Question

Factorise by using the factor theorem y3-2y2-29y-42

Answer 1

So this question is pretty easy and I will explain step by step ok but with a pen and sheet.

Answer 2

By factorization of $f(-2)=(-2)^{3}-2(-2)^{2}-29(-2)-42$, the factors are (y+3)(y-7)(y+2)

Solution:

By trial, f(1) ≠ 0,f(-1) ≠ 0,f(2) ≠ 0,f(-2) = 0

Hence y+2 is a factor.

$f(-2)=(-2)^{3}-2(-2)^{2}-29(-2)-42$

f(-2) = (-8) – 2(4) + 58 – 42  

f(-2) = -16 + 58 – 42  

f(-2) = 0

The long division process is shown in the below attached image.

$\left(y^{2}-4 y-21\right)(y+2)$ are the factors.

$=\left(y^{2}-7 y+3 y-21\right)(y+2)$

= y(y – 7) + 3(y – 7)( y+2)  

(y+3)(y-7)(y+2) are the factors.