Using the Remainder Theorem, factorise the expression 3x³ + 10x² + x - 6. Hence, solve the equation 3x³ + 10x² + x - 6 = 0.

For x = -1, the value of 3x³ + 10x² + x - 6,

= 3(-1)³ + 10(-1)² + (-1) - 6

= 3(-1) + 10(1) - 7

= 10 - 10

= 0.

Hence, (x + 1) is the factor of 3x³ + 10x² + x - 6.

On dividing 3x³ + 10x² + x - 6 by (x + 1),

$\begin{array}{l} \phantom{x + 1)}{3x^2 + 7x - 6} \ x + 1\overline{\smash{\big)}3x^3 + 10x^2 + x - 6} \ \phantom{x + 1}\underline{\underset{-}{}3x^3 \underset{-}{+} 3x^2} \ \phantom{{x + 1}3x^3+10}7x^2 + x \ \phantom{{x + 1}3x^3+1}\underline{\underset{-}{}7x^2 \underset{-}{+} 7x} \ \phantom{{x + 1}{3x^3+1}{+2x^2}}-6x - 6 \ \phantom{{x + 1}{2x^3+}{+2x^2}{41\space}}\underline{\underset{+}{-}6x \underset{+}{-} 6} \ \phantom{{x + 1}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, quotient = 3x² + 7x - 6

Factorising 3x² + 7x - 6,

= 3x² + 9x - 2x - 6

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

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

∴ 3x² + 7x - 6 = (3x - 2)(x + 3).

Hence, 3x³ + 10x² + x - 6 = (x + 1)(3x - 2)(x + 3).