Use the remainder theorem to factorise the expression 2x³ + 9x² + 7x - 6. Hence, solve the equation 2x³ + 9x² + 7x - 6.

For x = -2, the value of the given expression

= 2(-2)³ + 9(-2)² + 7(-2) - 6

= 2 × -8 + 9 × 4 - 14 - 6

= -16 + 36 - 14 - 6

= -36 + 36

= 0.

⇒ x + 2 is a factor of 2x³ + 9x² + 7x - 6.

Now dividing 2x³ + 9x² + 7x - 6 by (x + 2),

$\begin{array}{l} \phantom{x + 2)}{2x^2 + 5x - 3} \ x + 2\overline{\smash{\big)}2x^3 + 9x^2 + 7x - 6} \ \phantom{x + 2}\underline{\underset{-}{}2x^3 \underset{-}{+}4x^2} \ \phantom{{x + 2}2x^3+2}5x^2 + 7x - 6 \ \phantom{{x + 2}x^3+2}\underline{\underset{-}{}5x^2 \underset{-}{+} 10x} \ \phantom{{x + 2}x^3+2-5x^2}-3x - 6 \ \phantom{{x + 2}x^3+2-5x^23}\underline{\underset{+}{-}3x \underset{+}{-} 6} \ \phantom{{x + 2}x^3+2-5x^23-3}\times \end{array}$

we get quotient = 2x² + 5x - 3

∴ 2x³ + 9x² + 7x - 6 = (x + 2)(2x² + 5x - 3)

= (x + 2)(2x² + 6x - x - 3)

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

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

∴ (x + 2), (2x - 1) and (x + 3) are the factors of 2x³ + 9x² + 7x - 6.

⇒ x + 2 = 0

⇒ x = -2

⇒ 2x - 1 = 0

⇒ x = $\dfrac{1}{2}$

⇒ x + 3 = 0

⇒ x = -3.

Hence, 2x³ + 9x² + 7x - 6 = (x + 2)(2x - 1)(x + 3) and x = -2, -3, $\dfrac{1}{2}$.