Using factor theorem, show that:

(3x - 2) is a factor of (3x³ + x² - 20x + 12).

Let f(x) = (3x³ + x² - 20x + 12)

Given,

Divisor:

⇒ 3x - 2 = 0

⇒ 3x = 2

⇒ x = $\dfrac{2}{3}$

By factor theorem,

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

Substituting x = $\Big(\dfrac{2}{3}\Big)$ in f(x), we get :

$\Rightarrow f\Big(\dfrac{2}{3}\Big) = 3\Big(\dfrac{2}{3}\Big)^3 + \Big(\dfrac{2}{3}\Big)^2 - 20\Big(\dfrac{2}{3}\Big) + 12 \\[1em] = \Big(\dfrac{8}{9}\Big) + \Big(\dfrac{4}{9}\Big) - \Big(\dfrac{40}{3}\Big) + 12 \\[1em] = \Big(\dfrac{8 + 4 - 120 + 108}{9}\Big) \\[1em] = \Big(\dfrac{120 - 120}{9}\Big) \\[1em] = 0.$

Since, $f\Big(\dfrac{2}{3}\Big)$ = 0, thus (3x - 2) is a factor of f(x).

Hence, proved that (3x - 2) is factor of 3x³ + x² - 20x + 12.