Given that x - 2 and x + 1 are factors of f(x) = x³ + 3x² + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).

x - 2 = 0 ⇒ x = 2.

Since, x - 2 is a factor of x³ + 3x² + ax + b. Hence, on substituting x = 2 in above expression, remainder = 0.

⇒ (2)³ + 3(2)² + a(2) + b = 0

⇒ 8 + 12 + 2a + b = 0

⇒ 2a + b = -20

⇒ b = -20 - 2a ……..(i)

x + 1 = 0 ⇒ x = -1.

Since, x + 1 is a factor of x³ + 3x² + ax + b. Hence, on substituting x = -1 in above expression, remainder = 0.

⇒ (-1)³ + 3(-1)² + a(-1) + b = 0

⇒ -1 + 3 - a + b = 0

⇒ 2 - a + b = 0

⇒ b = a - 2 …….(ii)

From (i) and (ii) we get,

⇒ -20 - 2a = a - 2

⇒ a + 2a = -20 + 2

⇒ 3a = -18

⇒ a = -6.

Substituting value of a in (ii) we get,

⇒ b = a - 2 = -6 - 2 = -8.

∴ a = -6 and b = -8.

On dividing, x³ + 3x² - 6x - 8 by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{x^2 + 5x + 4} \ x - 2\overline{\smash{\big)}x^3 + 3x^2 - 6x - 8} \ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-} 2x^2} \ \phantom{{x - 2}3x^3+}5x^2 - 6x \ \phantom{{x - 2}3x^3\enspace\space}\underline{\underset{-}{}5x^2 \underset{+}{-} 10x} \ \phantom{{x - 2}{3x^3+1}{+2x^2}}4x - 8 \ \phantom{{x - 2}{2x^3+}{+2x^2}{4}}\underline{\underset{-}{}4x \underset{+}{-} 8} \ \phantom{{x - 2}{2x^3+}{+2x^2-}{4x}}\times \end{array}$

we get, quotient = x² + 5x + 4.

Factorising x² + 5x + 4,

= x² + 4x + x + 4

= x(x + 4) + 1(x + 4)

= (x + 1)(x + 4).

Hence, a = -6, b = -8 and f(x) = (x - 2)(x + 1)(x + 4).