Show that 2x + 7 is a factor of 2x³ + 5x² - 11x - 14. Hence, factorise the given expression completely, using the factor theorem.

By factor theorem, (x - a) is a factor of f(x), if f(a) = 0.

f(x) = 2x³ + 5x² - 11x - 14

(2x + 7) or $2(x - \big(-\dfrac{7}{2}\big))$ is a factor of f(x), if $f\big(-\dfrac{7}{2}\big)$ = 0

$f\big(-\dfrac{7}{2}\big) = 2\big(-\dfrac{7}{2}\big)^3 + 5\big(-\dfrac{7}{2}\big)^2 - 11\big(-\dfrac{7}{2}\big) -14 \\[1em] = 2\big(-\dfrac{343}{8}\big) + 5\big(\dfrac{49}{4}\big) + \big(\dfrac{77}{2}\big) - 14 \\[1em] = \big(-\dfrac{343}{4}\big) + \big(\dfrac{245}{4}\big) + \big(\dfrac{77}{2}\big) - 14$

On taking LCM,

$= \big(\dfrac{-343 + 245 + 154 - 56}{4}\big) \\[1em] = \big(\dfrac{-399 + 399}{4}\big) \\[1em] = 0$

Hence, (2x + 7) is the factor of f(x).

On dividing f(x) by (2x + 7),

$\begin{array}{l} \phantom{2x + 7)}{x^2 - x - 2} \ 2x + 7\overline{\smash{\big)}2x^3 + 5x^2 - 11x - 14} \ \phantom{2x + 7}\underline{\underset{-}{}2x^3 \underset{-}{+} 7x^2} \ \phantom{{2x + 7}2x^3+} -2x^2 - 11x \ \phantom{{2x + 7}2x^3+}\underline{\underset{+}{-}2x^2 \underset{+}{-} 7x} \ \phantom{{2x + 7}{2x^3+}{+2x^2-}}-4x - 14 \ \phantom{{2x + 7}{2x^3+}{+2x^2-}}\underline{\underset{+}{-}4x \underset{+}{-} 14} \ \phantom{{2x + 7}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, x² - x - 2 as the quotient and remainder = 0.

$\therefore 2x^3 + 5x^2 - 11x - 14 = (2x + 7)(x^2 - x - 2) \\[0.5em] = (2x + 7)(x^2 - 2x + x - 2) \\[0.5em] = (2x + 7)(x(x - 2) + 1(x - 2)) \\[0.5em] = (2x + 7)(x + 1)(x - 2)$

Hence, 2x³ + 5x² - 11x - 14 = (2x + 7)(x + 1)(x - 2)