If x³ - 2x² + px + q has a factor (x + 2) and leaves a remainder 9 when divided by (x + 1), find the values of p and q. With these values of p and q, factorise the given polynomial completely.

By factor theorem (x - b) is a factor of f(x), if f(b) = 0.

f(x) = x³ - 2x² + px + q

Given, (x + 2) or (x - (-2)) is a factor of f(x).

∴ f(-2) = 0

$\Rightarrow (-2)^3 - 2(-2)^2 + p(-2) + q = 0 \\[0.5em] \Rightarrow -8 - 8 - 2p + q = 0 \\[0.5em] \Rightarrow -2p + q = 16 \\[0.5em] q = 2p + 16 \text{ \space (Equation 1)}$

By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

∴ On dividing f(x) by (x + 1) or (x - (-1)), Remainder = f(-1)

Given, Remainder = 9

∴ f(-1) = 9

$\Rightarrow (-1)^3 - 2(-1)^2 + p(-1) + q = 9 \\[0.5em] \Rightarrow -1 - 2 - p + q = 9 \\[0.5em] \Rightarrow -3 - p + q = 9 \\[0.5em] \Rightarrow q - p = 9 + 3 \\[0.5em] \Rightarrow q - p = 12$

Putting value of q = 2p + 16 from equation 1,

$\Rightarrow 2p + 16 - p = 12 \\[0.5em] \Rightarrow p + 16 = 12 \\[0.5em] \Rightarrow p = 12 - 16 \\[0.5em] \Rightarrow p = -4 \\[0.5em] \text{ and } q = 2p + 16 = 2(-4) + 16 = -8 + 16 = 8$

Now putting p = -4 and q = 8 in f(x),

f(x) = x³ - 2x² - 4x + 8

Since, (x + 2) is a factor of f(x), on dividing f(x) by (x + 2),

$\begin{array}{l} \phantom{x + 2)}{x^2 - 4x + 4} \ x + 2\overline{\smash{\big)}x^3 - 2x^2 - 4x + 8} \ \phantom{x + 2}\underline{\underset{-}{ }x^3 \underset{-}{+} 2x^2} \ \phantom{{x + 2}{x^3+}}-4x^2 - 4x \ \phantom{{x + 2}x^3+}\underline{\underset{+}{-}4x^2 \underset{+}{-} 8x} \ \phantom{{x + 2}{-x^3+2x^21+}}4x + 8 \ \phantom{{x + 2}{-x^3+2x^2+}}\underline{\underset{-}{ }4x \underset{-}{+} 8} \ \phantom{{x + 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

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

$\therefore x^3 - 2x^2 - 4x + 8 = (x + 2)(x^2 - 4x + 4) \\[0.5em] =(x + 2)(x^2 - 2 \times 2 \times x + 2^2) \\[0.5em] = (x + 2)(x - 2)^2$

Hence, value of p = -4 and q = 8; x³ - 2x² - 4x + 8 = (x + 2) (x - 2)².