Answer
$\dfrac{4x^2+5x-9}{x^3-6x-9}=\dfrac{2}{x-3}+\dfrac{2x+5}{x^2+3x+3}$
Work Step by Step
We are given the fraction:
$\dfrac{4x^2+5x-9}{x^3-6x-9}$
Factor the denominator:
$\dfrac{4x^2+5x-9}{x^3-6x-9}=\dfrac{4x^2+5x-9}{x^3-3x^2+3x^2-9x+3x-9}=\dfrac{4x^2+5x-9}{x^2(x-3)+3x(x-3)+3(x-3)}=\dfrac{4x^2+5x-9}{(x-3)(x^2+3x+3)}$
Write the partial fraction decomposition:
$\dfrac{4x^2+5x-9}{(x-3)(x^2+3x+3)}=\dfrac{A}{x-3}+\dfrac{Bx+C}{x^2+3x+3}$
Multiply all terms by the least common denominator $(x-3)(x^2+3x+3)$:
$(x-3)(x^2+3x+3)\cdot\dfrac{4x^2+5x-9}{(x-3)(x^2+3x+3)}=(x-3)(x^2+3x+3)\cdot\dfrac{A}{x-3}+(x-3)(x^2+3x+3)\cdot\dfrac{Bx+C}{x^2+3x+3}$
$4x^2+5x-9=A(x^2+3x+3)+(Bx+C)(x-3)$
$4x^2+5x-9=Ax^2+3Ax+3A+Bx^2-3Bx+Cx-3C$
$4x^2+5x-9=(A+B)x^2+(3A-3B+C)x+(3A-3C)$
Identify the coefficients:
$\begin{cases}
A+B=4\\
3A-3B+C=5\\
3A-3C=-9
\end{cases}$
$\begin{cases}
A+B=4\\
3A-3B+C=5\\
A-C=-3
\end{cases}$
$\begin{cases}
A+B=4\\
3A-3B+C+A-C=5-3
\end{cases}$
$\begin{cases}
A+B=4\\
4A-3B=2
\end{cases}$
$\begin{cases}
3A+3B=3(4)\\
4A-3B=2
\end{cases}$
$3A+3B+4A-3B=12+2$
$7A=14$
$A=2$
$A+B=4$
$2+B=4$
$B=2$
$A-C=-3$
$2-C=-3$
$C=5$
The partial fraction decomposition is:
$\dfrac{4x^2+5x-9}{x^3-6x-9}=\dfrac{2}{x-3}+\dfrac{2x+5}{x^2+3x+3}$
