Answer
$\dfrac{5x^2-9x+19}{(x-4)(x^2+5)}=\dfrac{3}{x-4}+\dfrac{2x-1}{x^2+5}$
Work Step by Step
We are given the fraction:
$\dfrac{5x^2-9x+19}{(x-4)(x^2+5)}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{5x^2-9x+19}{(x-4)(x^2+5)}=\dfrac{A}{x-4}+\dfrac{Bx+C}{x^2+5}$
Multiply all terms by the least common denominator $(x-4)(x^2+5)$:
$(x-4)(x^2+5)\cdot\dfrac{5x^2-9x+19}{(x-4)(x^2+5)}=(x-4)(x^2+5)\cdot\dfrac{A}{x-4}+(x-4)(x^2+5)\cdot\dfrac{Bx+C}{x^2+5}$
$5x^2-9x+19=A(x^2+5)+(Bx+C)(x-4)$
$5x^2-9x+19=Ax^2+5A+Bx^2-4Bx+Cx-4C$
$5x^2-9x+19=(A+B)x^2+(-4B+C)x+(5A-4C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=5\\
-4B+C=-9\\
5A-4C=19
\end{cases}$
Solve the system: Multiply Equation 2 by 4 and add it to Equation 3:
$\begin{cases}
A+B=5\\
-16B+4C=-36\\
5A-4C=19
\end{cases}$
$\begin{cases}
A+B=5\\
-16B+4C+5A-4C=-36+19
\end{cases}$
$\begin{cases}
A+B=5\\
5A-16B=-17
\end{cases}$
$\begin{cases}
16A+16B=16(5)\\
5A-16B=-17
\end{cases}$
$16A+16B+5A-16B=80-17$
$21A=63$
$A=3$
$A+B=5$
$3+B=5$
$B=2$
$-4B+C=-9$
$-4(2)+C=-9$
$-8+C=-9$
$C=-1$
The partial fraction decomposition is:
$\dfrac{5x^2-9x+19}{(x-4)(x^2+5)}=\dfrac{3}{x-4}+\dfrac{2x-1}{x^2+5}$