Using remainder theorem, factorise 6x³ - 11x² - 3x + 2 completely.

Substituting, x = 2 in 6x³ - 11x² - 3x + 2, we get :

⇒ 6(2)³ - 11(2)² - 3(2) + 2

⇒ 6 × 8 - 11 × 4 - 6 + 2

⇒ 48 - 44 - 6 + 2

⇒ 6 - 6

⇒ 0.

∴ (x - 2) is a factor of 6x³ - 11x² - 3x + 2.

On dividing 6x³ - 11x² - 3x + 2 by (x - 2), we get:

$\begin{array}{l} \phantom{x - 2}{6x^2 + x - 1} \ x - 2 \overline{\smash{\big)}6x^3 - 11x^2 - 3x + 2} \ \phantom{x - 2}\underline{\underset{-}{}6x^3 \underset{+}{-}12x^2} \ \phantom{{x - 2}2x^3+}x^2 - 3x \ \phantom{{x - 2}x^3+}\underline{\underset{-}{}x^2 \underset{+}{-}2x} \ \phantom{{x - 2}x^3+2-5}-x + 2 \ \phantom{{x - 2}x^3+2-x1}\underline{\underset{+}{-}x\underset{-}{+}2} \ \phantom{{x - 2}x^3+2-5x^233} \times \end{array}$

∴ 6x³ - 11x² - 3x + 2 = (x - 2)(6x² + x - 1)

⇒ 6x³ - 11x² - 3x + 2 = (x - 2)(6x² + 3x - 2x - 1)

⇒ 6x³ - 11x² - 3x + 2 = (x - 2)[3x(2x + 1) - 1(2x + 1)]

⇒ 6x³ - 11x² - 3x + 2 = (x - 2)(3x - 1)(2x + 1).

Hence, 6x³ - 11x² - 3x + 2 = (x - 2)(3x - 1)(2x + 1).