Factorise: 2xcube-3x-17x+30

Let f(x) = 2x³ - 3x² - 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 x³ 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 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).
$\Rightarrow$ x² + x - 6 is a factor of f(x).
Let us noe divide f(x) = 2x³ - 3x² - 17x + 30 by x² + x - 6 to get the other factors of f(x).
Factors of f(x).
By long division, we have

$\therefore$ 2x³ - 3x² - 17x + 30 = (x² + x - 6)(2x - 5)
$\Rightarrow$ 2x³ - 3x² - 17x + 30 = (x - 2)(x + 3)(2x - 5)
Hence, 2x³ - 3x² - 17x + 30 = (x - 2)(x + 3)(2x - 5)