Using factor theorem, factorize of the polynomials:

Solution:

Given, $f(x)=y^{3}-2 y^{2}-29 y-42$

The constant in f(x) is - 42

The factors of -42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42

Let, y + 2 = 0

=> y = – 2

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

= -8 -8 + 58 – 42

= 0

So, (y + 2) is the factor of f(y)

Now, divide f(y) with (y + 2) to get other factors

By, long division

$y^{2}-4 y-21$

$y+2 y^{3}-2 y^{2}-29 y-42$

 

$y^{3}+2 y^{2}$

$(-) \quad(-)$

$-4 y^{2}-29 y$

 

$-4 y^{2}-8 y$

(+)      (+)

- 21y – 42

- 21y – 42

(+)        (+)

0

$\Rightarrow y^{3}-2 y^{2}-29 y-42=(y+2)\left(y^{2}-4 y-21\right)$

Now,

$y^{2}-4 y-21=y^{2}-7 y+3 y-21$

= y(y – 7) +3(y – 7)

= (y – 7)(y + 3)

Hence, $y^{3}-2 y^{2}-29 y-42=(y+2)(y-7)(y+3)$