Answer
$\dfrac{7x^2-7x+23}{(x-3)(x^2+4)}=\dfrac{5}{x-3}+\dfrac{2x-1}{x^2+4}$
Work Step by Step
We are given the fraction:
$\dfrac{7x^2-7x+23}{(x-3)(x^2+4)}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{7x^2-7x+23}{(x-3)(x^2+4)}=\dfrac{A}{x-3}+\dfrac{Bx+C}{x^2+4}$
Multiply all terms by the least common denominator $(x-3)(x^2+4)$:
$(x-3)(x^2+4)\cdot\dfrac{7x^2-7x+23}{(x-3)(x^2+4)}=(x-3)(x^2+4)\cdot\dfrac{A}{x-3}+(x-3)(x^2+4)\cdot\dfrac{Bx+C}{x^2+4}$
$7x^2-7x+23=A(x^2+4)+(Bx+C)(x-3)$
$7x^2-7x+23=Ax^2+4A+Bx^2-3Bx+Cx-3C$
$7x^2-7x+23=(A+B)x^2+(-3B+C)x+(4A-3C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=7\\
-3B+C=-7\\
4A-3C=23
\end{cases}$
Solve the system:
$\begin{cases}
A+B=7\\
3(-3B+C)=3(-7)\\
4A-3C=23
\end{cases}$
$\begin{cases}
A+B=7\\
-9B+3C=-21\\
4A-3C=23
\end{cases}$
$\begin{cases}
A+B=7\\
-9B+3C+4A-3C=-21+23
\end{cases}$
$\begin{cases}
A+B=7\\
4A-9B=2
\end{cases}$
$\begin{cases}
9A+9B=9(7)\\
4A-9B=2
\end{cases}$
$9A+9B+4A-9B=63+2$
$13A=65$
$A=5$
$A+B=7$
$5+B=7$
$B=2$
$4A-3C=23$
$4(5)-3C=23$
$20-3C=23$
$-3C=3$
$C=-1$
The partial fraction decomposition is:
$\dfrac{7x^2-7x+23}{(x-3)(x^2+4)}=\dfrac{5}{x-3}+\dfrac{2x-1}{x^2+4}$
