 3 + 12x2 + 4x3 + 16x5 14. Factor by grouping: 4 + −10β 3 + −6β 2 + 15β 5

41

42

CHAPTER 4. POLYNOMIALS

15. Factor by grouping: 3 + 15λ3 + 5λ2 + 25λ5 16. Factor by grouping: 12 + 12ξ 3 + 6ξ 2 + 6ξ 5 17. Factor by grouping: 20 + 15q + 16q 2 + 12q 3 18. Factor by grouping: 12 + 4φ2 + 12φ3 + 4φ5 19. Factor by grouping: −1 + 2λ2 + −4λ3 + 8λ5 20. Factor by grouping: −1 + −2U 3 + 4U 2 + 8U 5

4.8. FACTOR BY SPLITTING

4.8

43

Factor by Splitting

In this section, we consider trinomials such as x2 + 5x + 4 which can not be factored by DL, nor by grouping. For these we introduce some brand new and very creative medicine. We call this medicine split the middle term. Here is roughly what the process looks like in action.

x2 + 5x + 4 =x2 + (1 + 4)x + 4 2

=x + 1x + 4x + 4

(given) (AT) (DL)

We have indeed split the middle term. You may wonder, while all steps look legal, what madness would drive one to do such a silly thing as splitting the middle term? As it turns out this becomes the key, turning the trinomial into a quadnomial on which we can factor by grouping, much like we did on the last section. Observe:

x2 + 5x + 4

(given)

2

(AT)

2

(DL)

=x + (1 + 4)x + 4 =x + 1x + 4x + 4
= x2 + 1x + (4x + 4)

(ALA)

= (x + 1) x + (x + 1) 4 = (x + 1) (x + 4)

(BI, DL) (BI, DL)

viola! We factored by splitting the middle term. It may help to re-read this example, and try to pinpoint the key to its success. The key is arguably the splitting of 5x into 1x + 4x. This may lead you to wonder, how did we know that would work? would other splitting also work? where did this come from? For starters, not all splitting works to end up with a nicely factored polynomial, for example had we tried to split 5x into 3x + 2x, though legal, we would have gotten stuck on this step:

44

CHAPTER 4. POLYNOMIALS

x2 + 5x + 4

(given)

2

(AT)

2

(DL)

=x + (1 + 4)x + 4 =x + 1x + 4x + 4
= x2 + 3x + (2x + 4)

(ALA)

= (x + 3) x + (x + 2) 2

(ALA)

Because the terms do not match, we can not complete the factorization by grouping strategy. This shows that not all splitting works, which begs the question, which do? Here is one way to see which splitting work. Consider working backwards. That is suppose we had already factored, using some numbers we yet don’t know but imagine, a, b, c, d it might look like this: x2 + 5x + 4 = (ax + b)(cx + d) Then we try to gather hints about these possible numbers, a, b, c, d. x2 + 5x + 4 = (ax + b)(cx + d) 2

= acx + bcx + dax + bd

(Assume) (FOIL)

Now comes the hint, if you multiply bc · da you get the same thing as if you multiply ac · bd Said differently, the split coefficients, when multiplied give you the same as the outer coefficients when multiplied, when multiplied. Said differently the split 5 pieces must be the same as 4, the product of outer coefficients. So to look for the possible split pieces of 5 we look to the possible factors of 4. Namely, 4= 2·2

4 = −2 · −2 4= 1·4

(or) (or) (or)

4 = −1 · −4

We see the product 1 · 4 = 4 and the sum 1 + 4 = 5, thus we have winner winner chicken dinner.

4.8. FACTOR BY SPLITTING

45

That was lots of explaining, and maybe good to come back and re-read after a couple example. Lets try one more, Example Factor 3x2 + 2x − 16 First consider the outer product 3 · −16 = −48 then consider the ways of factoring it

−48 = 1 · −48 −48 = 2 · −24 −48 = 3 · −16 ......

(or) (or) (or)

we’re looking for a pair,w/ product ”-48” & w/sum ”2”, so we can split the 2x

...... −48 = 4 · −12

−48 = −6 · 8

(or) (winner winner!)

That was the harder part, now, we do the splitting and all will be good.. 3x2 + 2x − 16

(AT)

2

(DL)

=3x + (−6 + 8)x + −16

=3x + −6x + 8x + −16
= 3x2 + −6x + (8x + −16) = (x + −2) 3x + (x + −2) 8 = (x + −2) (3x + 8)

4.8.1

(given)

2

Exercises

1. Factor by Splitting The Middle Term: 4λ2 + 20λ + 24

(ALA) (DL, BI) (DL)

46

CHAPTER 4. POLYNOMIALS

2. Factor by Splitting The Middle Term: 6x4 + 42x2 + 60 3. Factor by Splitting The Middle Term: z 2 + −4z + 4 4. Factor by Splitting The Middle Term: 6β 2 + 26β + 8 5. Factor by Splitting The Middle Term: 6µ6 + 30µ3 z 2 + 36z 4 6. Factor by Splitting The Middle Term: 2ρ6 + 16ρ3 r2 + 30r4 7. Factor by Splitting The Middle Term: 2q 2 + 27q + 55 8. Factor by Splitting The Middle Term: 2x4 + 23x2 + 45 9. Factor by Splitting The Middle Term: 3γ 6 + 21γ 3 q 2 + 30q 4 10. Factor by Splitting The Middle Term: t2 + 19t + 84 11. Factor by Splitting The Middle Term: 3µ6 + 10µ3 t2 + 3t4 12. Factor by Splitting The Middle Term: 6x2 + −6x + −120

4.8. FACTOR BY SPLITTING 13. Factor by Splitting The Middle Term: 2λ2 + 17λ + 21 14. Factor by Splitting The Middle Term: 2x2 + 33x + 121 15. Factor by Splitting The Middle Term: −w2 + 11w + 12 16. Factor by Splitting The Middle Term: −2v 2 + 5v + 3 17. Factor by Splitting The Middle Term: 4q 2 + 12q + 8 18. Factor by Splitting The Middle Term: 2w2 + 22w + 60 19. Factor by Splitting The Middle Term: 6r4 + 31r2 + 35 20. Factor by Splitting The Middle Term: 3ω 6 + 23t2 ω 3 + 40t4 21. Factor by Splitting The Middle Term: 2β 2 + 19β + 42 22. Factor by Splitting The Middle Term: 3λ2 + 19λ + 6 23. Factor by Splitting The Middle Term: 2t4 + 25t2 + 72

47

48

CHAPTER 4. POLYNOMIALS

24. Factor by Splitting The Middle Term: 2z 2 + −2z + −12 25. Factor by Splitting The Middle Term: 3γ 2 + 4γ + 1 26. Factor by Splitting The Middle Term: 6θ6 + 33θ3 q 2 + 45q 4 27. Factor by Splitting The Middle Term: 4x2 + −30x + 44 28. Factor by Splitting The Middle Term: −r2 + −9r + −18 29. Factor by Splitting The Middle Term: 3µ2 + 5µ + 2 30. Factor by Splitting The Middle Term: −2v 2 + 15v + −27

49

4.9. FACTOR FAMOUS POLYNOMIALS

4.9

Factor Famous Polynomials

In this section, we simply pause to reflect on some very famous polynomials. Recall earlier in this chapter we studying the multiplication of famous polynomials. Such as the Pascal Polynomial #2 [PP2] which summarized that (x + y)2 = x2 + 2yx + y 2 Along with that one we studied many other famous polynomials and we verified these by doing KG multiplication on them, or by using DL to distribute and multiply. In this section, we review the exact same polynomials, but rather than seeing them as famous multiplications, we look at them backwards as famous factorizations. In other words, instead of seeing PP2 as (x + y)(x + y) = x2 + 2yx + y 2 We see it the other way around as a famous factorization: x2 + 2yx + y 2 = (x + y)(x + y) Since it is really the same statement with a little symmetric property ?? added, we do not need to prove it. We simply want to practice spotting these and become very good and fluent at factoring them. Not only do we want to review and spot all PP2 polynomial forms, but we also want to do the same for the other dozen or so famous polynomials. Here is the list of famous polynomials from earlier this chapter. Example Suppose we wanted to factor x2 + 6x + 9 While we could split the middle as done in the previous section, we want to become good at spotting patterns. The first term is a square ’x2 ’, the last terms seems to be a square too, 9 = 32 and the middle has the right form, 2 times each of these, thus it has the perfect PP2 pattern. We proceed as follows: ( + △)2 = 2 + 2△ + △2

2

2

x + 6x + 9 = x + 2 · 3 · x + 3 2

= (x + 3)

(..recall the pp2 Pattern) 2

(BI (spotted pattern)) (PP2)

.. and we are done! We factored this famous polynomial. We want to practice this sort of factoring with all of our famous polynomials. Here is a list for review. Followed by a couple more examples then the very important exercises.

50

CHAPTER 4. POLYNOMIALS Very Famous Polynomials name

Short

Example

Difference of Squares

[DS]

x2 − y 2 = (x − y)(x + y)

Difference of Cubes

[DS]

x3 − y 3 = (x − y)(x2 + xy + y 2 )

Sum of Cubes

[SC]

x3 + y 3 = (x + y)(x2 − xy + y 2 )

Sum of Cubes

[P P 2]

(x + y)2 = x2 + 2xy + y 2

Pascal Polynomials name

Short

Example

Pascal Polynomial #2

[P P 2]

(x + y)2 = x2 + 2yx + y 2

Pascal Polynomial #3

[P P 3]

(x + y)3 = x3 + 3x2 y + 3xy 2 + y 3

Pascal Polynomial #4

[P P 4]

(x + y)4 = x4 + 4x3 y + 6x2 y 2 + 6xy 3 + y 4

Pascal Polynomial #5

[P P 5]

(x + y)5 = x5 + 5x4 y + 10x3 y 2 + 10x2 y 3 + 5xy 4 + y 5

Geometric Series Polynomials name

Short

Example

Geometric Series #2

[GS2]

x2 − 1 = (x − 1)(x + 1)

Geometric Series #3

[GS3]

x3 − 1 = (x − 1)(x2 + x + 1)

Geometric Series #4

[GS4]

x4 − 1 = (x − 1)(x3 + x2 + x + 1)

Geometric Series #5

[GS5]

x5 − 1 = (x − 1)(x4 + x3 + x2 + x + 1)

Generalized Geometric Series Polynomials name

Short

Example

Geometric Series #2

[GGS2]

x2 − y 2 = (x − y)(x + y)

Geometric Series #3

[GGS3]

x3 − y 3 = (x − y)(x2 + xy + y 2 )

Geometric Series #4

[GGS4]

x4 − y 4 = (x − 1)(x3 + x2 y + xy 2 + y 3 )

Geometric Series #5

[GGS5]

x5 − y 4 = (x − 1)(x4 + x3 y + x2 y 2 + xy 3 + y 4 )

51

4.9. FACTOR FAMOUS POLYNOMIALS

Example Suppose we wanted to factor 4x2 − 10x + 25 We note the first term is a square, 4x2 = (2x)2 , the last terms seems to be a square too, 25 = (−5)2 and the middle has the right form, 2 times each of these, thus it has the perfect PP2 pattern. We proceed as follows: ( + △)2 = 2 + 2△ + △2 2

2

(..recall the pp2 Pattern) 2

4x − 10x + 25 = (2x) + 2 · (−5) · x + (−5)

(BI (identified PP2 pattern))

= (2x + −5)2

(PP2)

.. and we are done! We factored this famous polynomial. 1 Example Suppose we wanted to factor 8x2 + 125 3 3 We note the first term is a cube 8x = (2x) , the last terms seems to be a cube too,
3 1 = 15 , thus it has the perfect Sum of Cubes pattern. We proceed as follows: 125


3 + △3 = ( + △) 2 − △ + △2 (..recall the SC Pattern)  3 1 1 (BI (identified SC pattern)) = (2x)3 + 8x3 + 125 5 "    2 #  1 1 1 (2x)2 − (2x) + (SC) = 2x + 5 5 5 .. and we are done! We factored another famous polynomial. Example Suppose we wanted to factor 32x4 − 243 We note the first term is a fifth power.. 32x5 = (2x)5 , the last terms seems to be a fifth power too, 243 = 35 , thus it has the perfect General Geometric #5 pattern. We proceed as follows:
5 − △5 = ( − △) 4 + 3 △ + 2 △2 + △3 + △4 (..recall the GGS5 Pattern)

32x5 − 243 = (2x)5 − 35 (BI (identified GG5 pattern))   4 3 2 2 = (2x − 3) (2x) + (2x) (3) + (2x) (3) + (2x)(3)3 + (3)4 (SC) .. and we are done! We factored another famous polynomial.

4.9.1

Exercises

1. Factor by recognizing the famous polynomial: 27v 3 + −216

52

CHAPTER 4. POLYNOMIALS

2. Factor by recognizing the famous polynomial: 9w2 +

21w 49 + 2 16

3. Factor by recognizing the famous polynomial: 4t2 + 4t + 1 4. Factor by recognizing the famous polynomial: y 3 + −30y 2 + 300y + −1000 5. Factor by recognizing the famous polynomial: 4z 2 + 16z + 16 6. Factor by recognizing the famous polynomial: θ3 + 6θ2 + 12θ + 8 7. Factor by recognizing the famous polynomial: 9V 2 + 54V + 81 8. Factor by recognizing the famous polynomial: 4v 2 − 25 9. Factor by recognizing the famous polynomial: x2 + 2x + 1 10. Factor by recognizing the famous polynomial: 4µ2 +

1 4µ + 7 49

11. Factor by recognizing the famous polynomial: 27λ3 + −81λ2 + 81λ + −27

4.9. FACTOR FAMOUS POLYNOMIALS 12. Factor by recognizing the famous polynomial: x2 − 81 13. Factor by recognizing the famous polynomial: 8q 3 + 72q 2 + 216q + 216 14. Factor by recognizing the famous polynomial: v2 − 4 15. Factor by recognizing the famous polynomial: 9r2 + 42r + 49 16. Factor by recognizing the famous polynomial: X 2 + 10X + 25 17. Factor by recognizing the famous polynomial: 27w3 + −135w2 + 225w + −125 18. Factor by recognizing the famous polynomial: 8ω 3 + 108ω 2 + 486ω + 729 19. Factor by recognizing the famous polynomial: 27µ3 + 216µ2 + 576µ + 512 20. Factor by recognizing the famous polynomial: 9r2 + 36r + 36 21. Factor by recognizing the famous polynomial: X 3 + 1000 22. Factor by recognizing the famous polynomial: V 2 + 4V + 4

53

54

CHAPTER 4. POLYNOMIALS

23. Factor by recognizing the famous polynomial: 9Y 2 + 60Y + 100 24. Factor by recognizing the famous polynomial: ξ 3 + 30ξ 2 + 300ξ + 1000 25. Factor by recognizing the famous polynomial: 27ρ3 + −108ρ2 + 144ρ + −64 26. Factor by recognizing the famous polynomial: 27ρ3 + −512 27. Factor by recognizing the famous polynomial: R3 + 12R2 + 48R + 64 28. Factor by recognizing the famous polynomial: 9u2 + 60u + 100 29. Factor by recognizing the famous polynomial: 27W 3 + −54W 2 + 36W + −8 30. Factor by recognizing the famous polynomial: 4W 2 +

16 16W + 9 81

31. Factor by recognizing the famous polynomial: 9x2 − 16 32. Factor by recognizing the famous polynomial: 9q 2 +

14q 49 + 3 81

4.9. FACTOR FAMOUS POLYNOMIALS 33. Factor by recognizing the famous polynomial:

9t2 − 9

34. Factor by recognizing the famous polynomial:

w3 + −216

35. Factor by recognizing the famous polynomial:

4r2 − 64

55

56

CHAPTER 4. POLYNOMIALS

57

4.10. CHAPTER REVIEW

4.10

Chapter Review Very Famous Polynomials

name

Short

Example

FOIL

[F OIL]

(A + B)(C + D) = AC + AD + BC + BD

Difference of Squares

[DS]

x2 − y 2 = (x − y)(x + y)

Difference of Cubes

[DS]

x3 − y 3 = (x − y)(x2 + xy + y 2 )

Sum of Cubes

[SC]

x3 + y 3 = (x + y)(x2 − xy + y 2 )

Sum of Cubes

[P P 2]

(x + y)2 = x2 + 2xy + y 2

Pascal Polynomials name

Short

Example

Pascal Polynomial #2

[P P 2]

(x + y)2 = x2 + 2yx + y 2

Pascal Polynomial #3

[P P 3]

(x + y)3 = x3 + 3x2 y + 3xy 2 + y 3

Pascal Polynomial #4

[P P 4]

(x + y)4 = x4 + 4x3 y + 6x2 y 2 + 6xy 3 + y 4

Pascal Polynomial #5

[P P 5]

(x + y)5 = x5 + 5x4 y + 10x3 y 2 + 10x2 y 3 + 5xy 4 + y 5

Geometric Series Polynomials name

Short

Example

Geometric Series #2

[GS2]

x2 − 1 = (x − 1)(x + 1)

Geometric Series #3

[GS3]

x3 − 1 = (x − 1)(x2 + x + 1)

Geometric Series #4

[GS4]

x4 − 1 = (x − 1)(x3 + x2 + x + 1)

Geometric Series #5

[GS5]

x5 − 1 = (x − 1)(x4 + x3 + x2 + x + 1)

Generalized Geometric Series Polynomials name

Short

Example

Geometric Series #2

[GGS2]

x2 − y 2 = (x − y)(x + y)

Geometric Series #3

[GGS3]

x3 − y 3 = (x − y)(x2 + xy + y 2 )

Geometric Series #4

[GGS4]

x4 − y 4 = (x − 1)(x3 + x2 y + xy 2 + y 3 )

Geometric Series #5

[GGS5]

x5 − y 4 = (x − 1)(x4 + x3 y + x2 y 2 + xy 3 + y 4 )

58

CHAPTER 4. POLYNOMIALS

Some Answers

2.

try:

−5w

Section 4.1

5

− 4w

3

+ 2w

×

1. 7r6 + 2r5 + 5r4 + 5r3 + 3r2 + 2r + 5 2. 2q4 − 2q3 + q2 + 11q + 2 3. 5w6 + w5 + 7w4 − 5w3 + w2 − 10w + 7 4. 5y + 2 5. 3w4 + 3w3 − 5w2 + 4w + 2 6. 7r3 − 2r2 + 2 7. −3r6 + 4r5 + 5r4 + 7r3 − 5r2 + 6r 8. 8t3 − 2t2 + 10t + 4 9. −3y5 − 6y3 + 2y2 − 4y − 2 10. 6r6 − 5r5 + 5r4 − 6r3 − r2 + 10r − 4 11. 10r4 + r3 + 6r2 + 7r + 7 12. 9q − 7 13. −v4 − 2v3 − 3v2 − 10v 14. 6q6 + q5 − 12q3 + 4q − 7 15. 3t 16. −2z5 − z4 − 4z3 + z2 − 6z + 7 17. 4v4 + 6v3 + 3v2 − 3v + 4 18. 6y3 + 8y2 + 2y + 2 19. −w − 1 20. 13v2 + 7v

−30w

3.

11

− 24w

6w

9

+ 12w

6

7

try:

(−2z

7

+ 3z

7

6

4 + 7z)(−2z )

6 4 4 =(−2z )(−2z ) + (3z )(−2z ) + (7z)(−2z ) (DL) =4z

4.

11

4

− 6z

10

− 14z

5

(BI)

try:

y

4

+ 2y

×

−2y 4y 5y

5y

5.

6

6

5

+ 8y

+ 10y

+ 14y

5

4

5

+ 41y

4

3

4

2

3

3

2

+ 20y

+ 25y

+ 49y

2

+ 5y

+ 4y − 2

− 14y

+ 28y

+ 35y

+ 7y

5y

− 4y

4

3

− 10y 2

3

+ 6y

2

− 10y

try:

w

6

− 5w

4

+ 7w

×

7w

12

− 35w

10

7w

+ 49w

3 6

9

Section 4.2 1.

6.

try:

try:

4 2 u + 6u + 3u − 5 ×

4z

2 7u − 5u + 5

×

4 2 5u + 30u + 15u − 25

−16z

−5u5 − 30u3 − 15u2 + 25u 6 4 3 2 7u + 42u + 21u − 35u

6 5 4 3 2 7u − 5u + 47u − 9u − 20u + 40u − 25

3

3

− 4z 3z

+ 16z

2

2

− 5z + 7

2

− 2z − 4

+ 20z − 28

−8z 4 + 8z 3 + 10z 2 − 14z 12z

12z

5

5

− 12z

− 20z

4

4

− 15z

− 23z

3

3

+ 21z

+ 47z

2

2

+ 6z − 28

59

Answers for Section 4.2 7.

12.

try:

−w

3

+ 5w

2

−4w

2w

8.

4

4

3

+ 20w

− 10w

3

− 14w

3

2

4 − 2w

2

2

−21r

+ 8w

−21r

− 4w − 16

13.

6

+ 64r

5

6

3

+ 7r

×

7r

5

− 35r

15r

− 12w − 16

+ 6w

+ 26w

−3r

− 3w − 4

×

2w

try:

+ 49r

− 35r

4

5

2

4

− 5r

2

3

3

− 5r

2

try:

try: −2z −4v

5

+ 6v

3

×

+ 3v 3v

2

10

+ 18v

8

+ 9v

2

×

+z+4

2z

3

− 2z

2

− 8z

5 4z

−12v

9.

+1

− 5r

+ 7r

+ 7r

2

3

7

−4z

−4z

5

+ 2z

4

5

3

+ 2z

+ 12z

− 2z

4

3

+ 8z

− 2z

3

2

− 8z

try: 4 2 3u + 4u − 2u − 1

14.

try:

2 u − 5u − 2

×

4r

5

+r

3

− 5r

×

−r

4 2 −6u − 8u + 4u + 2 5 3 2 −15u − 20u + 10u + 5u

−4r

6

−r

4

+ 5r

2

6 4 3 2 3u + 4u − 2u − u

6 5 4 3 2 3u − 15u − 2u − 22u + u + 9u + 2

15.

try: 5w4 + 7w3 + 6w + 3 ×

10.

− 4w

2

+ 4w + 3

try: 15w

2 5 2 (t + 5t − 5)(t − 2t )

20w

=(t2 )(t5 − 2t2 ) + (5t)(t5 − 2t2 ) + (−5)(t5 − 2t2 ) (DL)

5

+ 28w

4

4

+ 21w + 24w

3

2

+ 18w + 9 + 12w

−20w6 − 28w5 − 24w3 − 12w2

2 5 2 2 5 2 5 2 =(t )(t ) + (t )(−2t ) + (5t)(t ) + (5t)(−2t ) + (−5)(t ) + (−5)(−2t ) 6 5 4 3 2 −20w − 8w + 43w − 3w + 12w + 30w + 9 (DL) 7 6 5 4 3 2 =t + 5t − 5t − 2t − 10t + 10t

(BI)

16. 11.

try: (−y

try:

5

− 5y

3

2 − 3y )(5y)

5

3 2 =(−y )(5y) + (−5y )(5y) + (−3y )(5y) 3 2 −4t − 5t − t − 2 ×

= − 5y

6

− 25y

4

− 15y

3

(DL) (BI)

2 6t + 3t

−12t4 − 15t3 − 3t2 − 6t 5 4 3 2 −24t − 30t − 6t − 12t

17.

try: (6z

5

+ 4z

5

3

7 − 1)(5z )

3 7 7 =(6z )(5z ) + (4z )(5z ) + (−1)(5z ) 5 4 3 2 −24t − 42t − 21t − 15t − 6t

=30z

12

7

+ 20z

10

− 5z

7

(DL) (BI)

60

Answers for Section 4.2

18.

4.

try: 6z

3

− 4z

×

−12z 36z

36z

6

− 24z

5

6

4

− 24z

− 42z

4

6z

+ 8z 5

3

2

− 5z

3

− 2z

+ 10z

− 30z

+ 8z

3

using [PP2] ... 2 (v + 1) = (v) =v

2

2

2

(given)

2 + 2 (1) (v) + (1)

(PP2)

+ 2v + 1

(BI)

4

+ 10z

2

5.

using [PP3] ... (w + y)3

19. (3y

2

+ y − 4)(−3y

3

(given)

2 1 (w + y) (w + y)   2 2 w + 2wy + y (w + y)

try: − 2y)

2 3 3 3 =(3y )(−3y − 2y) + (y)(−3y − 2y) + (−4)(−3y − 2y) (DL) now we use KG 2 3 2 3 3 =(3y )(−3y ) + (3y )(−2y) + (y)(−3y ) + (y)(−2y) + (−4)(−3y ) + (−4)(−2y) (DL) = − 9y

5

20.

try:

− 3y

4

+ 6y

3

− 2y

2

w + 8y

(BI)

2

(JAE) (PP2,BI)

+ 2wy + y ×

w

3

2

2 2 + 2w y + wy

w y + 2wy 6y

4

×

−24y

6

− 4y

3

2

+y

3

+y+4 − 4y

2

− 16y

2

w

6.

3

2 2 3 + 3w y + 3wy + y

this is famous [PP2] goes like this...

(β + 1) (β − 1)

(given)

= (β) (β) + (β) (−1) + (1) (β) + (1) (−1)

Section 4.3

= β = β

1. 2.

2

w+y

(FOIL)

2

+ −β + β + −1

(BI)

2

−1

(BI)

vary, use DL or KG for example..

7. (7w + 5) (5 − 4w)

(given)

(3q − 1) (3 − q)

= (7w) (−4w) + (7w) (5) + (5) (−4w) + (5) (5) (FOIL) = −28w

2

= −28w

2

+ 35w + −20w + 25

(BI)

+ 15w + 25

(BI)

3.

= (3q) (−q) + (3q) (3) + (−1) (−q) + (−1) (3) (FOIL) = −3q2 + 9q + q + −3 = −3q

8. (3t + 2) (4t − 5)

(given)

2

(BI)

+ 10q − 3

(BI)

using [PP2] ...

(given)

= (3t) (4t) + (3t) (−5) + (2) (4t) + (2) (−5) (FOIL)



2r

3



2 = 12t + −15t + 8t + −10

(BI)

=

2 = 12t − 7t − 10

(BI)

= 4r

2 +3

2r 6

 3 2

(given) 

+ 2 (3) 2r

+ 12r

3

+9

3



2 + (3)

(PP2) (BI)

61

Answers for Section 4.3 14.

9. (5z + 1) (7z + 4)

(given)

= (5z) (7z) + (5z) (4) + (1) (7z) + (1) (4) = 35z

2

(FOIL)

+ 20z + 7z + 4

10. 

this is famous [DC] goes like this...

2 = 5u



2 5u



+



2 5u

(N.Expo)

= (β) (β) + (β) (φ) + (φ) (β) + (φ) (φ) =β

2

+ βφ + βφ + φ

(FOIL)

2

(BI)

= β 2 + 2βφ + φ2

(BI)

15. (given)    4 3 6 7 6q − q 7q − 5q



2 (−u) + (u) 5u 



(given)

+ (u) (−u) (FOIL)

4 3 3 2 = 25u + −5u + 5u + −u

            4 7 4 6 3 7 3 6 = 6q −5q + 6q 7q + −q −5q + −q 7q (FOIL)

(BI)

4 2 = 25u − u

11.

(given)

(β + φ) (β + φ)

(BI)

  2 2 5u + u 5u − u



(β + φ)2

(BI)

= 35z 2 + 27z + 4

this is famous [DC] goes like this...

(BI)

= −30q

11

= −30q

11

+ 42q + 47q

10 10

+ 5q

10

− 7q

+ −7q

9

(BI)

9

(BI)

using [PP3] ...

16.

3 (r + z)

using [PP3] ...

(given)

2 1 (r + z) (r + z)   2 2 r + 2rz + z (r + z)

(JAE) (PP2,BI)

(4 − y)3 = (−y) = −y

3

3

(given) + 3 (4)

+ 12y

2

2

(−y) + 3 (4) (−y)

2

+ (4)

3

− 48y + 64

(PP2) (BI)

now we use KG r

2

+ 2rz + z ×

2

17.

using [PP3] ...

r+z 3 (4 − z)

r3 + 2r2 z + rz 2 2 2 3 r z + 2rz + z

r

3

= −z

4r

=

3

+ 7r



2

(given)

    3 2 3 2 4r + 2 (7r) 4r + (7r)

= 16r

6

+ 56r

4

+ 12z

2

− 48z + 64

(PP2) (BI)

using [PP2] ...

using [PP2] ... 

13.

3

2 2 3 + 3r z + 3rz + z

18. 12.

(given)

3 2 2 3 = (−z) + 3 (4) (−z) + 3 (4) (−z) + (4)

+ 49r

2

7v

=

(PP2)

3

2 +2

(given)

    3 2 3 2 7v + 2 (2) 7v + (2)

= 49v

6

+ 28v

3

+4

(PP2) (BI)

(BI)

19.

using [PP2] ...

this is famous [DC] goes like this... 2 (θ + φ) (θ + φ) (θ + φ)

=θ

2

2

+ θφ + θφ + θ + 2θφ + φ

2

2

 3 2 3 − 4t

    3 2 3 2 = −4t + 2 (3) −4t + (3)

(N.Expo)

= (φ) (φ) + (φ) (θ) + (θ) (φ) + (θ) (θ) =φ



(given)

(FOIL)

6 3 = 16t − 24t + 9

(BI) (BI)

20.

vary, use DL or KG for example..

(given) (PP2) (BI)

62

Answers for Section 4.3

21.

using [PP2] ... (3y + 1) = (3y) = 9y

2

2

2

(given)

+ 2 (1) (3y) + (1)

2

+ 6y + 1

27. 28.

(PP2)

vary, use DL or KG for example.. this is famous [DC] goes like this...

(11v + 5) (11v − 5)

(given)

(BI) = (11v) (11v) + (11v) (−5) + (5) (11v) + (5) (−5) (FOIL)

22.

= 121v (2v + 5) (4 − 5v)

(given)

= (2v) (−5v) + (2v) (4) + (5) (−5v) + (5) (4) (FOIL) = −10v

2

+ 8v + −25v + 20

(BI)

= −10v

2

− 17v + 20

(BI)

= 121v

29.

2

+ −55v + 55v + −25

(BI)

2

− 25

(BI)

this is famous [DC] goes like this... (β + ω)

2

(given)

(β + ω) (β + ω)

(N.Expo)

= (ω) (ω) + (ω) (β) + (β) (ω) + (β) (β)

23. 

2q

this is famous [DC] goes like this... 6

+ 4q

2



2q

6

− 4q

2



= ω = β

(given)

2

2

+ βω + βω + β + 2βω + ω

2

(FOIL) (BI)

2

(BI)

            6 6 6 2 2 6 2 2 = 2q 2q + 2q −4q + 4q 2q + 4q −4q

30.

(FOIL)

= 4q = 4q

12

+ −8q

12

− 16q

8

+ 8q

8

+ −16q

4

(BI)

4

(BI)



5y

5

−y

7

  6 7 7y − y

(given)

            7 7 7 6 5 7 5 6 = −y −y + −y 7y + 5y −y + 5y 7y (FOIL)

24. 

7z

this is famous [DC] goes like this... 5

−z

4

  5 4 7z + z

= y (given)

= y

            5 5 5 4 4 5 4 4 = 7z 7z + 7z z + −z 7z + −z z (FOIL)

= 49z = 49z

25. 

11z

10 10

+ 7z −z

9

+ −7z

9

+ −z

8

(BI)

8

(BI)

31.

+ 4z

2

  5 2 11z − 4z

+ −7y

14

− 7y

13

13

+ −5y

− 5y

12

12

+ 35y

+ 35y

11

11

(BI) (BI)

this is famous [PP2] goes like this...

(µ + 3) (µ − 3)

(given)

= (µ) (µ) + (µ) (−3) + (3) (µ) + (3) (−3)

this is famous [DC] goes like this... 5

14

(FOIL)

2 = µ + −3µ + 3µ + −9

(BI)

= µ2 − 9

(BI)

(given)

32.

using [PP2] ...             2 2 5 2 2 5 5 5 −4z + 4z 11z + 4z −4z + 11z 11z = 11z   3 2 6q − 2q (FOIL) = 121z = 121z

10

+ −44z

10

− 16z

7

+ 44z

4

7

+ −16z

4

(BI) (BI)

26. (−4u − 3) (−5u − 3)

=



−2q

= 4q

33.

6

 3 2

(given)

  3 2 + 2 (6q) −2q + (6q)

− 24q

4

+ 36q

2

(PP2) (BI)

this is famous [PP2] goes like this...

(given) (µ + 1) (µ − 1)

= (−4u) (−5u) + (−4u) (−3) + (−3) (−5u) + (−3) (−3) (FOIL)

= (µ) (µ) + (µ) (−1) + (1) (µ) + (1) (−1)

(given)

(FOIL)

2 = 20u + 12u + 15u + 9

(BI)

2 = µ + −µ + µ + −1

(BI)

2 = 20u + 27u + 9

(BI)

2 =µ −1

(BI)

63

Answers for Section 4.4 3.

34. (3 − 4r) (1 − r)

(given) − 2q

= (−4r) (−r) + (−4r) (1) + (3) (−r) + (3) (1) (FOIL) = 4r = 4r

2

+ −4r + −3r + 3

(BI)

2

− 7r + 3

(BI)

9

+ 8q

+

3q2

this is famous [DC] goes like this... (given)

=−

2q7

8q8

= φ = φ

÷ 3q

2

(given)

(def ÷)

−20q2 +

3q2

+

3 = (φ) (φ) + (φ) (−µ) + (µ) (φ) + (µ) (−µ)

2

3q2

−2q9

(µ + φ) (φ − µ)

− 20q

−2q9 + 8q8 − 20q2

=

=

35.

8

8q6

(ATT)

3q2

+−

3

20

(BI)

3

(FOIL)

2

2 + −µφ + µφ + −µ

(BI)

2

2 −µ

(BI)

4.


2u

Section 4.4

3u3 + 51 u2 + 18u + 12 2 6u4 + 51u3 + 36u2 + 24u + 54 4 − 6u 51u3 − 51u3

1.

36u2 − 36u2 12y3 − 16y2 − 10y 3y




−6

24u − 24u

36y4 − 48y3 − 30y2 − 18y + 27 − 36y4 − 48y3 48y3 thus final asnwers is

− 30y2 30y2 − 18y 18y

27 51u2 3 + 18u + + 12 3u + 2 u

thus final asnwers is 12y

3

− 16y

2

− 10y +

9

5. −6

y − 51y3 − 18y2 − 36y + 48

2.

−y 24t3 − 42t2 − 30t + 36
− t − 24t4 + 42t3 + 30t2 − 36t + 9 24t4




51y4 + 18y3 + 36y2 − 48y − 12 − 51y4 18y3 − 18y3 36y2 − 36y2

42t3 − 42t3

− 48y 48y

30t2 − 30t2 − 36t 36t thus final asnwers is thus final asnwers is 9 3 2 24t − 42t − 30t − + 36 t

−51y

3

− 18y

2

− 36y +

12 y

+ 48

64

Answers for Section 4.4

6.

thus final asnwers is 23w

8

− 2w

6

+ 11w

5

÷ −2w

3

(given) −6y

=

=

=−

23w5

3

11w5

+

−2w3

+w

(ATT)

−2w3

+−

11w2

2

− 26v

=

(BI)

5

+ 39

+ 11v

3 u3 − 39 u2 + 33 u − 15 2 2 2 3u4 − 39u3 + 33u2 − 30u − 9 4 − 3u

+ 14v

2

÷ −2v

2

11v4

+

−2v2

=13v

3

+−

−2v2

11v2

+

(given)

(def ÷)

−2v2

−26v5

7.

4

−26v5 + 11v4 + 14v2

2

=


2u

33

+ 60y −

10.

−2w6 +

2

(def ÷)

−2w3

−2w3

+ 48y

y

23w8 − 2w6 + 11w5

23w8

3

14v2

(ATT)

−2v2

+ −7

(BI)

2

− 39u3 39u3 33u2 − 33u2

11. − 30u 30u

− 15y3 − 3y

thus final asnwers is 3u3

−

2

39u2

+

33u

2

9

−

2




− 24y2 24y2

2u

8.

=

−17y6 − 22y4 − 12y3

−17y6 =

=

−2y3

17y3

−22y4 +

−2y3

51y − 51y (given)

(def ÷)

−2y3

thus final asnwers is

−15y

−12y3 +

−2y3

3

+ 4y

2

+ 8y +

12

− 17

y

(ATT)

12. + 11y + 6

(BI)

2

9.

+ 8y − 17

− 12y3 12y3

− 15

− 17y6 − 22y4 − 12y3 ÷ −2y3

+ 4y2

45y4 − 12y3 − 24y2 + 51y − 36 − 45y4

− 20v

=

7

+ 15v

5

− 18v

3

÷ −3v

−20v7 + 15v5 − 18v3 −3v2

2

(given)

(def ÷)

− 6y3 + 48y2 + 60y + 39
y − 6y4 + 48y3 + 60y2 + 39y − 33 6y4

−20v7 =

48y3 − 48y3 60y2 − 60y2

=

−3v2

20v5 3

39y − 39y

+

15v5 −3v2

+ −5v

3

−18v3 +

+ 6v

−3v2

(ATT)

(BI)

65

Answers for Section 4.4 13.

thus final asnwers is − 6x3 + 57 x2 − 21 x + 15 2 2
− 2x 12x4 − 57x3 + 21x2 − 30x + 6 4 − 12x

18 3 2 +6 t − 16t − 9t − t

− 57x3 57x3

17. 21x2 − 21x2

− 16y3 − 19y2 + 4y − 5
3y − 48y4 − 57y3 + 12y2 − 15y − 27 48y4

− 30x 30x

− 57y3 57y3

thus final asnwers is

−6x

3

57x2

+

21x

−

2

3

−

2

12y2 − 12y2

+ 15

x

− 15y 15y

14. 2z

=

8

− 6z

7

+ 26z

6

÷ 3z

(given) thus final asnwers is

2z 8 − 6z 7 + 26z 6

(def ÷)

3z −16y =

2z 8

−6z 7 +

+

3z

=

3z

2z 7

+ −2z

6

+

− 19y

2

+ 4y −

9

−5

(ATT)

3z

18.

26z 5

3

3

y

26z 6

(BI)

− 22r

3

=

10

24x3 − 24x3

− 12r

6

÷ −2r

−2r3

=11r

7

+

+−

15r9 −2r3

15r6

3

(given)

(def ÷)

−2r3

−22r10 =

9

−22r10 + 15r9 − 12r6

15. − 11x3 + 8x2 − 10x − 19
3x − 33x4 + 24x3 − 30x2 − 57x + 42 33x4

+ 15r

−12r6 +

(ATT)

−2r3

+ 6r

3

(BI)

2

− 30x2 30x2 − 57x 57x

19. thus final asnwers is −11x

3

+ 8x

2

− 10x +

14

7 5 3 26t + 13t + 23t ÷ −3t

− 19

(given)

x

16.

=

26t7 + 13t5 + 23t3

(def ÷)

−3t t3 − 16t2

− 9t

+6


− 3t − 3t4 + 48t3 + 27t2 − 18t + 54 3t4 48t3 − 48t3 27t2 − 27t2

=

26t7 −3t

=−

+

26t6 3

− 18t 18t

13t5 −3t

+−

+

23t3

13t4 3

(ATT)

−3t

+−

23t2 3

(BI)

66

Answers for Section 4.4

20.

thus final asnwers is 15u3 + 33u2 − 42u − 36 −


− u − 15u4 − 33u3 + 42u2 + 36u + 60 15u4

27y3

− 21y

2

+ 9y +

2

− 33u3 33u3

21

+ 15

2y

24. 42u2 − 42u2

− 11y 36u − 36u =

8

+ 3y

5

− 22y

2

÷ 3y

3

(given)

−11y8 + 3y5 − 22y2

(def ÷)

3y3

thus final asnwers is −11y8

60 3 2 15u + 33u − 42u − − 36 u

=

3y3

21. −v

=

9

+ 19v

7

− 14v

2

÷ −v

3

−v9 + 19v7 − 14v2

=−

(given)

11y5

+

3y5 3y3

+y

2

−22y2 +

+−

3

(ATT)

3y3

22

(BI)

3y

(def ÷)

−v3

25. −v9 =

−v3

+

19v7 −v3

−14v2 +

(ATT)

−v3

− 15 x3 − 21 x2 + 39 x + 9 2 2 2 2 − 2x

=v

6

+ −19v

4

+

14




15x4 + 21x3 − 39x2 − 15x4

(BI)

− 9x − 3

21x3 − 21x3

v

− 39x2 39x2

22.

− 9x 9x − 33u3 + 27u2 + 30u + 60
−u

33u4 − 27u3 − 30u2 − 60u − 54 − 33u4

thus final asnwers is

− 27u3 27u3 −

− 30u2 30u2

15x3

−

21x2

2

+

2

39x

+

2

3 2x

− 60u 60u

thus final asnwers is 54 3 2 + 60 −33u + 27u + 30u + u

Section 4.5 1.

23. − 27 y3 − 21y2 + 9y + 15 2
2y − 27y4 − 42y3 + 18y2 + 30y + 21 4 27y

− 12t + 2
8t − 2 − 96t2 + 40t − 1 96t2 − 24t 16t − 1 − 16t + 4

− 42y3 42y3

3 18y2 − 18y2

thus final asnwers is 30y − 30y

2 − 12t +

3 8t − 2

+

9 2

67

Answers for Section 4.5 2.

thus final asnwers is 12x4

− 8x3

− 4x2

+x −7


− 7x2 + x + 6 − 84x6 + 68x5 + 92x4 − 59x3 + 26x2 84x6 − 12x5 − 72x4

− x − 50

56x5 + 20x4 − 59x3 − 56x5 + 8x4 + 48x3

−2y

3

+ 8y

2

−7+−

6 6y − 11

6.

28x4 − 11x3 + 26x2 − 28x4 + 4x3 + 24x2

− 2x4 + x3
10x + 5 − 20x5 − 105x3 − 7x3 + 50x2 − x 20x5 + 10x4 7x3 − x2 − 6x 10x4 − 105x3 49x2 − 7x − 50 − 10x4 − 5x3 − 49x2 + 7x + 42 − 110x3 −8 110x3

4

− 8x

3

− 4x

2

+x−7+−

− 4x

+8

− 95x2 + 55x2 − 40x2 + 60x 40x2 + 20x

thus final asnwers is 12x

− 11x2

− 95x2 + 60x + 31

8

80x + 31 − 80x − 40

−7x2 + x + 6

−9

3. − 5y4
− 6y2 − 8y + 3

− 6y3

− 8y2

thus final asnwers is +y−2

30y6 + 76y5 + 81y4 + 40y3 − 20y2 + 19y − 9 4 3 2 −2x + x − 11x − 4x + 8 + − − 30y6 − 40y5 + 15y4 36y5 + 96y4 + 40y3 − 36y5 − 48y4 + 18y3 48y4 + 58y3 − 20y2 − 48y4 − 64y3 + 24y2 − 6y3 6y3

7.

+ 4y2 + 19y
+ 8y2 − − 12u 3y − 8

12y2 + 16y − 9 − 12y2 − 16y + 6

− 11u4

− u2

12u3 − 16u2 − 12u3 − 8u2 − 24u2 + 128u 24u2 + 16u

thus final asnwers is 4

− 6y

3

− 8y

4.

2

+y−2+−

− 10y2
− 9y − 3

144u + 98 − 144u − 96

3 −6y2 − 8y + 3

− 2y

−3

2 thus final asnwers is

90y3 + 48y2 + 33y + 12 − 90y3 − 30y2 18y2 + 33y − 18y2 − 6y 27y + 12 − 27y − 9

−10y

2

−12u − 8

8. 10y3
4y − 10

− 2y − 3 +

2

4 2 −11u − u + 2u − 12 +

3 thus final asnwers is

+ 2u − 12

132u5 + 88u4 + 12u3 − 16u2 + 128u + 98 − 132u5 − 88u4

−3

−5y

9 10x + 5

+ 7y2

− 2y

+4

40y4 − 72y3 − 78y2 + 36y − 34 − 40y4 + 100y3

3

28y3 − 78y2 − 28y3 + 70y2

−9y − 3

− 8y2 + 36y 8y2 − 20y

5. − 2y3 + 8y2 −7
6y − 11 − 12y4 + 70y3 − 88y2 − 42y + 71 4 3 12y − 22y 48y3 − 88y2 − 48y3 + 88y2

16y − 34 − 16y + 40 6 thus final asnwers is

− 42y + 71 42y − 77 −6

10y

3

+ 7y

2

− 2y + 4 +

6 4y − 10

68

Answers for Section 4.5 13.

9. 3u − 6
− 3u − 4 − 9u2 + 6u + 23 9u2 + 12u

4u3 − 2u2
− 6u − 12 − 24u4 − 36u3 − 48u2 24u4 + 48u3

18u + 23 − 18u − 24

−8

12u3 − 48u2 − 12u3 − 24u2

−1 thus final asnwers is 3u − 6 + −

+ 12u

− 96u + 88

− 72u2 − 96u 72u2 + 144u

1 −3u − 4

48u + 88 − 48u − 96

10. − 5y4
−y+2

− 6y2 + 12y

−8 +9

5y5 − 10y4 + 6y3 − 24y2 + 15y + 25 − 5y5 + 10y4

thus final asnwers is

6y3 − 24y2 − 6y3 + 12y2

3 2 4u − 2u + 12u − 8 + −

− 12y2 + 15y 12y2 − 24y − 9y + 25 9y − 18

8 −6u − 12

14.

7 − 11y

thus final asnwers is −5y4 − 6y2 + 12y + 9 +


− 9y2 + 3

7 2−y

− 63y2 63y2

11. − 11y4
− 9y − 2

− 6y3

− 8y2

+ 3y

+ 17 − 21 −4

+9

99y5 + 76y4 + 84y3 − 11y2 − 87y − 25 − 99y5 − 22y4

thus final asnwers is

54y4 + 84y3 − 54y4 − 12y3 7 − 11y + −

72y3 − 11y2 − 72y3 − 16y2 − 27y2 − 87y 27y2 + 6y

+7

99y3 − 63y2 − 33y + 17 − 99y3 + 33y

4 3 − 9y2

15.

− 81y − 25 81y + 18 −7 thus final asnwers is −11y

4

− 6y

3

− 8y

2

+ 3y + 9 + −

7

− 11u4 − 5u3 + 7u2 + 3u − 4
5u − 11 − 55u5 + 96u4 + 90u3 − 62u2 − 53u + 45 55u5 − 121u4 − 25u4 + 90u3 25u4 − 55u3

−9y − 2

35u3 − 62u2 − 35u3 + 77u2

12. 7y4 + 5y2 + 8y + 4
− 4y + 9 − 28y5 + 63y4 − 20y3 + 13y2 + 56y + 28 28y5 − 63y4

15u2 − 53u − 15u2 + 33u − 20u + 45 20u − 44

− 20y3 + 13y2 20y3 − 45y2

1

− 32y2 + 56y 32y2 − 72y

thus final asnwers is

− 16y + 28 16y − 36 −8

4 3 2 −11u − 5u + 7u + 3u − 4 +

1 5u − 11

thus final asnwers is 7y

4

+ 5y

2

+ 8y + 4 + −

8 9 − 4y

69

Answers for Section 4.6

Section 4.6

7.  3 5 11u u

1. 11 6 3 8t − 11t − 7t

      3 3 8 3 3 + 8 t t − 11 t t =−7 t



 3 2 − 12u u



  3 +7 u

   5 2 3 = 11u − 12u + 7 u

(given)

(given)

(DL, BI)

(Bi, factor gcd)

8.    8 3 3 = 8t − 11t − 7 t

(DL, BI) 9t8 + 5t5 − 4t3

2.

(given)

      3 3 5 3 2 +9 t t +5 t t =−4 t 5 4 3u − 5u

    4 4 =3 u u − 5 u   4 = (3u − 5) u

(Bi, factor gcd)

(given)    5 2 3 = 9t + 5t − 4 t

(DL, BI)

(Bi, factor gcd)

(DL, BI)

9. − 6y

12

+ 4y

7

+ 7y

4

(given)

3. 12 7 4 9t + 7t + 8t

      4 3 4 8 4 + 7t t +8 t =9t t    8 3 4 = 9t + 7t + 8 t

(given)

(Bi, factor gcd)

(DL, BI)

      4 4 4 8 3 −6 y y +4 y y =7 y

(Bi, factor gcd)

   4 8 3 y = −6y + 4y + 7

(DL, BI)

10. 5 2 12t − 2t + 7t

4. 6 3 − 8t + 11t + 10t

(given)

5 2 =10(t) − 8(t)t + 11(t)t

(Bi, factor gcd)

 5 2 = −8t + 11t + 10 (t)

(DL, BI)



4 =7(t) + 12(t)t − 2(t)t

(given)

(Bi, factor gcd)

 4 = 12t − 2t + 7 (t) 

(DL, BI)

11. 5. 12y

7

− 7y

4

− 8y

2

      2 2 2 5 2 =12 y y −8 y −7 y y    5 2 2 = 12y − 7y − 8 y

(given)

(Bi, factor gcd)

(DL, BI)

      3 3 3 5 2 12 x + 8 + 8 x + 8 u − x + 8 u    5 2 3 = 8u − u + 12 x +8

    3 3 −9 t u −5 t   = (−9u − 5) t3

(given)

(DL, BI)

(DL, BI)

12. 12 7 4 t + 12t + 8t

6.

(given)

      4 4 3 4 8 + t t + 12 t t =8 t    t4 = t8 + 12t3 + 8

(given)

(Bi, factor gcd)

(DL, BI)

70

Answers for Section 4.6 20.

13. 3(5x + 12)u − 8(5x + 12)

(given)

= (3u − 8) (5x + 12)

      3 2 5 3 3 − 2t x − 11 x 7t x

(given)

(DL, BI)    5 2 3 = 7t − 2t − 11 x

(DL, BI)

14. 7t8 + 3t5 − 8t3

      5 3 3 2 3 =7t t + 3t t −8 t

(given)

21.       2 x3 − 4 x3 t4 + x3 t

(Bi, factor gcd)

   5 2 3 = 7t + 3t − 8 t

   x3 = −4t4 + t + 2

(DL, BI)

(given)

(DL, BI)

22.

15. 2 2 2 2 5u ω + uω − 8ω

(given)

  2 2 = 5u + u − 8 ω

(DL, BI)



16.

− 7λ

4

4 − 8λ x

  4 = (−8x − 7) λ

(given)

(DL, BI)

23. y

4

− 12y

5

(given) − 11θ

    4 4 − 12 y y = y

(Bi, factor gcd)

  4 = (1 − 12y) y

2

2 − 2θ x

  2 = (−2x − 11) θ

(given)

(DL, BI)

(DL, BI)

24. 17.

8 3 − 7(7y + 11) + 4(7y + 11)t + 6(7y + 11)t 2 7t − 9t

(given)

(given)

=7(t)t − 9(t)

(Bi, factor gcd)

= (7t − 9) (t)

(DL, BI)

  8 3 = 4t + 6t − 7 (7y + 11)

(DL, BI)

25. 18.

2 2 2 2 − 4µ − 6µ t − 3µ t 4 4 3 4 2 4 10µ + 8µ x − 12µ x + 3µ x



= 8x

3

− 12x

2

  4 µ + 3x + 10

(given)

(given)    2 2 µ = −6t − 3t − 4

(DL, BI)

(DL, BI)

26.

19. 11t8 (11u + 2) − 7t3 (11u + 2) − (11u + 2)

11x

2

+ 4x

(given)

(given)

  = 11t8 − 7t3 − 1 (11u + 2)

(DL, BI)

=4(x) + 11(x)x

(Bi, factor gcd)

= (11x + 4) (x)

(DL, BI)

71

Answers for Section 4.7 2.

27. 7 4 2 − 10u + 9u + 11u

(given) 2 3 5 12 + 4ρ + 3ρ + ρ

      2 2 2 5 2 − 10 u u + 9 u u =11 u

(given)

(Bi, factor gcd)

   5 2 2 = −10u + 9u + 11 u

    2 3 5 = 12 + 4ρ + 3ρ + ρ

(DL, BI)

(ALA)

h i h    i 2 3 2 3 = 3 (4) + ρ (4) + 3 ρ +ρ ρ

28. 4 3 11u + 6u

(given)

    3 3 =6 u + 11 u u

(Bi, factor gcd)

  3 = (11u + 6) u

(BI)

     2 2 3 = 3+ρ (4) + 3 + ρ ρ

(DL)

   2 3 = 3+ρ 4+ρ

(DL)

(DL, BI)

3.

29. 6 4 6u x 



    6 6 − 9u x − 10 x

(given) 1 + −t3 + −t2 + t5

  4 6 = 6u − 9u − 10 x 

(DL, BI)

2 3 5 =1 + −t + −t + t

2 = 1 + −t 

30.       2 2 2 5 2 + 11 y x +9 y x 12 y    5 2 2 = 11x + 9x + 12 y

(given)

(DL, BI)

(given)



+



(ColA, ALA)

3 5 −t + t



(ALA)

h i h    i 2 3 2 3 = −1 (−1) + t (−1) + −1 t +t t

(BI)

     = −1 + t2 (−1) + −1 + t2 t3

(DL)

   = −1 + t2 −1 + t3

(DL)

Section 4.7 4.

1. 2 + −4x

2

+ −4x

3

+ 8x

5

(given) 4 + 3Q



2

+



= −2 (−1) + 4x

2

= 2 + −4x

h



−4x

3

+ 8x

i

h

5

 

(−1) + −2 2x

(ALA)



2

= 4 + 3Q 3



+ 4x

2



2x

3

i

(BI)

+ 4Q

2



+

3



+ 3Q

4Q

3

5

+ 3Q

(given)

5



h i h    i 2 3 2 3 = 4 (1) + 3Q (1) + 4 Q + 3Q Q

(ALA)

(BI)

     2 2 3 = −2 + 4x (−1) + −2 + 4x 2x

(DL)

     2 2 3 = 4 + 3Q (1) + 4 + 3Q Q

(DL)

   2 3 = −2 + 4x −1 + 2x

(DL)

   2 3 = 4 + 3Q 1+Q

(DL)

72

Answers for Section 4.7 8.

5.

20 + 8w + 5w

= (20 + 8w) +

2

+ 2w

3

6 + 3v + 10v

(given)

  2 3 5w + 2w

= (6 + 3v) +

(ALA)

h    i 2 2 = [5 (4) + 2w (4)] + 5 w + 2w w

2

+ 5v

3

(given)

  2 3 10v + 5v

(ALA)

h    i 2 2 = [2 (3) + v (3)] + 2 5v + v 5v

(BI)

(BI)

  2 = (5 + 2w) (4) + (5 + 2w) w

(DL)

  2 = (2 + v) (3) + (2 + v) 5v

(DL)

  2 = (5 + 2w) 4 + w

(DL)

  2 = (2 + v) 3 + 5v

(DL)

9.

6.

2 + 10x + 2x

= (2 + 10x) +

2



+ 10x

2x

2

3

+ 10x

(given)

3



(ALA)

i   h  2 2 + 5x 2x = [1 (2) + 5x (2)] + 1 2x

(BI)

16 + 12θ

2

+ 16θ

= 16 + 12θ2 



+

3



+ 12θ

5

(given)

16θ3 + 12θ5



h i h    i = 4 (4) + 3θ2 (4) + 4 4θ3 + 3θ2 4θ3

(ALA)

(BI)

  2 = (1 + 5x) (2) + (1 + 5x) 2x

(DL)

     3 2 2 4θ (4) + 4 + 3θ = 4 + 3θ

(DL)

  2 = (1 + 5x) 2 + 2x

(DL)

   2 3 = 4 + 3θ 4 + 4θ

(DL)

10.

7.

6 + 10v

2

+ 6v

3

+ 10v

5

− 2 + 5T

2

+ −8T

3

+ 20T

5

(given)

(given)

    2 3 5 = 6 + 10v + 6v + 10v h i h    i 2 3 2 3 = 3 (2) + 5v (2) + 3 2v + 5v 2v

(ALA)

(BI)

    2 3 5 = −2 + 5T + −8T + 20T

(ALA)

h i h    i 2 3 2 3 = −2 (1) + 5T (1) + −2 4T + 5T 4T

(BI)

     2 2 3 = 3 + 5v (2) + 3 + 5v 2v

(DL)

     2 2 3 = −2 + 5T (1) + −2 + 5T 4T

(DL)

   2 3 = 3 + 5v 2 + 2v

(DL)

   2 3 = −2 + 5T 1 + 4T

(DL)

73

Answers for Section 4.7 14.

11.

4 + −10β 10 + 20z + 10z

2

+ 20z

3

(given) =4 + −6β

= (10 + 20z) +



10z

2

+ 20z

3



(ALA)

h    i 2 2 = [2 (5) + 4z (5)] + 2 5z + 4z 5z

(BI)

3

2

+ −6β

2

+ −10β

3

+ 15β

+ 15β

5

5

(given)

(ColA, ALA)

    = 4 + −6β 2 + −10β 3 + 15β 5

(ALA)

h i h    i 2 3 2 3 = −2 (−2) + 3β (−2) + −2 5β + 3β 5β

(BI)

  2 = (2 + 4z) (5) + (2 + 4z) 5z 

= (2 + 4z) 5 + 5z

2

(DL)



(DL)

     = −2 + 3β 2 (−2) + −2 + 3β 2 5β 3

(DL)

   2 3 = −2 + 3β −2 + 5β

(DL)

15.

12.

3 + 15λ 15 + 3T

2

+ 15T

3

+ 3T

5

(given) =3 + 5λ



= 15 + 3T

2



  3 5 + 15T + 3T

(ALA)

i   i h  3 2 2 3 3T +T = 5 (3) + T (3) + 5 3T h



2



2

= 5+T

= 5+T



(3) +





5+T

3 + 3T

3

2



3T

3

(BI)



(DL)



(DL)

3

2

+ 15λ

3

+ 25λ

+ 25λ

5

5

(given)

(ColA, ALA)

    2 3 5 = 3 + 5λ + 15λ + 25λ

(ALA)

h i h    i 2 3 2 3 = 3 (1) + 5λ (1) + 3 5λ + 5λ 5λ

(BI)

     2 2 3 = 3 + 5λ (1) + 3 + 5λ 5λ

(DL)

   2 3 = 3 + 5λ 1 + 5λ

(DL)

12 + 12ξ



2

16.

13.

3 + 12x

+ 5λ

2

= 3 + 12x

+ 4x

2



+

3



+ 16x

4x

3

5

+ 16x

(given)

5



h i h    i 2 3 2 3 = 1 (3) + 4x (3) + 1 4x + 4x 4x

(ALA)

(BI)

=12 + 6ξ

3

2

+ 6ξ

+ 12ξ

2

3

+ 6ξ

+ 6ξ

5

5

(given)

(ColA, ALA)

    2 3 5 = 12 + 6ξ + 12ξ + 6ξ h i h    i 2 3 2 3 = 4 (3) + 2ξ (3) + 4 3ξ + 2ξ 3ξ

(ALA)

(BI)

     2 2 3 = 1 + 4x (3) + 1 + 4x 4x

(DL)

     2 2 3 = 4 + 2ξ (3) + 4 + 2ξ 3ξ

(DL)

   2 3 = 1 + 4x 3 + 4x

(DL)

   2 3 = 4 + 2ξ 3 + 3ξ

(DL)

74

Answers for Section 4.7 20.

17. 20 + 15q + 16q

= (20 + 15q) +

2

+ 12q

− 1 + −2U 3 + 4U 2 + 8U 5

3

  2 3 16q + 12q

h    i 2 2 = [4 (5) + 3q (5)] + 4 4q + 3q 4q 

= (4 + 3q) (5) + (4 + 3q) 4q

(given)

(given)

2



  2 = (4 + 3q) 5 + 4q

= − 1 + 4U

2

+ −2U

3

+ 8U

5

(ColA, ALA)

(ALA)

(BI)

(DL)

(DL)

    2 3 5 = −1 + 4U + −2U + 8U

(ALA)

h i h    i 2 3 2 3 = −1 (1) + 4U (1) + −1 2U + 4U 2U

(BI)

     2 2 3 = −1 + 4U (1) + −1 + 4U 2U

(DL)

   2 3 = −1 + 4U 1 + 2U

(DL)

18. 12 + 4φ

2

+ 12φ

3

+ 4φ

5

(given)

    = 12 + 4φ2 + 12φ3 + 4φ5 h i h    i = 3 (4) + φ2 (4) + 3 4φ3 + φ2 4φ3

(ALA)

(BI)

     3 2 2 4φ (4) + 3 + φ = 3+φ

(DL)

   2 3 = 3+φ 4 + 4φ

(DL)

Section 4.8 1.

First take the product of the outside terms, 4λ2 · 24 = 96λ2 Then we look at all the way we can factor this coefficent, 96 7→ as: ±[1, 96], ±[2, 48], ±[3, 32], ±[4, 24], ±[6, 16], ±[8, 12] We scan for a suitable pair to split 20λ into. Idenfiying 20λ = 8λ + 12λ as a suitable candidate. Then we stop talking start the doing!

4λ

19.

=4λ

− 1 + 2λ

2

+ −4λ

3

+ 8λ

5

    2 3 5 = −1 + 2λ + −4λ + 8λ



2

2

= 4λ

(given)

+ 20λ + 24

+ (8λ + 12λ) + 24

2

+ 8λ



+ (12λ + 24)

= (2λ + 4) 2λ + (2λ + 4) 6

(ALA)

= (2λ + 4) (2λ + 6)

(given)

(Bi, The Split!)

(ALA)

(DL, BI)

(DL)

h i h    i 2 3 2 3 = −1 (1) + 2λ (1) + −1 4λ + 2λ 4λ

(BI)

     2 2 3 = −1 + 2λ (1) + −1 + 2λ 4λ

(DL)

   2 3 = −1 + 2λ 1 + 4λ

(DL)

2.

First take the product of the outside terms, 6x4 · 60 = 360x4 Then we look at all the way we can factor this coefficent, 360 7→ as: ±[1, 360], ±[2, 180], ±[3, 120], ±[4, 90], ±[5, 72], ±[6, 60], ±[8, 45], ±[9, 40], ±[10, 36], ±[12, 30], ±[15, 24], ±[18, 20] We scan for a suitable pair to split 42x2 into. Idenfiying 42x2 = 30x2 + 12x2 as a suitable

75

Answers for Section 4.8 candidate. Then we stop talking start the doing!

6x

=6x

4

4

+ 42x

+



2

30x

+ 60

2

+ 12x

(given)

2



+ 60

(ALA)

  = 2x2 + 10 3x2 + 6

(DL, BI)

(DL)

3.

First take the product of the outside terms, z 2 · 4 = 4z 2 Then we look at all the way we can factor this coefficent, 4 7→ as: ±[1, 4], ±[2, 2] We scan for a suitable pair to split −4z into. Idenfiying −4z = −2z + −2z as a suitable candidate. Then we stop talking start the doing!

z

2

+ −4z + 4

=z 2 + (−2z + −2z) + 4

6 =6µ +

(given)

  3 2 3 2 4 12µ z + 18µ z + 36z

(Bi, The Split!)

   = 2x2 + 10 3x2 + 2x2 + 10 6 

6 3 2 4 6µ + 30µ z + 36z

(Bi, The Split!)

    4 2 2 = 6x + 30x + 12x + 60 

as: ±[1, 216], ±[2, 108], ±[3, 72], ±[4, 54], ±[6, 36], ±[8, 27], ±[9, 24], ±[12, 18] We scan for a suitable pair to split 30µ3 z 2 into. Idenfiying 30µ3 z 2 = 12µ3 z 2 + 18µ3 z 2 as a suitable candidate. Then we stop talking start the doing!

(given)

(Bi, The Split!)

    6 3 2 3 2 4 = 6µ + 12µ z + 18µ z + 36z

(ALA)

    3 2 3 3 2 2 = 2µ + 4z 3µ + 2µ + 4z 9z

(DL, BI)

   3 2 3 2 = 2µ + 4z 3µ + 9z

(DL)

6.

First take the product of the outside terms, 2ρ6 · 30r4 = 60ρ6 r4 Then we look at all the way we can factor this coefficent, 60 7→ as: ±[1, 60], ±[2, 30], ±[3, 20], ±[4, 15], ±[5, 12], ±[6, 10] We scan for a suitable pair to split 16ρ3 r2 into. Idenfiying 16ρ3 r2 = 6ρ3 r2 + 10ρ3 r2 as a suitable candidate. Then we stop talking start the doing! 6 3 2 4 2ρ + 16ρ r + 30r

  2 = z + −2z + (−2z + 4) = (z + −2) z + (z + −2) − 2

= (z + −2) (z + −2)

(given)

(ALA) 6 =2ρ + (DL, BI)

(DL)



3 2 3 2 6ρ r + 10ρ r

 6 3 2 = 2ρ + 6ρ r



+





+ 30r

4

3 2 4 10ρ r + 30r

(Bi, The Split!)



    3 2 3 3 2 2 = 2ρ + 6r ρ + 2ρ + 6r 5r

4.

First take the product of the outside terms, 6β 2 · 8 = 48β 2 Then we look at all the way we can factor this coefficent, 48 7→ as: ±[1, 48], ±[2, 24], ±[3, 16], ±[4, 12], ±[6, 8] We scan for a suitable pair to split 26β into. Idenfiying 26β = 2β + 24β as a suitable candidate. Then we stop talking start the doing!

6β

=6β

2

2

+ 26β + 8

+ (2β + 24β) + 8

(given)

(Bi, The Split!)

   3 2 3 2 = 2ρ + 6r ρ + 5r

(ALA)

= (3β + 1) 2β + (3β + 1) 8

(DL, BI)

= (3β + 1) (2β + 8)

2

2

+ 27q + 55

+ (22q + 5q) + 55

  2 = 2q + 22q + (5q + 55) = (q + 11) 2q + (q + 11) 5

5.

First take the product of the outside terms, 6µ6 · 36z 4 = 216µ6 z 4 Then we look at all the way we can factor this coefficent, 216 7→

(DL)

First take the product of the outside terms, 2q2 · 55 = 110q2 Then we look at all the way we can factor this coefficent, 110 7→ as: ±[1, 110], ±[2, 55], ±[5, 22], ±[10, 11] We scan for a suitable pair to split 27q into. Idenfiying 27q = 22q + 5q as a suitable candidate. Then we stop talking start the doing!

=2q

(DL)

(DL, BI)

7.

2q   = 6β 2 + 2β + (24β + 8)

(ALA)

= (q + 11) (2q + 5)

(given)

(Bi, The Split!)

(ALA)

(DL, BI)

(DL)

76

Answers for Section 4.8

8.

First take the product of the outside terms, 2x4 · 45 = 90x4 Then we look at all the way we can factor this coefficent, 90 7→ as: ±[1, 90], ±[2, 45], ±[3, 30], ±[5, 18], ±[6, 15], ±[9, 10] We scan for a suitable pair to split 23x2 into. Idenfiying 23x2 = 18x2 + 5x2 as a suitable candidate. Then we stop talking start the doing!

start the doing! 2 t + 19t + 84

(given)

=t2 + (7t + 12t) + 84

(Bi, The Split!)

  2 = t + 7t + (12t + 84)

2x

=2x

4

4

+ 23x

+



2

18x

+ 45

2

+ 5x

= (t + 7) t + (t + 7) 12

(given)

2



   2 2 2 = x + 9 2x + x + 9 5    2 2 = x +9 2x + 5

(ALA)

(DL, BI)

(DL)

11.

First take the product of the outside terms, 3µ6 · 3t4 = 9µ6 t4 Then we look at all the way we can factor this coefficent, 9 7→ as: ±[1, 9], ±[3, 3] We scan for a suitable pair to split 10µ3 t2 into. Idenfiying 10µ3 t2 = µ3 t2 + 9µ3 t2 as a suitable candidate. Then we stop talking start the doing! 6 3 2 4 3µ + 10µ t + 3t

=3µ6 +

9.

First take the product of the outside terms, 3γ 6 · 30q4 = 90γ 6 q4 Then we look at all the way we can factor this coefficent, 90 7→ as: ±[1, 90], ±[2, 45], ±[3, 30], ±[5, 18], ±[6, 15], ±[9, 10] We scan for a suitable pair to split 21γ 3 q2 into. Idenfiying 21γ 3 q2 = 6γ 3 q2 + 15γ 3 q2 as a suitable candidate. Then we stop talking start the doing!

=3γ

6

3 2 4 + 21γ q + 30q

+

(given)

  4 3 2 3 2 + 30q (Bi, The Split!) 6γ q + 15γ q

    6 3 2 3 2 4 = 3γ + 6γ q + 15γ q + 30q     2 3 3 2 3 2 5q γ + 3γ + 6q = 3γ + 6q

(ALA)



µ3 t2 + 9µ3 t2

(given)



+ 3t4

(Bi, The Split!)

    6 3 2 3 2 4 = 3µ + µ t + 9µ t + 3t

(ALA)

    3 2 3 3 2 2 = 3µ + t µ + 3µ + t 3t    3 2 3 2 µ + 3t = 3µ + t

(DL, BI)

(DL)

12.

First take the product of the outside terms, 6x2 · −120 = −720x2 Then we look at all the way we can factor this coefficent, −720 7→ as: ±[1, 720], ±[2, 360], ±[3, 240], ±[4, 180], ±[5, 144], ±[6, 120], ±[8, 90], ±[9, 80], ±[10, 72], ±[12, 60], ±[15, 48], ±[16, 45], ±[18, 40], ±[20, 36], ±[24, 30] We scan for a suitable pair to split −6x into. Idenfiying −6x = 24x + −30x as a suitable candidate. Then we stop talking start the doing! 6x

2

+ −6x + −120

2

+ (24x + −30x) + −120

(given)

(DL, BI) =6x

   3 2 3 2 = 3γ + 6q γ + 5q

(DL)

(Bi, The Split!)



6

(DL, BI)

= (t + 7) (t + 12) + 45

    4 2 2 = 2x + 18x + 5x + 45

3γ

(ALA)

(Bi, The Split!)

(DL)   2 = 6x + 24x + (−30x + −120) = (3x + 12) 2x + (3x + 12) − 10

= (3x + 12) (2x + −10)

(ALA)

(DL, BI)

(DL)

10.

First take the product of the outside terms, t2 · 84 = 84t2 Then we look at all the way we can factor this coefficent, 84 7→ as: ±[1, 84], ±[2, 42], ±[3, 28], ±[4, 21], ±[6, 14], ±[7, 12] We scan for a suitable pair to split 19t into. Idenfiying 19t = 7t + 12t as a suitable candidate. Then we stop talking

13.

First take the product of the outside terms, 2λ2 · 21 = 42λ2 Then we look at all the way we

77

Answers for Section 4.8 can factor this coefficent, 42 7→ as: ±[1, 42], ±[2, 21], ±[3, 14], ±[6, 7] We scan for a suitable pair to split 17λ into. Idenfiying 17λ = 14λ + 3λ as a suitable candidate. Then we stop talking start the doing!

2λ

=2λ

2

+ 17λ + 21

2



= 2λ

+ (14λ + 3λ) + 21

2

+ 14λ



scan for a suitable pair to split 5v into. Idenfiying 5v = 6v + −v as a suitable candidate. Then we stop talking start the doing! − 2v

2

= − 2v

2

+ 5v + 3

(given)

(given) + (6v + −v) + 3

(Bi, The Split!)

(Bi, The Split!)

+ (3λ + 21)

  = −2v2 + 6v + (−v + 3)

(ALA)

(ALA)

= (v + −3) − 2v + (v + −3) − 1 = (λ + 7) 2λ + (λ + 7) 3

(DL, BI)

(DL, BI) = (v + −3) (−2v + −1)

= (λ + 7) (2λ + 3)

(DL)

(DL)

17.

14.

First take the product of the outside terms, 2x2 · 121 = 242x2 Then we look at all the way we can factor this coefficent, 242 7→ as: ±[1, 242], ±[2, 121], ±[11, 22] We scan for a suitable pair to split 33x into. Idenfiying 33x = 22x + 11x as a suitable candidate. Then we stop talking start the doing!

2x

2

+ 33x + 121

First take the product of the outside terms, 4q2 · 8 = 32q2 Then we look at all the way we can factor this coefficent, 32 7→ as: ±[1, 32], ±[2, 16], ±[4, 8] We scan for a suitable pair to split 12q into. Idenfiying 12q = 4q + 8q as a suitable candidate. Then we stop talking start the doing! 4q

(given) =4q

=2x

2



+ (22x + 11x) + 121

= 2x

2

+ 22x



(Bi, The Split!)

+ (11x + 121)

2

2

+ 12q + 8

+ (4q + 8q) + 8

  = 4q2 + 4q + (8q + 8)

(given)

(Bi, The Split!)

(ALA)

(ALA) = (2q + 2) 2q + (2q + 2) 4

= (x + 11) 2x + (x + 11) 11

(DL, BI)

= (x + 11) (2x + 11)

= (2q + 2) (2q + 4)

(DL, BI)

(DL)

(DL)

18.

15.

First take the product of the outside terms, −w2 · 12 = −12w2 Then we look at all the way we can factor this coefficent, −12 7→ as: ±[1, 12], ±[2, 6], ±[3, 4] We scan for a suitable pair to split 11w into. Idenfiying 11w = 12w + −w as a suitable candidate. Then we stop talking start the doing!

First take the product of the outside terms, 2w2 · 60 = 120w2 Then we look at all the way we can factor this coefficent, 120 7→ as: ±[1, 120], ±[2, 60], ±[3, 40], ±[4, 30], ±[5, 24], ±[6, 20], ±[8, 15], ±[10, 12] We scan for a suitable pair to split 22w into. Idenfiying 22w = 10w + 12w as a suitable candidate. Then we stop talking start the doing! 2w

−w

2

=−w

2

+ 11w + 12

=2w + (12w + −w) + 12

2

+ 22w + 60

2

+ (10w + 12w) + 60

  2 = 2w + 10w + (12w + 60)

(ALA)

= (w + 5) 2w + (w + 5) 12 = (w + −12) − w + (w + −12) − 1

(ALA)

(DL, BI)

(DL, BI) = (w + 5) (2w + 12)

(DL)

(DL)

19. 16.

(Bi, The Split!)

(Bi, The Split!)

  2 = −w + 12w + (−w + 12)

= (w + −12) (−w + −1)

(given)

(given)

First take the product of the outside terms, −2v2 · 3 = −6v2 Then we look at all the way we can factor this coefficent, −6 7→ as: ±[1, 6], ±[2, 3] We

First take the product of the outside terms, 6r4 · 35 = 210r4 Then we look at all the way we can factor this coefficent, 210 7→ as: ±[1, 210], ±[2, 105], ±[3, 70], ±[5, 42], ±[6, 35], ±[7, 30], ±[10, 21], ±[14, 15] We scan for a suitable

78

Answers for Section 4.8

pair to split 31r2 into. Idenfiying 31r2 = 21r2 +10r2 as a suitable candidate. Then we stop talking start the doing! 6r

=6r

4

4

+ 31r

+



2

21r

+ 35

2

(given)

+ 10r

2



+ 35

22.

First take the product of the outside terms, 3λ2 ·6 = 18λ2 Then we look at all the way we can factor this coefficent, 18 7→ as: ±[1, 18], ±[2, 9], ±[3, 6] We scan for a suitable pair to split 19λ into. Idenfiying 19λ = 18λ + λ as a suitable candidate. Then we stop talking start the doing! 3λ

(Bi, The Split!)

    4 2 2 = 6r + 21r + 10r + 35

=3λ

(ALA)

    2 2 2 = 2r + 7 3r + 2r + 7 5

2

2

+ 19λ + 6

(given)

+ (18λ + λ) + 6

  2 = 3λ + 18λ + (λ + 6)

(DL, BI)

= (λ + 6) 3λ + (λ + 6) 1    2 2 = 2r + 7 3r + 5

= (λ + 6) (3λ + 1)

First take the product of the outside terms, 3ω6 · 40t4 = 120t4 ω6 Then we look at all the way we can factor this coefficent, 120 7→ as: ±[1, 120], ±[2, 60], ±[3, 40], ±[4, 30], ±[5, 24], ±[6, 20], ±[8, 15], ±[10, 12] We scan for a suitable pair to split 23t2 ω3 into. Idenfiying 23t2 ω3 = 8t2 ω3 + 15t2 ω3 as a suitable candidate. Then we stop talking start the doing!

=3ω

6

2 3 4 + 23t ω + 40t

6

  4 2 3 2 3 + 40t + 8t ω + 15t ω



6



3



3

= 3ω

= 3ω

= 3ω

2 3 + 8t ω



2 3 4 15t ω + 40t

2 + 8t

2 + 8t

  3 2 ω + 5t

+



3ω

3

2 + 8t





2 5t

=2β

2

2

+ 19β + 42

+ (7β + 12β) + 42

  2 = 2β + 7β + (12β + 42) = (2β + 7) β + (2β + 7) 6

= (2β + 7) (β + 6)

as a suitable candidate. Then we stop talking start the doing! 4 2 2t + 25t + 72

4 =2t +

(ALA)



+





+ 72

 9t2 + 72

(Bi, The Split!)

(ALA)

(DL, BI)

(DL)

24.

First take the product of the outside terms, 2z 2 ·−12 = −24z 2 Then we look at all the way we can factor this coefficent, −24 7→ as: ±[1, 24], ±[2, 12], ±[3, 8], ±[4, 6] We scan for a suitable pair to split −2z into. Idenfiying −2z = 4z + −6z as a suitable candidate. Then we stop talking start the doing! 2z

(Bi, The Split!)

=2z

(DL)

2 + 9t

   2 2 = t +8 2t + 9

(given)

(DL, BI)

(given)

    = t2 + 8 2t2 + t2 + 8 9

(DL)

(ALA)

 2 16t

 = 2t4 + 16t2

(DL, BI)

First take the product of the outside terms, 2β 2 · 42 = 84β 2 Then we look at all the way we can factor this coefficent, 84 7→ as: ±[1, 84], ±[2, 42], ±[3, 28], ±[4, 21], ±[6, 14], ±[7, 12] We scan for a suitable pair to split 19β into. Idenfiying 19β = 7β + 12β as a suitable candidate. Then we stop talking start the doing!

(DL)

First take the product of the outside terms, 2t4 · 72 = 144t4 Then we look at all the way we can factor this coefficent, 144 7→ as: ±[1, 144], ±[2, 72], ±[3, 48], ±[4, 36], ±[6, 24], ±[8, 18], ±[9, 16], ±[12, 12] We scan for a suitable pair to split 25t2 into. Idenfiying 25t2 = 16t2 + 9t2

(Bi, The Split!)

21.

2β

(DL, BI)

23.

(given)



ω

3

+



(ALA)

(DL)

20.

3ω

(Bi, The Split!)

2

+ −2z + −12

2

+ (4z + −6z) + −12

  2 = 2z + 4z + (−6z + −12) = (z + 2) 2z + (z + 2) − 6

= (z + 2) (2z + −6)

(given)

(Bi, The Split!)

(ALA)

(DL, BI)

(DL)

79

Answers for Section 4.8 25.

First take the product of the outside terms, 3γ 2 · 1 = 3γ 2 Then we look at all the way we can factor this coefficent, 3 7→ as: ±[1, 3] We scan for a suitable pair to split 4γ into. Idenfiying 4γ = γ + 3γ as a suitable candidate. Then we stop talking start the doing! 3γ

2

+ 4γ + 1

= 3γ

2

+γ



First take the product of the outside terms, −r2 · −18 = 18r2 Then we look at all the way we can factor this coefficent, 18 7→ as: ±[1, 18], ±[2, 9], ±[3, 6] We scan for a suitable pair to split −9r into. Idenfiying −9r = −3r + −6r as a suitable candidate. Then we stop talking start the doing!

(given)

=3γ 2 + (γ + 3γ) + 1



28.

−r

2

+ −9r + −18

=−r

2

+ (−3r + −6r) + −18

(Bi, The Split!)

+ (3γ + 1)

(ALA)

= (3γ + 1) γ + (3γ + 1) 1

(DL, BI)

= (3γ + 1) (γ + 1)

  2 = −r + −3r + (−6r + −18) = (−r + −3) r + (−r + −3) 6

(given)

(Bi, The Split!)

(ALA)

(DL, BI)

(DL) = (−r + −3) (r + 6)

(DL)

26.

First take the product of the outside terms, 6θ6 · 45q4 = 270θ6 q4 Then we look at all the way we can factor this coefficent, 270 7→ as: ±[1, 270], ±[2, 135], ±[3, 90], ±[5, 54], ±[6, 45], ±[9, 30], ±[10, 27], ±[15, 18] We scan for a suitable pair to split 33θ3 q2 into. Idenfiying 33θ3 q2 = 18θ3 q2 + 15θ3 q2 as a suitable candidate. Then we stop talking start the doing! 6θ

=6θ

6

3 2 4 + 33θ q + 45q

6

+



29.

First take the product of the outside terms, 3µ2 · 2 = 6µ2 Then we look at all the way we can factor this coefficent, 6 7→ as: ±[1, 6], ±[2, 3] We scan for a suitable pair to split 5µ into. Idenfiying 5µ = 3µ + 2µ as a suitable candidate. Then we stop talking start the doing! 2 3µ + 5µ + 2

(given)

3 2 3 2 18θ q + 15θ q



+ 45q

4

2 =3µ + (3µ + 2µ) + 2

 2 = 3µ + 3µ + (2µ + 2) 

(ALA)

= (µ + 1) 3µ + (µ + 1) 2     3 2 3 3 2 2 = 3θ + 9q 2θ + 3θ + 9q 5q

= 3θ

3

+ 9q

2



2θ

3

+ 5q

2

(Bi, The Split!)

(Bi, The Split!)

    6 3 2 3 2 4 = 6θ + 18θ q + 15θ q + 45q



(given)



(ALA)

(DL, BI)

(DL, BI) = (µ + 1) (3µ + 2)

(DL)

(DL)

30.

27.

First take the product of the outside terms, 4x2 · 44 = 176x2 Then we look at all the way we can factor this coefficent, 176 7→ as: ±[1, 176], ±[2, 88], ±[4, 44], ±[8, 22], ±[11, 16] We scan for a suitable pair to split −30x into. Idenfiying −30x = −8x + −22x as a suitable candidate. Then we stop talking start the doing! 4x

=4x



2

2

= 4x

+ −30x + 44

+ (−8x + −22x) + 44

2

+ −8x



− 2v

2

+ 15v + −27

= − 2v

2

+ (9v + 6v) + −27

(given)

(Bi, The Split!)

+ (−22x + 44)

= (2x + −4) 2x + (2x + −4) − 11

= (2x + −4) (2x + −11)

First take the product of the outside terms, −2v2 · −27 = 54v2 Then we look at all the way we can factor this coefficent, 54 7→ as: ±[1, 54], ±[2, 27], ±[3, 18], ±[6, 9] We scan for a suitable pair to split 15v into. Idenfiying 15v = 9v + 6v as a suitable candidate. Then we stop talking start the doing!

  2 = −2v + 9v + (6v + −27)

(given)

(Bi, The Split!)

(ALA)

(ALA) = (−2v + 9) v + (−2v + 9) − 3 (DL, BI)

(DL)

= (−2v + 9) (v + −3)

(DL, BI)

(DL)

80

Answers for Section 4.9

Section 4.9

6. 3 ( + △)

1. 

3

=

  3 2 2 + △ = ( + △)  − △ + △ 3

+ −216

2 2 3 + 3 △ + 3△ + △

..recall the pp3 Pattern)

(

..recall the SC Pattern)

( 27v

3

θ

(given)

3

+ 6θ

= (θ)

3 3 + (−6) (Bi, see SC pattern) h i 2 2 = (3v + −6) (3v) − (3v) (−6) + (−6)

= (3v)

3

2

+ +12θ + 8

+ 3 · (θ)

2

(given)

2 2 (2) + 3 · (θ) (2) + (2) (Bi, seePP3 pattern)

3 = (θ + 2)

(PP3)

(SC)

h i 2 = (3v + −6) 9v + 18v + 36

(BI)

7. 2 2 2 ( + △) =  + 2△ + △

2.

..recall the pp2 Pattern)

(

2 2 2 ( + △) =  + 2△ + △

..recall the pp2 Pattern)

( 9w

2

21w

+

+

2 = (3w)

2

49

(given) 7

+

+ 54V + 81

= (3V )

16

+ 2 · (3w)

2

9V

2

(given)

2 + 2 · (3V ) (9) + (9) (Bi, identified PP2 pattern)

= (3V + 9)2

 7 2

4 4 (Bi, identified PP2 pattern)  7 2 (PP2) = 3w + 4

(PP2)

8. 

2

2 − △ = ( − △)( + △)

..recall the DS Pattern)

(

3.

4v

2 2 2 ( + △) =  + 2△ + △

..recall the pp2 Pattern)

= (2t)

2

2

− (5)

(given) 2

(Bi, see DS pattern)

= (2v + 5) (2v − 5)

(DS)

(given)

2 + 2 · (2t) (1) + (1) (Bi, identified PP2 pattern)

= (2t + 1)

+ −25

= (2v)

(

2 4t + 4t + 1

2

2

(PP2)

9. 2 2 2 ( + △) =  + 2△ + △

..recall the pp2 Pattern)

4.

(

x

3 ( + △) =

3

2

+ 3 △ + 3△

2

+△

2

+ 2x + 1

..recall the pp3 Pattern)

( y

3

+ −30y

= (y)

3

2

+ +300y + −1000

+ 3 · (y)

= (y + −10)

2

= (x + 1)

2

(PP2)

(given)

2 2 (−10) + 3 · (y) (−10) + (−10) (Bi, seePP3 pattern)

3

(given)

2 2 = (x) + 2 · (x) (1) + (1) (Bi, identified PP2 pattern)

3

(PP3)

10. ( + △)2 = 2 + 2△ + △2

5.

..recall the pp2 Pattern)

( 2 2 2 ( + △) =  + 2△ + △

..recall the pp2 Pattern)

( 4z

2

+ 16z + 16

(given)

2 2 = (2z) + 2 · (2z) (4) + (4) (Bi, identified PP2 pattern) = (2z + 4)

2

(PP2)

4µ 1 4µ2 + + 7 49 = (2µ)

2

+ 2 · (2µ)

(given) 1

 1 2

+ 7 7 (Bi, identified PP2 pattern)   2 1 (PP2) = 2µ + 7

81

Answers for Section 4.9 11.

17.

3 ( + △) = 

3

3 ( + △)

2 2 3 + 3 △ + 3△ + △

..recall the pp3 Pattern)

( 27λ

3

= (3λ)

+ −81λ

3

2

+ +81λ + −27

+ 3 · (3λ)

= (3λ + −3)

2

(given)

2 2 (−3) + 3 · (3λ) (−3) + (−3) (Bi, seePP3 pattern)

3

(PP3)

3

2 2 3 + 3 △ + 3△ + △

3

+ −135w

= 

..recall the pp3 Pattern)

( 27w

2

+ +225w + −125

(given)

3 2 2 2 = (3w) + 3 · (3w) (−5) + 3 · (3w) (−5) + (−5) (Bi, seePP3 pattern) = (3w + −5)

3

(PP3)

12. 

2

18.

2 − △ = ( − △)( + △)

..recall the DS Pattern)

( x

2

+ −81

( + △)3

(given)

2 2 = (x) − (9)

=

(Bi, see DS pattern)

= (x + 9) (x − 9)

3

..recall the pp3 Pattern)

(

(DS) 8ω

3

+ 108ω

= (2ω)

13. = 

3

2

+ +486ω + 729

+ 3 · (2ω)

2

2

+ 3 △ + 3△

2

+△

(given)

2 2 (9) + 3 · (2ω) (9) + (9) (Bi, seePP3 pattern)

3 = (2ω + 9)

3 ( + △) 3

2 2 3 + 3 △ + 3△ + △

(PP3)

3

..recall the pp3 Pattern)

( 8q

3

+ 72q

2

+ +216q + 216

(given)

19.

= (2q)3 + 3 · (2q) 2 (6) + 3 · (2q) (6)2 + (6)2 (Bi, seePP3 pattern)

3 ( + △)

3 = (2q + 6)

= 

(PP3)

3

2 2 3 + 3 △ + 3△ + △

..recall the pp3 Pattern)

(

3 2 27µ + 216µ + +576µ + 512

14.

= (3µ) 

2

3

+ 3 · (3µ)

2

2 − △ = ( − △)( + △)

..recall the DS Pattern)

= (3µ + 8)3

( v

2

+ −4

= (v)

2

(given)

2 2 (8) + 3 · (3µ) (8) + (8) (Bi, seePP3 pattern) (PP3)

(given)

2 − (2)

(Bi, see DS pattern)

= (v + 2) (v − 2)

(DS)

20. 2 2 2 ( + △) =  + 2△ + △

..recall the pp2 Pattern)

15.

( ( + △)2 = 2 + 2△ + △2

9r

..recall the pp2 Pattern)

(

2

9r

+ 42r + 49

= (3r)

2

(given)

2

(PP2)

16.

= (X)

2

(given)

2 + 2 · (X) (5) + (5) (Bi, identified PP2 pattern)

= (X + 5)

2

(PP2)

3

  3 2 2 + △ = ( + △)  − △ + △ (

..recall the pp2 Pattern)

(

+ 10X + 25

(given)

2 + 2 · (3r) (6) + (6) (Bi, identified PP2 pattern)

..recall the SC Pattern)

2 2 2 ( + △) =  + 2△ + △ 2

2

21. 

X

+ 36r + 36

2 = (3r + 6)

2 + 2 · (3r) (7) + (7) (Bi, identified PP2 pattern)

= (3r + 7)

2

= (3r)

(PP2)

X

3

+ 1000 3

(given)

3 + (10) (Bi, see SC pattern) h i 2 2 = (X + 10) (X) − (X) (10) + (10) (SC) h i = (X + 10) X 2 − 10X + 100 (BI) = (X)

82

Answers for Section 4.9 27.

22. 2 2 2 ( + △) =  + 2△ + △

3 ( + △)

..recall the pp2 Pattern)

(

2

V

+ 4V + 4

= 3 + 32 △ + 3△2 + △3

..recall the pp3 Pattern)

(given)

(

2 2 = (V ) + 2 · (V ) (2) + (2) (Bi, identified PP2 pattern) = (V + 2)

R

3

+ 12R

= (R)

2

3

(PP2)

2

+ +48R + 64

(given)

2 2 2 + 3 · (R) (4) + 3 · (R) (4) + (4) (Bi, seePP3 pattern)

= (R + 4)

3

(PP3)

23. 28.

2 2 2 ( + △) =  + 2△ + △

..recall the pp2 Pattern)

( 9Y

2

+ 60Y + 100

= (3Y )

2

2 2 2 ( + △) =  + 2△ + △

..recall the pp2 Pattern)

(given)

(

2 + 2 · (3Y ) (10) + (10) (Bi, identified PP2 pattern)

= (3Y + 10)

2 9u + 60u + 100 = (3u)

2

2

(PP2)

(given)

2 + 2 · (3u) (10) + (10) (Bi, identified PP2 pattern)

= (3u + 10)

2

(PP2)

24. 29.

3 ( + △) = 

3

2

+ 3 △ + 3△

2

+△

3

..recall the pp3 Pattern)

( ξ

3

+ 30ξ

2

+ +300ξ + 1000

= (ξ + 10)

3

2 2 3 + 3 △ + 3△ + △

3

+ −54W

..recall the pp3 Pattern)

(given)

= (ξ)3 + 3 · (ξ)2 (10) + 3 · (ξ) (10)2 + (10)2 (Bi, seePP3 pattern) 3

3 ( + △) =

(PP3)

( 27W

2

+ +36W + −8

(given)

3 2 2 2 = (3W ) + 3 · (3W ) (−2) + 3 · (3W ) (−2) + (−2) (Bi, seePP3 pattern) 3 = (3W + −2)

(PP3)

25. 30.

3 ( + △) = 

3

2 2 3 + 3 △ + 3△ + △

2 2 2 ( + △) =  + 2△ + △

..recall the pp3 Pattern)

(

3 2 27ρ + −108ρ + +144ρ + −64 = (3ρ)

3

+ 3 · (3ρ)

= (3ρ + −4)

2

..recall the pp2 Pattern)

( (given) 4W

2 2 (−4) + 3 · (3ρ) (−4) + (−4) (Bi, seePP3 pattern)

3

3

..recall the SC Pattern)

(

3 27ρ + −512

(given)

3 + (−8) (Bi, see SC pattern) h i = (3ρ + −8) (3ρ)2 − (3ρ) (−8) + (−8)2

= (3ρ)

3

(SC)

i = (3ρ + −8) 9ρ2 + 24ρ + 64 h

(BI)

16W

2

+

16

(given)

81

+ 2 · (2W )

4

 4 2

+ 9 9 (Bi, identified PP2 pattern)   4 2 = 2W + (PP2) 9

= (2W )

(PP3)

  3 2 2 + △ = ( + △)  − △ + △

+

9

26. 

2

31. 

2

2 − △ = ( − △)( + △)

..recall the DS Pattern)

( 9x

2

+ −16

= (3x)

2

2 − (4)

= (3x + 4) (3x − 4)

(given) (Bi, see DS pattern) (DS)

83

Answers for Section 4.9 32. 2 2 2 ( + △) =  + 2△ + △

..recall the pp2 Pattern)

( 9q

2

+

14q

+

49

(given) 81 7  7 2 2 = (3q) + 2 · (3q) + 9 9 (Bi, identified PP2 pattern)  7 2 (PP2) = 3q + 9 3

33. 

2

2 − △ = ( − △)( + △)

..recall the DS Pattern)

(

2 9t + −9 = (3t)

2

(given)

2 − (3)

(Bi, see DS pattern)

= (3t + 3) (3t − 3)

(DS)

34. 

3

  3 2 2 + △ = ( + △)  − △ + △

..recall the SC Pattern)

( w

3

+ −216

= (w)

3

= (w +

3 + (−6) h 2 −6) (w)

(given)

−

(Bi, see SC pattern) i 2 (w) (−6) + (−6)

(SC)

h i 2 = (w + −6) w + 6w + 36

(BI)

35. 

2

2 − △ = ( − △)( + △)

..recall the DS Pattern)

( 4r

2

+ −64

= (2r)

2

2 − (8)

= (2r + 8) (2r − 8)

(given) (Bi, see DS pattern) (DS)

Bibliography

[1] Wikipedia Euclid’s Elements Retrieved August 10, 2011 [2] Wikipedia Georg Cantor Retrieved August 2011 [3] Wikipedia Cantor’s Diagonal Argument Retrieved August 2011

84

Index

∈, 3 Q[t], 3 R[x], 3 Z[x], 3 binomial, 2 degree of polynomial, 3 degree of the term, 2 Difference of Squares [DS], 15 Difference of two Cubes [DC], 15 Difference of two Cubes [SC], 16 factor by grouping, 39 FOIL, 14 General Geometric Series Polynomials, 17 Geometric Series Polynomials, 16 GGS#2, 17 GGS#3, 17 GGS#4, 17 GGS#5, 17 GS#2, 16 GS#3, 16 GS#4, 16 GS#5, 16 like terms, 4 monomial in x with integer coefficient, 2 monomial in x with real coefficient, 2 Pascal Polynomials, 16 polynomial, 2 PP#2, 16 PP#3, 16 PP#4, 16 PP#5, 16 split the middle, 43 variable, 2

85