Using factor theorem, factorize the following:

2x³ + x² - 13x + 6

Let, f(x) = 2x³ + x² - 13x + 6.

Substituting, x = 2 in f(x) we get :

f(2) = 2(2)³ + (2)² - 13(2) + 6

= 2(8) + 4 - 26 + 6

= 16 + 4 - 26 + 6

= 0.

Since, f(2) = 0, thus (x - 2) is factor of f(x).

Dividing, f(x) by (x - 2), we get :

$\begin{array}{l} \phantom{x - ]3)}{2x^2 + 5x - 3} \ x - 2\overline{\smash{\big)}2x^3 + x^2 - 13x + 6} \ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}4x^2} \ \phantom{{x - 2}x^,,,3-}5x^2 - 13x \ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}5x^2 \underset{+}{-}10x} \ \phantom{{x - 2]euo[ki]}x^3o.\space}{-3x + 6} \ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{+}{-}3x\underset{-}{+}6} \ \phantom{{x - 2}{x^,jo-k2x^2\space}{-9x}}\times \end{array}$

∴ 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).