Answer
$\dfrac{2x+1}{(x-2)^2}=\dfrac{2}{x-2}+\dfrac{5}{(x-2)^2}$
Work Step by Step
We are given the fraction:
$\dfrac{2x+1}{(x-2)^2}$
As the denominator is factored, we write the partial fraction decomposition:
$\dfrac{2x+1}{(x-2)^2}=\dfrac{A}{x-2}+\dfrac{B}{(x-2)^2}$
Multiply each term by the least common denominator $(x-2)^2$:
$(x-2)^2\cdot\dfrac{2x+1}{(x-2)^2}=(x-2)^2\cdot\dfrac{A}{x-2}+(x-2)^2\cdot\dfrac{B}{(x-2)^2}$
$2x+1=A(x-2)+B$
$2x+1=Ax-2A+B$
$2x+1=Ax+(-2A+B)$
Identify the coefficients:
$\begin{cases}
A=2\\
-2A+B=1
\end{cases}$
Solve the system:
$A=2$
$-2A+B=1$
$-2(2)+B=1$
$-4+B=1$
$B=5$
The partial fraction decomposition is:
$\dfrac{2x+1}{(x-2)^2}=\dfrac{2}{x-2}+\dfrac{5}{(x-2)^2}$
