Use factor theorem to factorise the following polynomials completely :

(i) 4x³ + 4x² - 9x - 9

(ii) x³ - 19x - 30

(iii) 2x³ - x² - 13x - 6

(i) f(x) = 4x³ + 4x² - 9x - 9

Let x = -1, substituting the value of x in f(x),

$f(-1) = 4(-1)^3 + 4(-1)^2 - 9(-1) - 9 \\[0.5em] = -4 + 4 + 9 - 9 \\[0.5em] = 0$

Since, f(-1) = 0 hence, (x + 1) is a factor of 4x³ + 4x² - 9x - 9.

On dividing, 4x³ + 4x² - 9x - 9 by (x + 1),

$\begin{array}{l} \phantom{x + 1)}{4x^2 - 9} \ x + 1\overline{\smash{\big)}4x^3 + 4x^2 - 9x - 9} \ \phantom{x + 1}\underline{\underset{-}{}4x^3 \underset{-}{+}4x^2} \ \phantom{{x + 1}{4x^3}{4x^2-9}}-9x - 9 \ \phantom{{x + 1}{4x^3}{4x^2-9x}}\underline{\underset{+}{-}9x \underset{+}{-} 9} \ \phantom{{x + 1}{4x^3}{4x^2-9}{-9x}}\times \end{array}$

we get (4x² - 9) as quotient and remainder = 0.

$\therefore 4x^3 + 4x^2 - 9x - 9 = (x + 1)(4x^2 - 9) \\[0.5em] = (x + 1)((2x)^2 - (3)^2) \\[0.5em] = (x + 1)(2x - 3)(2x + 3)$

Hence, 4x³ + 4x² - 9x - 9 = (x + 1) (2x - 3)(2x + 3).

(ii) f(x) = x³ - 19x - 30

Let x = -2, substituting the value of x in f(x),

$f(-2) = (-2)^3 - 19(-2) - 30 \\[0.5em] = -8 + 38 - 30 \\[0.5em] = 0$

Since, f(-2) = 0 hence, (x + 2) is a factor of x³ - 19x - 30.

On dividing, x³ - 19x - 30 by (x + 2),

$\begin{array}{l} \phantom{x + 2)}{x^2 - 2x - 15} \ x + 2\overline{\smash{\big)}x^3 - 19x - 30} \ \phantom{x + 2}\underline{\underset{-}{ }x^3 \underset{-}{+} 2x^2} \ \phantom{{x + 2}{x^3+}}-2x^2 - 19x \ \phantom{{x + 2}x^3+}\underline{\underset{+}{-}2x^2 \underset{+}{-} 4x} \ \phantom{{x + 2}{-x^3+2x^2}}-15x - 30 \ \phantom{{x + 2}{-x^3+2x^2+}}\underline{\underset{+}{-}15x \underset{+}{-} 30} \ \phantom{{x + 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get x² - 2x - 15 as quotient and remainder = 0.

$\therefore x^3 - 19x - 30 = (x + 2)(x^2 - 2x - 15) \\[0.5em] = (x + 2)(x^2 - 5x + 3x - 15) \\[0.5em] = (x + 2)(x(x - 5) + 3(x - 5)) \\[0.5em] = (x + 2)(x + 3)(x - 5)$

Hence, x³ - 19x - 30 = (x + 2) (x + 3)(x - 5).

(iii) f(x) = 2x³ - x² - 13x - 6

Substituting x = -2 in 2x³ - x² - 13x - 6, we get :

⇒ 2(-2)³ - (-2)² - 13(-2) - 6

⇒ 2(-8) - 4 + 26 - 6

⇒ -16 - 4 + 20

⇒ -20 + 20

⇒ 0.

∴ x + 2 is a factor of the polynomial 2x³ - x² - 13x - 6.

Dividing, 2x³ - x² - 13x - 6 by x + 2, we get :

$\begin{array}{l} \phantom{x + 2)}{\quad 2x^2 -5x - 3} \ x + 2\overline{\smash{\big)}\quad 2x^3 - x^2 - 13x - 6} \ \phantom{x + 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{-}{+}4x^2} \ \phantom{{x + 2}x^3-2}-5x^2 - 13x \ \phantom{{x + 2)}x^3-2}\underline{\underset{+}{-}5x^2 \underset{+}{-} 10x} \ \phantom{{x + 2)}{x^3-2x^{2}(3)}}-3x - 6 \ \phantom{{x + 2)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}3x \underset{+}{-} 6} \ \phantom{{x + 2)}{x^3-2x^{2}(31)}{-2x}}\times \end{array}$

∴ 2x³ - x² - 13x - 6 = (x + 2)(2x² - 5x - 3)

= (x + 2)(2x² - 6x + x - 3)

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

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

Hence, 2x³ - x² - 13x - 6 = (x + 2)(2x + 1)(x - 3).