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

3x³ + 2x² - 19x + 6

For x = 2, the value of 3x³ + 2x² - 19x + 6,

= 3(2)³ + 2(2)² - 19(2) + 6

= 3(8) + 2(4) - 38 + 6

= 38 - 38

= 0.

Hence, (x - 2) is the factor of 3x³ + 2x² - 19x + 6.

On dividing 3x³ + 2x² - 19x + 6 by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{3x^2 + 8x - 3} \ x - 2\overline{\smash{\big)}3x^3 + 2x^2 - 19x + 6} \ \phantom{x - 2}\underline{\underset{-}{}3x^3 \underset{+}{-} 6x^2} \ \phantom{{x - 2}2x^3+4}8x^2 - 19x \ \phantom{{x - 2}2x^3+}\underline{\underset{-}{}8x^2 \underset{-}{+} 16x} \ \phantom{{x - 2}{2x^3+}{+2x^2}}-3x + 6 \ \phantom{{x - 2}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}3x \underset{-}{+} 6} \ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, quotient = 3x² + 8x - 3

Factorising, 3x² + 8x - 3

= 3x² + 9x - x - 3

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

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

∴ 3x² + 8x - 3 = (3x - 1)(x + 3).

Hence, 3x³ + 2x² - 19x + 6 = (x - 2)(3x - 1)(x + 3).