Using the remainder theorem, factorise each of the following completely :

x³ + x² - 4x - 4

For x = -1 the value of x³ + x² - 4x - 4

= (-1)³ + (-1)² - 4(-1) - 4

= -1 + 1 + 4 - 4

= 0.

Hence, (x + 1) is the factor of x³ + x² - 4x - 4.

On dividing x³ + x² - 4x - 4 by (x + 1),

$\begin{array}{l} \phantom{x + 1)}{x^2 - 4} \ x + 1\overline{\smash{\big)}x^3 + x^2 - 4x - 4} \ \phantom{x + 1}\underline{\underset{-}{}x^3 \underset{-}{+} x^2} \ \phantom{{x + 1}x^3+x^2-}-4x - 4 \ \phantom{{x + 1}x^3+x^2-}\underline{\underset{+}{-}4x \underset{+}{-} 4} \ \phantom{{x + 1}x^3+x^2-4x\enspace} \times \end{array}$

we get, quotient = x² - 4

Factorising x² - 4,

= (x)² - 4

= (x + 2)(x - 2)

∴ x² - 4 = (x - 2)(x + 2)

Hence, x³ + x² - 4x - 4 = (x + 1)(x + 2)(x - 2).