Factorise completely $2x^3+7x^2+4x-4$. Express $\dfrac{12x^2+32x+31}{2x^3+7x^2+4x-4}$ in partial fractions.

1. Use the Factor Theorem to find a linear factor of the polynomial.

Let $P(x) = 2x^3+7x^2+4x-4$.
Try $x = -2$:
$P(-2) = 2(-2)^3 + 7(-2)^2 + 4(-2) - 4$
$P(-2) = -16 + 28 - 8 - 4 = 0$
Since $P(-2) = 0$, $(x+2)$ is a factor.

2. Find the quadratic factor by comparing coefficients or long division.

Let $2x^3+7x^2+4x-4 = (x+2)(ax^2+bx+c)$.
By comparing the first and last terms, $a = 2$ and $c = -2$.
So, $2x^3+7x^2+4x-4 = (x+2)(2x^2+bx-2)$.
Comparing the coefficient of $x^2$:
$2b + 4 = 7 \implies b = 3$
Thus, the quadratic factor is $2x^2+3x-2$.

3. Factorise the quadratic factor completely.

$2x^2+3x-2 = (2x-1)(x+2)$
Therefore, $2x^3+7x^2+4x-4 = (x+2)(2x-1)(x+2) = (x+2)^2(2x-1)$.

1. Factorise the denominator using the result from part (i)

From part (i), $2x^3+7x^2+4x-4 = (x+2)^2(2x-1)$.
So, $\dfrac{12x^2+32x+31}{2x^3+7x^2+4x-4} = \dfrac{12x^2+32x+31}{(x+2)^2(2x-1)}$

2. Set up the partial fraction decomposition

$\dfrac{12x^2+32x+31}{(x+2)^2(2x-1)} = \dfrac{A}{x+2} + \dfrac{B}{(x+2)^2} + \dfrac{C}{2x-1}$
$12x^2+32x+31 = A(x+2)(2x-1) + B(2x-1) + C(x+2)^2$

3. Find the value of B by substituting $x = -2$

Let $x = -2$:
$12(-2)^2 + 32(-2) + 31 = B(2(-2)-1)$
$48 - 64 + 31 = -5B$
$15 = -5B \implies B = -3$

4. Find the value of C by substituting $x = \frac{1}{2}$

Let $x = \frac{1}{2}$:
$12\left(\frac{1}{2}\right)^2 + 32\left(\frac{1}{2}\right) + 31 = C\left(\frac{1}{2}+2\right)^2$
$3 + 16 + 31 = C\left(\frac{5}{2}\right)^2$
$50 = \frac{25}{4}C \implies C = 8$

5. Find the value of A by comparing coefficients of $x^2$

Compare coefficients of $x^2$:
$12 = 2A + C$
$12 = 2A + 8 \implies 2A = 4 \implies A = 2$

Therefore, the partial fractions are $\dfrac{2}{x+2} - \dfrac{3}{(x+2)^2} + \dfrac{8}{2x-1}$.