# Using factor theorem, factorize each of the following polynomials:$2y^3 - 5y^2 - 19y + 42$

Given:

Given expression is $2y^3 - 5y^2 - 19y + 42$.

To do:

We have to find the given polynomial using factor theorem.

Solution:

Let $f(y)=2 y^{3}-5 y^{2}-19 y+42$

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,

$f(1)=2(1)^{3}-5(1)^{2}-19(1)+42$

$=2-5-19+42$

$=44-24$

$=20 \ eq 0$

Therefore, $y-1$ is not a factor of $f(x)$.

Let $y=2$, this implies,

$y(2)=2(2)^{3}-5(2)^{2} - 19(2)+42$

$=16-20-38+42$

$=58-58$

$=0$

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

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

$y - 2$ )$2 y ^ { 3 } - 5 y ^ { 2 } - 1 9 y + 4 2$ ( $2 y ^ { 2 } - y - 2 1$

$2 y^{3}-4 y^{2}$

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$-y^{2}-19 y$

$-y^{2}+2 y$

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$-21 y+42$
$-21 y+42$

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0
Therefore,

$2 y^{3}-5 y^{2}-19 y+42=(y-2)(2 y^{2}-7 y+6 y-21)$

$=(y-2)[y(2 y-7)+3(2 y-7)]$

$=(y-2)(2 y-7)(y+3)$

Therefore, $2y^{3}-5y^{2}-19 y+42=(y-2)(2y-7)(y+3)$.

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Updated on: 10-Oct-2022

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