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