Using factor theorem, factorize each of the following polynomials:$y^3 - 2y^2- 29y - 42$

Given:

Given expression is $y^3 - 2y^2- 29y - 42$.

To do:

We have to find the given polynomial using factor theorem.

Solution:

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

The factors of the constant term $-42$ are $\pm 1, \pm 2, \pm 3, \pm 6, \pm 7, \pm 14, \pm 21, \pm 42$

Let $y=-1$, this implies,

$f(-1)=(-1)^{3}-2(-1)^{2}-29(-1)-42$

$=-1-2+29-42$

$=29-45$

$=-36 \
eq 0$

Therefore, $y \
eq 1$ is not its factor.

Let $y=-2$, this implies,

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

$=-8-8+58-42$

$=58-58$

$=0$

Therefore, $y+2$ is a factor of $f(x)$.

Dividing $f(y)$ by $y+2$, we have,

$y+2$) $y^{3}-2 y^{2}-29 y-42$($y^2-4y-21$

              $y^{3}+2 y^{2}$

           ---------------------------

                         $-4 y^{2}-29 y-42$

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

                      --------------------------

                                       $-21y-42$

                                       $-21y-42$ 

                                   ------------------

                                               0   

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

$=y(y-7)+3(y-7)$

$=(y-7)(y+3)$

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

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Updated on: 10-Oct-2022

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