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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,
$y - 2$ )$2 y ^ { 3 } - 5 y ^ { 2 } - 1 9 y + 4 2$ ( $2 y ^ { 2 } - y - 2 1$
$2 y^{3}-4 y^{2}$
-----------------------
$-y^{2}-19 y$
$-y^{2}+2 y$
--------------------
$-21 y+42$
$-21 y+42$
-------------------
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, $2y^{3}-5y^{2}-19 y+42=(y-2)(2y-7)(y+3)$.