Factorise the expression

f(x) = 2x³ - 7x² - 3x + 18.

Hence, find all possible values of x for which f(x) = 0.

For x = 2, the value of 2x³ - 7x² - 3x + 18,

= 2(2)³ - 7(2)² - 3(2) + 18

= 16 - 28 - 6 + 18

= 34 - 34

= 0.

Hence, (x - 2) is the factor of 2x³ - 7x² - 3x + 18.

On dividing, 2x³ - 7x² - 3x + 18 by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{2x^2 - 3x - 9} \ x - 2\overline{\smash{\big)}2x^3 - 7x^2 - 3x + 18} \ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-} 4x^2} \ \phantom{{x - 2}3x^3+}-3x^2 - 3x \ \phantom{{x - 2}3x^3+}\underline{\underset{+}{-}3x^2 \underset{-}{+} 6x} \ \phantom{{x - 2}{3x^3+1}{+2x^2}}-9x + 18 \ \phantom{{x - 2}{2x^3+}{+2x^2}{41\space}}\underline{\underset{+}{-}9x \underset{-}{+} 18} \ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get quotient = 2x² - 3x - 9.

Factorising 2x² - 3x - 9,

= 2x² - 6x + 3x - 9

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

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

∴ 2x² - 3x - 9 = (2x + 3)(x - 3).

∴ f(x) = 2x³ - 7x² - 3x + 18 = (x - 2)(2x + 3)(x - 3).

f(x) = 0, if (x - 2) = 0, (2x + 3) = 0 or x - 3 = 0.

x - 2 = 0 ⇒ x =2,

2x + 3 = 0 ⇒ x = -$\dfrac{3}{2}$,

x - 3 = 0 ⇒ x = 3.

Hence, 2x³ - 7x² - 3x + 18 = (x - 2)(2x + 3)(x - 3), values for which f(x) = 0 are 2, 3, $-\dfrac{3}{2}$.