Using factor theorem, factorize the following:

2x³ − x² − 13x − 6

Let, f(x) = 2x³ − x² − 13x − 6.

Substituting, x = 3 in f(x) we get :

f(3) = 2(3)³ − (3)² − 13(3) − 6

= 2(27) − 9 − 39 − 6

= 54 − 9 − 39 − 6

= 0.

Since, f(3) = 0, thus (x − 3) is factor of f(x).

Dividing, f(x) by (x − 3), we get :

$\begin{array}{l} \phantom{x - ]3)}{2x^2 + 5x + 2} \ x - 3\overline{\smash{\big)}2x^3 - x^2 - 13x - 6} \ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}6x^2} \ \phantom{{x - 2}x^,,,3-}5x^2 - 13x \ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}5x^2 \underset{+}{-}15x} \ \phantom{{x - 2]euo[ki]}x^3okk\space}{2x - 6} \ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}2x\underset{+}{-}6} \ \phantom{{x - 2}{x^,jo-k2x^2\space}{-9x}}\times \end{array}$

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

= (x − 3)(2x² + 4x + x + 2)

= (x − 3)[2(x + 2) + 1(x + 2)]

= (x − 3)(2x + 1)(x + 2).

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