If (x + 1) and (x - 2) are factors of x³ + (a + 1)x² - (b - 2)x - 6, find the values of a and b. And then, factorise the given expression completely.

Given,

(x + 1) and (x - 2) are factors of x³ + (a + 1)x² - (b - 2)x - 6

x + 1 = 0 ⇒ x = -1.

Since, x + 1 is factor of x³ + (a + 1)x² - (b - 2)x - 6. Hence, on substituting x = -1 in expression, remainder = 0.

⇒ (-1)³ + (a + 1)(-1)² - (b - 2)(-1) - 6 = 0

-1 + (a + 1)(1) - (-b + 2) - 6 = 0

-1 + a + 1 + b - 2 - 6 = 0

a + b = 8

b = 8 - a ……..(i)

x - 2 = 0 ⇒ x = 2.

Since, x - 2 is factor of x³ + (a + 1)x² - (b - 2)x - 6. Hence, on substituting x = 2 in expression, remainder = 0.

⇒ (2)³ + (a + 1)(2)² - (b - 2)(2) - 6 = 0

⇒ 8 + (a + 1)(4) - (2b - 4) - 6 = 0

⇒ 8 + 4a + 4 - 2b + 4 - 6 = 0

⇒ 10 + 4a - 2b = 0

⇒ 2b = 4a + 10

⇒ b = 2a + 5 ……..(ii)

From (i) and (ii) we get,

⇒ 8 - a = 2a + 5

⇒ 2a + a = 8 - 5

⇒ 3a = 3.

⇒ a = 1.

Substituting a = 1, in (i) we get,

⇒ b = 8 - 1 = 7.

Substituting a = 1, b = 7 in expression we get,

Expression = x³ + (1 + 1)x² - (7 - 2)x - 6 = x³ + 2x² - 5x - 6.

On dividing, x³ + 2x² - 5x - 6 by (x + 1),

$\begin{array}{l} \phantom{x + 1)}{x^2 + x - 6} \ x + 1\overline{\smash{\big)}x^3 + 2x^2 - 5x - 6} \ \phantom{x + 1}\underline{\underset{-}{}x^3 \underset{+}{-} x^2} \ \phantom{{x + 1}3x^3+}x^2 - 5x \ \phantom{{x + 1}3x^3\enspace}\underline{\underset{-}{}x^2 \underset{-}{+} x} \ \phantom{{x + 1}{x^32x^2}{+2}}-6x - 6 \ \phantom{{x + 1}{x^32x^2}{+2x}}\underline{\underset{+}{-}6x \underset{+}{-} 6} \ \phantom{{x + 1}{x^32x^2}{+2x^2-}}\times \end{array}$

we get, quotient = x² + x - 6.

Factorising, x² + x - 6

= x² + 3x - 2x - 6

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

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

Hence, a = 1, b = 7 and x³ + 2x² - 5x - 6 = (x + 1)(x - 2)(x + 3).