Using factor theorem, factorize the following:

6x³ - 7x² - 11x + 12

Let, f(x) = 6x³ - 7x² - 11x + 12.

Substituting, x = 1 in f(x), we get :

f(1) = 6(1)³ - 7(1)² - 11(1) + 12

= 6 - 7 - 11 + 12

= 0

Since, f(1) = 0, (x - 1) is a factor of f(x).

Dividing f(x) by (x - 1), we get :

$\begin{array}{l} \phantom{x - ]3)}{6x^2 - x - 12} \ x - 1\overline{\smash{\big)}6x^3 - 7x^2 - 11x + 12} \ \phantom{x - 2l}\underline{\underset{-}{}6x^3 \underset{+}{-}6x^2} \ \phantom{{x - 2}x^,,,3-}-x^2 - 11x \ \phantom{{x -k.l2}fx^3]\space}\underline{\underset{+}{-}x^2 \underset{-}{+}x} \ \phantom{{x - 2]euo[ki]}x^3okk\space}{-12x + 12} \ \phantom{{x - 2}x^3o;llk]lmk,\space}\underline{\underset{+}{-}12x\underset{-}{+}12} \ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

∴ 6x³ − 7x² − 11x + 12 = (x − 1)(6x² − x − 12)

= (x − 1)(6x² − 9x + 8x − 12)

= (x − 1)[3x(2x − 3) + 4(2x − 3)]

= (x − 1)(2x − 3)(3x + 4)

Hence, 6x³ − 7x² − 11x + 12 = (x − 1)(2x − 3)(3x + 4).