Factorize completely using factor theorem : 2x³ - x² - 13x - 6.

Substituting x = -2 in 2x³ - x² - 13x - 6, we get :

⇒ 2(-2)³ - (-2)² - 13(-2) - 6

⇒ 2(-8) - 4 + 26 - 6

⇒ -16 - 4 + 20

⇒ -20 + 20

⇒ 0.

∴ x + 2 is a factor of the polynomial 2x³ - x² - 13x - 6.

Dividing, 2x³ - x² - 13x - 6 by x + 2, we get :

$\begin{array}{l} \phantom{x + 2)}{\quad 2x^2 -5x - 3} \ x + 2\overline{\smash{\big)}\quad 2x^3 - x^2 - 13x - 6} \ \phantom{x + 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{-}{+}4x^2} \ \phantom{{x + 2}x^3-2}-5x^2 - 13x \ \phantom{{x + 2)}x^3-2}\underline{\underset{+}{-}5x^2 \underset{+}{-} 10x} \ \phantom{{x + 2)}{x^3-2x^{2}(3)}}-3x - 6 \ \phantom{{x + 2)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}3x \underset{+}{-} 6} \ \phantom{{x + 2)}{x^3-2x^{2}(31)}{-2x}}\times \end{array}$

∴ 2x³ - x² - 13x - 6 = (x + 2)(2x² - 5x - 3)

= (x + 2)(2x² - 6x + x - 3)

= (x + 2)[2x(x - 3) + 1(x - 3)]

= (x + 2)(2x + 1)(x - 3).

Hence, 2x³ - x² - 13x - 6 = (x + 2)(2x + 1)(x - 3).