Answer
$\dfrac{2x^2+8x+3}{(x+1)^3}=\dfrac{2}{x+1}+\dfrac{4}{(x+1)^2}-\dfrac{3}{(x+1)^3}$
Work Step by Step
We are given the fraction:
$\dfrac{2x^2+8x+3}{(x+1)^3}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{2x^2+8x+3}{(x+1)^3}=\dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}+\dfrac{C}{(x+1)^3}$
Multiply all terms by the least common denominator $(x+1)^3$:
$(x+1)^3\cdot\dfrac{2x^2+8x+3}{(x+1)^3}=(x+1)^3\cdot\dfrac{A}{x+1}+(x+1)^3\cdot\dfrac{B}{(x+1)^2}+(x+1)^3\cdot\dfrac{C}{(x+1)^3}$
$2x^2+8x+3=A(x+1)^2+B(x+1)+C$
$2x^2+8x+3=Ax^2+2Ax+A+Bx+B+C$
$2x^2+8x+3=Ax^2+(2A+B)x+(A+B+C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A=2\\
2A+B=8\\
A+B+C=3
\end{cases}$
Solve the system:
$A=2$
$2(2)+B=8$
$B=4$
$2+4+C=3$
$6+C=3$
$C=-3$
The partial fraction decomposition is:
$\dfrac{2x^2+8x+3}{(x+1)^3}=\dfrac{2}{x+1}+\dfrac{4}{(x+1)^2}-\dfrac{3}{(x+1)^3}$
