Let $f(x)=2 x^{3}-3 x^{2}-17 x+30$ 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 $f ( x )$ 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 have $\quad f(2)=2(2)^{3}-3(2)^{2}-17(2)+30$

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

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

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

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

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

So, $(x-2)$ and $(x+3)$ are factors of $f(x)$ $\Rightarrow \quad x^{2}+x-6$ is a factor of $f(x)$

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

Factors of $f(x).$

By long division, we have

$\begin{array}{l}x ^ { 2 } + x – 6 |\overline {2 x ^ { 3 } – 3 x ^ { 2 } – 1 7 x + 3 0} (2x-5)\\ \;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\; 2 x^{3}+2 x^{2}-12 x\;\;\;\;\;\;\; \\ \hline \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-5 x^{2}-5 x+30 \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\,-5 x^{2}-5 x+30 \\ \hline \;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\; 0 \end{array}$

$\therefore \quad 2 x^{3}-3 x^{2}-17 x+30=\left(x^{2}+x-6\right)(2 x-5)$

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

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