Answer
$\dfrac{x^2+4x-23}{(x+3)(x^2+4)}=-\dfrac{2}{x+3}+\dfrac{3x-5}{x^2+4}$
Work Step by Step
We are given the fraction:
$\dfrac{x^2+4x-23}{(x+3)(x^2+4)}$
Decompose the fraction:
$\dfrac{x^2+4x-23}{(x+3)(x^2+4)}=\dfrac{A}{x+3}+\dfrac{Bx+C}{x^2+4}$
Bring the fractions at the same denominator:
$\dfrac{x^2+4x-23}{(x+3)(x^2+4)}=\dfrac{A(x^2+4)}{(x+3)(x^2+4)}+\dfrac{(Bx+C)(x+3)}{(x+3)(x^2+4)}$
$x^2+4x-23=A(x^2+4)+(Bx+C)(x+3)$
$x^2+4x-23=Ax^2+4A+Bx^2+3Bx+Cx+3C$
$x^2+4x-23=(A+B)x^2+(3B+C)x+(4A+3C)$
Identify the coefficients and build the system:
$\begin{cases}
A+B=1\\
3B+C=4\\
4A+3C=-23
\end{cases}$
Multiply Equation 1 by -4 and add it to Equation 3:
$\begin{cases}
-4A-4B=-4\\
3B+C=4\\
4A+3C=-23
\end{cases}$
$\begin{cases}
-4A-4B+4A+3C=-4-23\\
3B+C=4
\end{cases}$
$\begin{cases}
-4B+3C=-27\\
3B+C=4
\end{cases}$
Multiply Equation 2 by -3 and add it to Equation 1:
$\begin{cases}
-4B+3C=-27\\
-9B-3C=-12
\end{cases}$
Add the two equations:
$-4B+3C-9B-3C=-27-12$
$-13B=-39$
$B=3$
Determine $C$:
$3B+C=4$
$3(3)+C=4$
$C=4-9$
$C=-5$
Determine $A$:
$A+B=1$
$A+3=1$
$A=1-3$
$A=-2$
Write the decomposed fraction:
$\dfrac{x^2+4x-23}{(x+3)(x^2+4)}=-\dfrac{2}{x+3}+\dfrac{3x-5}{x^2+4}$
