If 2x³ + ax² + bx - 6 has a factor (2x + 1) and leaves a remainder 12 when divided by (x + 2). Calculate the values of a and b hence factorise the given expression completely.

By factor theorem,

A polynomial f(x) has a factor (x - a), if and only if, f(a) = 0.

⇒ 2x + 1 = 0

⇒ 2x = -1

⇒ x = -$\dfrac{1}{2}$.

Since, (2x + 1) is a factor of 2x³ + ax² + bx - 6.

$\therefore 2\Big(-\dfrac{1}{2}\Big)^3 + a\Big(-\dfrac{1}{2}\Big)^2 + b\Big(-\dfrac{1}{2}\Big) - 6 = 0 \\[1em] \Rightarrow 2 \times -\dfrac{1}{8} + a \times \dfrac{1}{4} - \dfrac{b}{2} - 6 = 0 \\[1em] \Rightarrow -\dfrac{1}{4} + \dfrac{a}{4} - \dfrac{b}{2} - 6 = 0 \\[1em] \Rightarrow \dfrac{-1 + a - 2b - 24}{4} = 0 \\[1em] \Rightarrow a - 2b - 25 = 0 \\[1em] \Rightarrow a = 2b + 25 \text{ …………..(Eq. 1)}.$

By remainder theorem,

If f(x), a polynomial in x, is divided by (x - a), the remainder = f(a).

⇒ x + 2 = 0

⇒ x = -2.

Given,

2x³ + ax² + bx - 6 leaves a remainder 12 when divided by (x + 2).

∴ 2(-2)³ + a(-2)² + b(-2) - 6 = 12

⇒ -2(8) + 4a - 2b - 6 = 12

⇒ -16 + 4a - 2b - 6 = 12

⇒ 4a - 2b = 12 + 16 + 6

⇒ 4a - 2b = 34

⇒ 2(2a - b) = 34

⇒ 2a - b = 17

Substituting value of a in above equation from Eq. 1 :

⇒ 2(2b + 25) - b = 17

⇒ 4b + 50 - b = 17

⇒ 3b = 17 - 50

⇒ 3b = -33

⇒ b = -11.

Substituting value of b in equation (1), we get :

⇒ a = 2b + 25

⇒ a = 2(-11) + 25 = -22 + 25 = 3.

Substituting value of a and b in 2x³ + ax² + bx - 6, we get :

⇒ 2x³ + (3)x² + (-11)x - 6

⇒ 2x³ + 3x² - 11x - 6

Since, (2x + 1) is the factor. On dividing 2x³ + 3x² - 11x - 6 by (2x + 1), we get :

$\begin{array}{l} \phantom{2x + 12}{x^2 + x - 6} \ 2x + 1 \overline{\smash{\big)}2x^3 + 3x^2 - 11x - 6} \ \phantom{2x + 1}\underline{\underset{-}{}2x^3 \underset{-}{+}x^2} \ \phantom{{2x + 1}2x^3+}2x^2 - 11x \ \phantom{{2x + 1}x^3+}\underline{\underset{-}{}2x^2 \underset{-}{+}x} \ \phantom{{2x + 1}x^3+2-5}-12x - 6 \ \phantom{{2x + 1}x^3+2-x1}\underline{\underset{+}{-}12x\underset{+}{-}6} \ \phantom{{2x + 1}x^3+2-5x^233} \times \end{array}$

⇒ 2x³ + 3x² - 11x - 6 = (2x + 1)(x² + x - 6)

⇒ 2x³ + 3x² - 11x - 6 = (2x + 1)(x² + 3x - 2x - 6)

⇒ 2x³ + 3x² - 11x - 6 = (2x + 1)[x(x + 3) - 2(x + 3)]

⇒ 2x³ + 3x² - 11x - 6 = (2x + 1)(x - 2)(x + 3).

Hence, a = 3, b = -11 and 2x³ + 3x² - 11x - 6 = (2x + 1)(x - 2)(x + 3).