Express $\frac{-3x^2 + 12x - 1}{(3x + 1)(x - 1)^2}$ in partial fractions.

1. State the appropriate form for the partial fractions

Let $\frac{-3x^2 + 12x - 1}{(3x + 1)(x - 1)^2} = \frac{A}{3x + 1} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}$

2. Form the basic algebraic identity

Multiplying throughout by $(3x + 1)(x - 1)^2$ gives:
$-3x^2 + 12x - 1 = A(x - 1)^2 + B(3x + 1)(x - 1) + C(3x + 1)$

3. Substitute $x = 1$ to find $C$

Let $x = 1$:
$-3(1)^2 + 12(1) - 1 = C(3(1) + 1)$
$8 = 4C \implies C = 2$

4. Substitute $x = -\frac{1}{3}$ to find $A$

Let $x = -\frac{1}{3}$:
$-3\left(-\frac{1}{3}\right)^2 + 12\left(-\frac{1}{3}\right) - 1 = A\left(-\frac{1}{3} - 1\right)^2$
$-3\left(\frac{1}{9}\right) - 4 - 1 = A\left(-\frac{4}{3}\right)^2$
$-\frac{16}{3} = \frac{16}{9}A \implies A = -3$

5. Compare coefficients of $x^2$ to find $B$ and state the final answer

Comparing coefficients of $x^2$:
$-3 = A + 3B$
$-3 = -3 + 3B \implies 3B = 0 \implies B = 0$

Therefore, the partial fractions are:
$\frac{-3}{3x + 1} + \frac{2}{(x - 1)^2}$