Question

Factorize:
$2x^3-3x^2-17x+30$

Answer 1

Let $f\left(x\right)=2x^3-3x^2-17x+30$ be the given polynomial. The factors of the constant term $+30$ are $\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30$. The factor of coefficient of $x^3$ is $2$.

Hence, possible rational roots of $f\left(x\right)$ are:

$\pm 1,\pm 3,\pm 5,\pm 15,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2}$.

We $f\left(2\right)=2(2{)}^3-3(2{)}^2-17\left(2\right)+30$

$=2\left(8\right)-3\left(4\right)-17\left(2\right)+30$

$=16-12-34+30=0$

And $f(-3)=2(-3{)}^3-3(-3{)}^2-17(-3)+30$

$=2(-27)-3\left(9\right)-17(-3)+30$

$=-54-27+51+30=0$

So, $(x-2)$ and $(x+3)$ are factors of $f\left(x\right)$.

$\Rightarrow x^2+x-6$ is a factor of $f\left(x\right)$.

Let us now divide $f\left(x\right)=2x^3-3x^2-17x+30$ by $x^2+x-6$ to get the other factors of $f\left(x\right)$.

Factors of $f\left(x\right)$.

By long division, we have

REF. IMAGE

$\therefore 2x^3-3x^2-17x+30=(x^2+x-6)(2x-5)$

$\Rightarrow 2x^3-3x^2-17x+30=(x-2)(x+3)(2x-5)$

Hence, $2x^3-3x^2-17x+30=(x-2)(x+3)(2x-5)$