Express $\frac{7x + 8}{(2x + 1)(x - 1)^2}$ in partial fractions.

1. State the correct form for the partial fractions

Let $\frac{7x + 8}{(2x + 1)(x - 1)^2} = \frac{A}{2x + 1} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}$

2. Form the basic algebraic identity

$7x + 8 = A(x - 1)^2 + B(2x + 1)(x - 1) + C(2x + 1)$

3. Find the value of $C$ by substituting $x = 1$

Substitute $x = 1$:
$7(1) + 8 = C(2(1) + 1)$
$15 = 3C$
$C = 5$

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

Substitute $x = -\frac{1}{2}$:
$7\left(-\frac{1}{2}\right) + 8 = A\left(-\frac{1}{2} - 1\right)^2$
$\frac{9}{2} = A\left(-\frac{3}{2}\right)^2$
$\frac{9}{2} = \frac{9}{4}A$
$A = 2$

5. Find the value of $B$ by substituting $x = 0$

Substitute $x = 0$, $A = 2$, and $C = 5$:
$8 = A(-1)^2 + B(1)(-1) + C(1)$
$8 = 2 - B + 5$
$8 = 7 - B$
$B = -1$