Answer
$\dfrac{4x^2-3x-4}{x(x+2)(x-1)}=\dfrac{2}{x}+\dfrac{3}{x+2}-\dfrac{1}{x-1}$
Work Step by Step
We are given the fraction:
$\dfrac{4x^2-3x-4}{x(x+2)(x-1)}$
As the denominator is factored, we can write the partial fraction decomposition:
$\dfrac{4x^2-3x-4}{x(x+2)(x-1)}=\dfrac{A}{x}+\dfrac{B}{x+2}+\dfrac{C}{x-1}$
Multiply each term by the least common denominator $x(x+2)(x-1)$:
$x(x+2)(x-1)\cdot\dfrac{4x^2-3x-4}{x(x+2)(x-1)}=x(x+2)(x-1)\cdot\dfrac{A}{x}+x(x+2)(x-1)\cdot\dfrac{B}{x+2}+x(x+2)(x-1)\cdot\dfrac{C}{x-1}$
$4x^2-3x-4=A(x+2)(x-1)+Bx(x-1)+Cx(x+2)$
$4x^2-3x-4=A(x^2+x-2)+Bx(x-1)+Cx(x+2)$
$4x^2-3x-4=Ax^2+Ax-2A+Bx^2-Bx+Cx^2+2Cx$
$4x^2-3x-4=(A+B+C)x^2+(A-B+2C)x+(-2A)$
Identify the coefficients:
$\begin{cases}
A+B+C=4\\
A-B+2C=-3\\
-2A=-4
\end{cases}$
Solve the system:
$A=2$
$\begin{cases}
2+B+C=4\\
2-B+2C=-3
\end{cases}$
$\begin{cases}
B+C=2\\
-B+2C=-5
\end{cases}$
$B+C-B+2C=2-5$
$3C=-3$
$C=-1$
$A+B+C=4$
$2+B-1=4$
$1+B=4$
$B=3$
The partial fraction decomposition is:
$\dfrac{4x^2-3x-4}{x(x+2)(x-1)}=\dfrac{2}{x}+\dfrac{3}{x+2}-\dfrac{1}{x-1}$
