Express $\frac{14x^2 - 15x - 29}{(x-1)^2(2x+3)}$ in partial fractions.

1. State the correct form for the partial fractions

Let $\frac{14x^2 - 15x - 29}{(x-1)^2(2x+3)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{2x+3}$

2. Form the basic algebraic identity

$14x^2 - 15x - 29 = A(x-1)(2x+3) + B(2x+3) + C(x-1)^2$

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

Let $x = 1$:
$14(1)^2 - 15(1) - 29 = B(2(1)+3)$
$-30 = 5B$
$B = -6$

4. Substitute $x = -\frac{3}{2}$ to find $C$

Let $x = -\frac{3}{2}$:
$14\left(-\frac{3}{2}\right)^2 - 15\left(-\frac{3}{2}\right) - 29 = C\left(-\frac{3}{2}-1\right)^2$
$14\left(\frac{9}{4}\right) + \frac{45}{2} - 29 = C\left(-\frac{5}{2}\right)^2$
$\frac{63}{2} + \frac{45}{2} - 29 = \frac{25}{4}C$
$54 - 29 = \frac{25}{4}C$
$25 = \frac{25}{4}C$
$C = 4$

5. Compare coefficients of $x^2$ to find $A$

Comparing coefficients of $x^2$:
$14 = 2A + C$
$14 = 2A + 4$
$2A = 10$
$A = 5$

6. Write the final partial fraction expression

$\frac{14x^2 - 15x - 29}{(x-1)^2(2x+3)} = \frac{5}{x-1} - \frac{6}{(x-1)^2} + \frac{4}{2x+3}$