Using the remainder theorem, factorise each of the following completely :

2x³ + x² - 13x + 6

For x = 2, the value of 2x³ + x² - 13x + 6

= 2(2)³ + (2)² - 13(2) + 6

= 2(8) + 4 - 26 + 6

= 26 - 26

= 0.

On dividing 2x³ + x² - 13x + 6 by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{2x^2 + 5x - 3} \ x - 2\overline{\smash{\big)}2x^3 + x^2 - 13x + 6} \ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-} 4x^2} \ \phantom{{x - 2}2x^3+4}5x^2 - 13x \ \phantom{{x - 2}2x^3+}\underline{\underset{-}{}5x^2 \underset{+}{-} 10x} \ \phantom{{x - 2}{2x^3+}{+2x^2}}-3x + 6 \ \phantom{{x - 2}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}3x \underset{-}{+} 6} \ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, quotient = 2x² + 5x - 3

Factorising, 2x² + 5x - 3

= 2x² + 6x - x - 3

= 2x(x + 3) - 1(x + 3)

= (2x - 1)(x + 3)

∴ 2x² + 5x - 3 = (2x - 1)(x + 3)

Hence, 2x³ + x² - 13x + 6 = (x - 2)(2x - 1)(x + 3).