Show that (x - 3) is a factor of (2x³ - 3x² - 11x + 6) and hence factorize (2x³ - 3x² - 11x + 6).

Let f(x) = 2x³ - 3x² - 11x + 6

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

f(3) = 2(3)³ - 3(3)² - 11(3) + 6

= 54 - 27 - 33 + 6

= 60 - 60

= 0.

Since f(3) = 0, thus (x − 3) is a factor of 2x³ - 3x² - 11x + 6.

Now, dividing f(x) by x - 3,

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

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

= (x - 3)(2x² + 4x - x - 2)

= (x - 3)[2x(x + 2) - 1(x + 2)]

= (x - 3)(2x - 1)(x + 2).

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