Use factor theorem to show that (x + 2) and (2x - 3) are factors of (2x² + x - 6).

Let f(x) = 2x² + x - 6

Given,

Factors : (x + 2) and (2x - 3)

⇒ x + 2 = 0 and 2x - 3 = 0

⇒ x = -2 and 2x = 3

⇒ x = -2 and x = $\dfrac{3}{2}$

By factor theorem,

(x - a) is a factor of f(x), if f(a) = 0.

Thus, (x + 2) and (2x - 3) are factors of f(x), if f(-2) = 0 and f$\Big(\dfrac{3}{2}\Big)$ = 0.

On dividing 2x² + x - 6 by (x + 2), we get :

⇒ f(-2) = 2(-2)² + (-2) - 6

= 2(4) - 2 - 6

= 8 - 2 - 6

= 8 - 8

= 0.

On dividing 2x² + x - 6 by (2x - 3), we get :

$\Rightarrow f\Big(\dfrac{3}{2}\Big) = 2\Big(\dfrac{3}{2}\Big)^2 + \Big(\dfrac{3}{2}\Big) - 6 \\[1em] = 2\Big(\dfrac{9}{4}\Big) + \Big(\dfrac{3}{2}\Big) - 6 \\[1em] = \Big(\dfrac{9}{2}\Big) + \Big(\dfrac{3}{2}\Big) - 6 \\[1em] = \Big(\dfrac{9 + 3}{2}\Big) - 6 \\[1em] = \Big(\dfrac{12}{2}\Big) - 6 \\[1em] = 6 - 6 \\[1em] = 0.$

Since, f(-2) = $f\Big(\dfrac{3}{2}\Big)$ = 0.

Hence, proved that x + 2 and 2x - 3 are factors of 2x² + x - 6.