Answer
$\dfrac{3x^2-2x+8}{(x+2)(x^2+4)}=\dfrac{3}{x+2}-\dfrac{2}{x^2+4}$
Work Step by Step
We are given the fraction:
$\dfrac{3x^2-2x+8}{x^3+2x^2+4x+4}$
Factor the denominator:
$\dfrac{3x^2-2x+8}{x^3+2x^2+4x+8}=\dfrac{3x^2-2x+8}{x^2(x+2)+4(x+2)}=\dfrac{3x^2-2x+8}{(x+2)(x^2+4)}$
We can write the partial fraction decomposition:
$\dfrac{3x^2-2x+8}{(x+2)(x^2+4)}=\dfrac{A}{x+2}+\dfrac{Bx+C}{x^2+4}$
Multiply all terms by the least common denominator $(x+2)(x^2+4)$:
$(x+2)(x^2+4)\cdot\dfrac{3x^2-2x+8}{(x+2)(x^2+4)}=(x+2)(x^2+4)\cdot\dfrac{A}{x+2}+(x+2)(x^2+4)\cdot\dfrac{Bx+C}{x^2+4}$
$3x^2-2x+8=A(x^2+4)+(Bx+C)(x+2)$
$3x^2-2x+8=Ax^2+4A+Bx^2+2Bx+Cx+2C$
$3x^2-2x+8=(A+B)x^2+(2B+C)x+(4A+2C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=3\\
2B+C=-2\\
4A+2C=8
\end{cases}$
Solve the system: multiply Equation 2 by -2 and add it to Equation 3:
$\begin{cases}
A+B=3\\
-2(2B+C)=-2(-2)\\
4A+2C=8
\end{cases}$
$\begin{cases}
A+B=3\\
-4B-2C=4\\
4A+2C=8
\end{cases}$
$\begin{cases}
A+B=3\\
4A+2C-4B-2C=8+4
\end{cases}$
$\begin{cases}
A+B=3\\
4A-4B=12
\end{cases}$
$\begin{cases}
A+B=3\\
A-B=3
\end{cases}$
$A+B+A-B=3+3$
$2A=6$
$A=3$
$A+B=3$
$3+B=3$
$B=0$
$4A+2C=8$
$4(3)+2C=8$
$12+2C=8$
$2C=-4$
$C=-2$
The partial fraction decomposition is:
$\dfrac{3x^2-2x+8}{(x+2)(x^2+4)}=\dfrac{3}{x+2}-\dfrac{2}{x^2+4}$
