Use factor theorem to factorise the following polynomials completely :

(i) x³ + 2x² - 5x - 6

(ii) x³ - 13x - 12

(iii) 6x³ + 17x² + 4x - 12

(i) f(x) = x³ + 2x² - 5x - 6

Putting, x = -1 in f(x)

$f(-1) = (-1)^3 + 2(-1)^2 - 5(-1) - 6 \\[0.5em] = -1 + 2 + 5 - 6 \\[0.5em] = 7 - 7 \\[0.5em] = 0$

Since, f(-1) = 0, hence (x + 1) is factor of f(x) by factor theorem.

Now, dividing f(x) by (x + 1),

we get x² + x - 6 as the quotient and remainder = 0.

$\therefore x^3 + 2x^2 - 5x - 6 = (x + 1)(x^2 + x - 6) \\[0.5em] = (x + 1)(x^2 + 3x - 2x - 6) \\[0.5em] = (x + 1)(x(x + 3) - 2(x + 3)) \\[0.5em] = (x + 1)(x - 2)(x + 3)$

Hence, x³ + 2x² - 5x - 6 = (x + 1)(x - 2)(x + 3).

(ii) Let f(x) = x³ - 13x - 12

Putting, x = 4 in f(x)

$f(4) = (4)^3 - 13(4) - 12 \\[0.5em] = 64 - 52 - 12 \\[0.5em] = 64 - 64 \\[0.5em] = 0$

Since, f(4) = 0, hence (x - 4) is factor of f(x) by factor theorem.

Now, dividing f(x) by (x - 4),

we get x² + 4x + 3 as quotient and remainder = 0.

$\therefore x^3 - 13x - 12 = (x - 4)(x^2 + 4x + 3) \\[0.5em] = (x - 4)(x^2 + 3x + x + 3) \\[0.5em] = (x - 4)(x(x + 3) + 1(x + 3)) \\[0.5em] = (x - 4)(x + 1)(x + 3)$

Hence, x³ - 13x - 12 = (x - 4)(x + 1)(x + 3).

(iii) Given,

f(x) = 6x³ + 17x² + 4x - 12

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

f(-2) = 6(-2)³ + 17(-2)² + 4(-2) - 12

= 6 × -8 + 17 × 4 - 8 - 12

= -48 + 68 - 8 - 12

= -68 + 68

= 0.

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

Dividing f(x) by (x + 2), we get :

$\begin{array}{l} \phantom{x - 3x )}{\quad 6x^2 + 5x - 6} \ x + 2\overline{\smash{\big)}\quad 6x^3 + 17x^2 + 4x - 12} \ \phantom{x^2 + 4)}\phantom{)}\underline{\underset{-}{+}6x^3 \underset{-}{+} 12x^2 } \ \phantom{{x^2 + 3x - 5 =d)}} 5x^2 + 4x - 12 \ \phantom{{x^2 + 3x - 54 =)}}\underline{\underset{-}{+}5x^2 \underset{-}{+}10x } \ \phantom{{x^2 + 3x - 54)} + 2x + 3 }-6x - 12\ \phantom{{x^2 + 3x - 54 =zccdvz)}}\underline {\underset{+}{-}6x \underset{+}{-}12 } \ \phantom{{x^2 + 3x - 54)} + 2x + 3 + dc}\times\ \end{array}$

We get 6x² + 5x - 6 as the quotient and remainder = 0.

∴ 6x³ + 17x² + 4x - 12 = (x + 2)(6x² + 5x - 6)

= (x + 2)(6x² + 9x - 4x - 6)

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

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

Hence,6x³ + 17x² + 4x - 12 = (x + 2)(2x + 3)(3x - 2).