Factor and Remainder Theorem Revision 5
Ewan Calder
Factor theorem To factorise cubic polynomial would be very difficult
without using it: x³+4x²-‐11x-‐30 into (x )(x ) (x ). However we can work out one of the factors to work out the rest. As we know the constants have to multiply to give thirty we know the constants have to have one of the following combinations: 5,3,2 3,1,10 6,5,1 30,1,1 or 15,2,1 If one of these numbers is a factor, when put into the equation, the equation must equal 0.
Con-nued.. Using the rule if f(a)=0 then (x-‐a) is a factor of f(x). Using the pervious example:
f(1)= 1+4-‐11-‐30 ≠ 0 f(-‐1) = -‐1+4+11-‐30 ≠ 0 f(2) = 8+16-‐22-‐30 ≠ 0 f(-‐2) = -‐8+16+22-‐30 = 0 Therefore (x+2) if a factor. Then could be 15,2,1 or 3, 2, 5 as factors however it must be 3,2,5 as factors as they add to make 11.
Con-nued Now to factorise:
x³+4x²-‐11x-‐30 (x+2)(Ax²+Bx+C) = Ax³+Bx²+Cx+2Ax²+2Bx+2C Therefore by equating coefficients, A=1, B+2A = 4 and by subbing in A=1 B= 2, C+2B= -‐11 C+4=-‐11 and C=-‐15. Therefore (x+2)(x²+2x-‐15) By factorising the second bracket: (x+2)(x+5)(x-‐3) Therefore the roots are: X = -‐2, -‐5 and 3. Also the y intercept is at (0,-‐30).
Remainder theorem Used to find the value of R when a value of x is
substituted into an equation of the form: (x+c)(Ax²+Bx+C)+R where (Ax²+Bx+C) is the quotient and R is the remainder. E.g. is (x+2) a factor of x³+3x²-‐6x+2 F(-‐2) = -‐8+12+12+2 = 18 and as F(x) ≠ 0 (x+2) id not a factor and R = 18. The next slides shows how to prove this using factorisation.
Con-nued... x³+3x²-‐6x+2 (x+2)(Ax²+Bx+C) + R
= Ax³+Bx²+Cx+2Ax²+2Bx+2C+R Therefore, A=1, 2A+B=3 and B=1, C+2B=-‐6 and C=-‐8. 2C+R = 2 -‐16+R= 2 R=18 This method may be needed to be used depending on the information given in the question.
Factor theorem To factorise cubic polynomial would be very difficult
without using it: x³+4x²-‐11x-‐30 into (x )(x ) (x ). However we can work out one of the factors to work out the rest. As we know the constants have to multiply to give thirty we know the constants have to have one of the following combinations: 5,3,2 3,1,10 6,5,1 30,1,1 or 15,2,1 If one of these numbers is a factor, when put into the equation, the equation must equal 0.
Con-nued.. Using the rule if f(a)=0 then (x-‐a) is a factor of f(x). Using the pervious example:
f(1)= 1+4-‐11-‐30 ≠ 0 f(-‐1) = -‐1+4+11-‐30 ≠ 0 f(2) = 8+16-‐22-‐30 ≠ 0 f(-‐2) = -‐8+16+22-‐30 = 0 Therefore (x+2) if a factor. Then could be 15,2,1 or 3, 2, 5 as factors however it must be 3,2,5 as factors as they add to make 11.
Con-nued Now to factorise:
x³+4x²-‐11x-‐30 (x+2)(Ax²+Bx+C) = Ax³+Bx²+Cx+2Ax²+2Bx+2C Therefore by equating coefficients, A=1, B+2A = 4 and by subbing in A=1 B= 2, C+2B= -‐11 C+4=-‐11 and C=-‐15. Therefore (x+2)(x²+2x-‐15) By factorising the second bracket: (x+2)(x+5)(x-‐3) Therefore the roots are: X = -‐2, -‐5 and 3. Also the y intercept is at (0,-‐30).
Remainder theorem Used to find the value of R when a value of x is
substituted into an equation of the form: (x+c)(Ax²+Bx+C)+R where (Ax²+Bx+C) is the quotient and R is the remainder. E.g. is (x+2) a factor of x³+3x²-‐6x+2 F(-‐2) = -‐8+12+12+2 = 18 and as F(x) ≠ 0 (x+2) id not a factor and R = 18. The next slides shows how to prove this using factorisation.
Con-nued... x³+3x²-‐6x+2 (x+2)(Ax²+Bx+C) + R
= Ax³+Bx²+Cx+2Ax²+2Bx+2C+R Therefore, A=1, 2A+B=3 and B=1, C+2B=-‐6 and C=-‐8. 2C+R = 2 -‐16+R= 2 R=18 This method may be needed to be used depending on the information given in the question.
