Question

factorize using factor theorem: x^3-23x^2+32x+20

Answer 1

Step-by-step explanation:

let p(x)=x³-23x²+32x+20

prime factor of constant term 20=1*2*2*5

putting x=2

p(2)=2³-23*4+32*2+20

=8-92+64+20

=92-92

=0

(x-2)is a factor of p(x)

by long division

x-2)x³-23x²+32x+20(x²-21x-10

x³-2x²

-21x²+32x

-21x²+42x

- 10x+20

-10x+20

0

since

x3-23x²+32x+20

=(x-2)(x²-21x-10)

=(x-2)[x²- (10+1)x-10

(x-2)(x²-10x+x-10)

(x-2)(x(x-10)+1(x-10)

(x-1)(x-10)(x+1)

HOPE IT HELPS YOU