Answer
$\dfrac{ax+b}{(x-c)^2}=\dfrac{a}{x-c}+\dfrac{ac+b}{(x-c)^2}$
Work Step by Step
We are given the fraction:
$\dfrac{ax+b}{(x-c)^2}$
As the denominator is already factored we can write the partial fraction decomposition:
$\dfrac{ax+b}{(x-c)^2}=\dfrac{A}{x-c}+\dfrac{B}{(x-c)^2}$
Multiply all terms by the least common denominator $(x-c)^2$:
$(x-c)^2\cdot\dfrac{ax+b}{(x-c)^2}=(x-c)^2\cdot\dfrac{A}{x-c}+(x-c)^2\cdot\dfrac{B}{(x-c)^2}$
$ax+b=A(x-c)+B$
$ax+b=Ax-Ac+B$
$ax+b=Ax+(-Ac+B)$
Identify the coefficients:
$\begin{cases}
A=a\\
-Ac+B=b
\end{cases}$
$A=a$
$-ac+B=b$
$B=ac+b$
The partial fraction decomposition is:
$\dfrac{ax+b}{(x-c)^2}=\dfrac{a}{x-c}+\dfrac{ac+b}{(x-c)^2}$