If ax³ + 3x² + bx - 3 has a factor (2x + 3) and leaves remainder -3 when divided by (x + 2), find the values of a and b. With these values of a and b, factorise the given expression.

Let f(x) = ax³ + 3x² + bx - 3

Given, (2x + 3) or $2(x - \big(-\dfrac{3}{2}\big))$ is factor of f(x), hence, f$\big(-\dfrac{3}{2}\big)$ = 0 by factor's theorem

$\therefore a\big(-\dfrac{3}{2}\big)^3 + 3\big(-\dfrac{3}{2}\big)^2 + b\big(-\dfrac{3}{2}\big) - 3 = 0 \\[1em] \Rightarrow -\dfrac{27a}{8} + \dfrac{27}{4} - \dfrac{3b}{2} - 3 = 0 \\[1em] \Rightarrow \big(\dfrac{-27a + 54 - 12b - 24}{8}\big) = 0$

On cross multiplication,

$\Rightarrow -27a - 12b + 30 = 0$

On dividing the equation by 3,

$\Rightarrow -9a - 4b + 10 = 0 \\[1em] \Rightarrow 4b + 9a = 10 \text{( Equation 1)}$

Given, on dividing f(x) by (x + 2) remainder left is -3

By remainder theorem, remainder = f(-2)

$\therefore a(-2)^3 + 3(-2)^2 + b(-2) - 3 = -3 \\[0.5em] \Rightarrow -8a + 12 - 2b - 3 = -3 \\[0.5em] \Rightarrow -8a - 2b + 9 = -3 \\[0.5em] \Rightarrow -8a - 2b = -12$

On dividing equation by -2,

$\Rightarrow -4a - b = -6 \\[0.5em] \Rightarrow 4a + b = 6$

Multiplying equation by 4,

$\Rightarrow 16a + 4b = 24$

Subtracting above equation from equation 1,

$\Rightarrow 4b + 9a - 16a - 4b = 10 - 24 \\[0.5em] \Rightarrow -7a = -14 \\[0.5em] \Rightarrow a = 2 \\[0.5em] \text{and } b = 6 - 4a = 6 - 8 = -2.$

Putting value of a and b in f(x) we get,

$f(x) = 2x^3 + 3x^2 - 2x - 3 \\[0.5em] = x^2(2x + 3) -1(2x + 3) \\[0.5em] = (x^2 - 1)(2x + 3) \\[0.5em] = ((x)^2 - (1)^2)(2x + 3) \\[0.5em] = (x - 1)(x + 1)(2x + 3)$

Hence, the value of a = 2 and b = -2 ; 2x³ + 3x² - 2x - 3 = (x - 1)(x + 1)(2x + 3).