Factor and Remainder Theorem Revision 6
Factors of Equations By Charlie Williams
Two ways of solving a cubic once you have one factor
Method 1: Equating coefficients This is the method we have done in class and involves equating coefficients and comparing them to find the answer. This method is good because it is easy to learn and it’s easier to understand what you’re doing, although it is not as fast and can be easy to make mistakes so you need to check your work carefully. There is an example of how to do it on the next slide
Example Take the equation x³+6x²+11x+6=0 and you have been given or worked out that (x+1) is a factor. The first step is you say: x³+6x²+11x+6 ≡ (x+1)(ax²+bx+c) You then expand the brackets on the right hand side and simplify which will give you (the brackets are added to make it clearer): x³+6x²+11x+6 ≡ ax³+(a+b)x²+(b+c)x+c This is where you actually compare coefficients by looking at both sides and seeing what matches up, and you are able to make some easily solvable linear equations which will give you a, b and c: a=1 a+b=6 b+c=11 c=6 (continues on next slide)
From this you can quickly work out that a is 1, b is 5 and c is 6. This means that if you divide the cubic equation x³+6x²+11x+6 by (x+1) you will be left with the quadratic x²+5x+6, which you write as: (x+1)(x²+5x+6) This can quickly be factorised again as it is a quadratic giving: (x+1)(x+2)(x+3) And therefore the roots of the equation are -1, -2 and -3. Sometimes you will not be able to factorise the quadratic, in which case leaving it in that form is ok.
Algebraic Long Division This is the other method of solving a cubic once you have one factor The advantages of this are it is faster, a lot neater and you are very unlikely to make mistake, although it is complicated to learn. There are a couple of examples of how to do it on the next few slides
Example You are given the equation x³-x²-10x-8 and you know (x-2) is a factor It is quite complicated so I will show the working then explain it on the next slides. (x+2)
x³
-
x²(x+2)
x³
+
-3x(x+2)
x²
-
10x
-
8
2x² -3x²
-
10x
-3x²
-
6x -4x
-
8
-4x
-
8
-4(x+2)
Giving you (x+2)(x²-3x-4) which factorises to (x+2)(x+1)(x-4) So your roots are -2, -1 and 4
The first thing you do is write out the factor on the left and the cubic on the right, leave it quite spread out to give you room for working You then see what you would have to multiply the x term in the factor (the thing in the brackets), which is just here, by to get the first thing in the cubic, in this case x² as x²*x=x³
Write this outside the bracket and then expand the bracket which here gives x³+2x², and write it under the original cubic.
Now draw a line under it, and minus 2x² from -x², giving you -3x² and write it underneath. Also bring down the x term in the original cubic, and write it next to what you’ve minused
Again see what you would need to multiply the first part of the factor by to get, in this case, -3x², which is -3x (-3x*x=-3x²) and write it next to the factor and expand the brackets, giving -3x²-6x Minus -6x from -10x, and due to the double negative this gives you -4x, write it below the line and bring down the constant (here -8) and write it next to it. Look to see what you need to multiply the x term in the factor by to get -4x, which is -4, write this next to the factor and expand the brackets. To get your final answer look down the left hand side, at the numbers outside the brackets which gives you x²-3x-4, so x³-x²-10x-8 = (x+2)(x²-3x-4) which factorises to (x+2)(x-4)(x+1), so your roots are -2, 4 and -1
(One good thing about this method is if these two don’t match up you know you’ve gone wrong, so it’s harder to make a mistake)
Here’s another example: You have the cubic x³+2x²-5x-6 and know (x+3) is a factor (x+3)
x³
+
2x²
x²(x+3)
x³
+
3x²
-x(x+3) -2(x+3)
-x² -x²
-
5x -
-
6
-
6 6
5x 3x -2x -2x
These two match up so you know you’re right. This gives you (x+2x²-5x-6) = (x+3)(x²-x-2) which factorises to (x+3)(x+1)(x-2) so your roots are 2, -1 and -3.
Two ways of solving a cubic once you have one factor
Method 1: Equating coefficients This is the method we have done in class and involves equating coefficients and comparing them to find the answer. This method is good because it is easy to learn and it’s easier to understand what you’re doing, although it is not as fast and can be easy to make mistakes so you need to check your work carefully. There is an example of how to do it on the next slide
Example Take the equation x³+6x²+11x+6=0 and you have been given or worked out that (x+1) is a factor. The first step is you say: x³+6x²+11x+6 ≡ (x+1)(ax²+bx+c) You then expand the brackets on the right hand side and simplify which will give you (the brackets are added to make it clearer): x³+6x²+11x+6 ≡ ax³+(a+b)x²+(b+c)x+c This is where you actually compare coefficients by looking at both sides and seeing what matches up, and you are able to make some easily solvable linear equations which will give you a, b and c: a=1 a+b=6 b+c=11 c=6 (continues on next slide)
From this you can quickly work out that a is 1, b is 5 and c is 6. This means that if you divide the cubic equation x³+6x²+11x+6 by (x+1) you will be left with the quadratic x²+5x+6, which you write as: (x+1)(x²+5x+6) This can quickly be factorised again as it is a quadratic giving: (x+1)(x+2)(x+3) And therefore the roots of the equation are -1, -2 and -3. Sometimes you will not be able to factorise the quadratic, in which case leaving it in that form is ok.
Algebraic Long Division This is the other method of solving a cubic once you have one factor The advantages of this are it is faster, a lot neater and you are very unlikely to make mistake, although it is complicated to learn. There are a couple of examples of how to do it on the next few slides
Example You are given the equation x³-x²-10x-8 and you know (x-2) is a factor It is quite complicated so I will show the working then explain it on the next slides. (x+2)
x³
-
x²(x+2)
x³
+
-3x(x+2)
x²
-
10x
-
8
2x² -3x²
-
10x
-3x²
-
6x -4x
-
8
-4x
-
8
-4(x+2)
Giving you (x+2)(x²-3x-4) which factorises to (x+2)(x+1)(x-4) So your roots are -2, -1 and 4
The first thing you do is write out the factor on the left and the cubic on the right, leave it quite spread out to give you room for working You then see what you would have to multiply the x term in the factor (the thing in the brackets), which is just here, by to get the first thing in the cubic, in this case x² as x²*x=x³
Write this outside the bracket and then expand the bracket which here gives x³+2x², and write it under the original cubic.
Now draw a line under it, and minus 2x² from -x², giving you -3x² and write it underneath. Also bring down the x term in the original cubic, and write it next to what you’ve minused
Again see what you would need to multiply the first part of the factor by to get, in this case, -3x², which is -3x (-3x*x=-3x²) and write it next to the factor and expand the brackets, giving -3x²-6x Minus -6x from -10x, and due to the double negative this gives you -4x, write it below the line and bring down the constant (here -8) and write it next to it. Look to see what you need to multiply the x term in the factor by to get -4x, which is -4, write this next to the factor and expand the brackets. To get your final answer look down the left hand side, at the numbers outside the brackets which gives you x²-3x-4, so x³-x²-10x-8 = (x+2)(x²-3x-4) which factorises to (x+2)(x-4)(x+1), so your roots are -2, 4 and -1
(One good thing about this method is if these two don’t match up you know you’ve gone wrong, so it’s harder to make a mistake)
Here’s another example: You have the cubic x³+2x²-5x-6 and know (x+3) is a factor (x+3)
x³
+
2x²
x²(x+3)
x³
+
3x²
-x(x+3) -2(x+3)
-x² -x²
-
5x -
-
6
-
6 6
5x 3x -2x -2x
These two match up so you know you’re right. This gives you (x+2x²-5x-6) = (x+3)(x²-x-2) which factorises to (x+3)(x+1)(x-2) so your roots are 2, -1 and -3.
