Using Factor Theorem, show that :

(x - 2) is a factor of x³ - 2x² - 9x + 18. Hence, factorise the expression x³ - 2x² - 9x + 18 completely.

x - 2 = 0 ⇒ x = 2.

Remainder = The value of x³ - 2x² - 9x + 18 at x = 2.

= (2)³ - 2(2)² - 9(2) + 18

= 8 - 8 - 18 + 18

= 0.

Hence, (x - 2) is a factor of x³ - 2x² - 9x + 18.

Now dividing x³ - 2x² - 9x + 18 by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{x^2 - 9} \ x - 2\overline{\smash{\big)}x^3 - 2x^2 - 9x + 18} \ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-}2x^2} \ \phantom{{x - 2}x^3-2x^2}-9x + 18 \ \phantom{{x - 2}x^3-2x^2\space}\underline{\underset{+}{-}9x \underset{-}{+} 18} \ \phantom{{x - 2}{x^3-2x^2\space}{-9x}}\times \end{array}$

we get quotient = x² - 9

∴ x³ - 2x² - 9x + 18 = (x - 2)(x² - 9) = (x - 2)(x - 3)(x + 3).

Hence, x³ - 2x² - 9x + 18 = (x - 2)(x - 3)(x + 3).