If (x + 2) and (x - 3) are the factors of x³ + ax + b, find the values of a and b. With these values of a and b, factorise the given expression.

f(x) = x³ + ax + b

If (x + 2) or (x - (-2)) and (x - 3) are factors of f(x) then, f(-2) and f(3) = 0.

$\therefore f(-2) = (-2)^3 + (-2)a + b = 0 \\[0.5em] \Rightarrow -8 - 2a + b = 0 \\[0.5em] \Rightarrow b = 2a + 8 \text{ \space (Equation 1)}$

$\therefore f(3) = (3)^3 + (3)a + b = 0 \\[0.5em] \Rightarrow 27 + 3a + b = 0$

Putting value of b = 2a + 8 from equation 1,

$\Rightarrow 27 + 3a + 2a + 8 = 0 \\[0.5em] \Rightarrow 35 + 5a = 0 \\[0.5em] \Rightarrow 5a = -35 \\[0.5em] \Rightarrow a = -7 \\[0.5em] \therefore b = 2a + 8 = -14 + 8 = -6$

Putting the values of a and b in f(x) we get,

f(x) = x³ - 7x - 6

Since, (x + 2) and (x - 3) are factors of f(x) hence, (x + 2)(x - 3) = (x² - x - 6) is also the factor.

On dividing f(x) by x² - x - 6,

$\begin{array}{l} \phantom{x^2 - x - 6)}{x + 1} \ x^2 - x - 6\overline{\smash{\big)}x^3 - 7x - 6} \ \phantom{x^2 - x - 6}\underline{\underset{-}{ }x^3 \underset{+}{-} x^2 \underset{+}{-} 6x} \ \phantom{{x^2 - x - 6}2x^3+4}x^2 - x - 6 \ \phantom{{x^2 - x - 6}2x^3+}\underline{\underset{-}{}x^2 \underset{+}{-} x \underset{+}{-} 6} \ \phantom{{x^2 - x - 6}{2x^3+}{+2x^2-}}\times \end{array}$

we get, (x + 1) as quotient and remainder = 0.

$\therefore x^3 - 7x - 6 = (x^2 - x - 6)(x + 1) \\[0.5em] = (x^2- 3x + 2x - 6)(x + 1) \\[0.5em] = (x(x - 3) + 2(x - 3))(x + 1) \\[0.5em] = (x + 2)(x - 3)(x + 1)$

Hence, the value of a = -7 and b = -6; x³ - 7x - 6 = (x + 2)(x - 3)(x + 1).