Using the factor theorem, show that (x - 2) is a factor of x³ + x² - 4x - 4. Hence factorize the polynomial completely.

Let, f(x) = x³ + x² - 4x - 4.

Factor :

⇒ x - 2 = 0

⇒ x = 2.

Substituting x = 2 in f(x), we get :

⇒ f(2) = (2)³ + (2)² - 4(2) - 4

= 8 + 4 - 8 - 4

= 12 - 12

= 0.

Since, f(2) = 0, thus (x - 2) is a factor of (x³ + x² - 4x - 4).

Now, dividing f(x) by (x - 2), we get :

$\begin{array}{l} \phantom{x -13}{x^2 + 3x + 2} \ x - 2\overline{\smash{\big)}x^3 + x^2 - 4x - 4} \ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-}2x^2} \ \phantom{{x - 2}x^,.3-}3x^2 - 4x \ \phantom{{x -l2}fx^l.\space}\underline{\underset{-}{+}3x^2 \underset{+}{-}6x} \ \phantom{{x - 2]euo[ki]}x^3okk\space}{2x - 4} \ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}2x\underset{+}{-}4} \ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

∴ x³ + x² - 4x - 4 = (x - 2)(x² + 3x + 2)

= (x - 2)(x² + x + 2x + 2)

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

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

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