Answer
$\dfrac{x^4+2x^3-4x^2+x-3}{x^2-x-2}=x^2+3x+1+\dfrac{3}{x+1}+\dfrac{5}{x-2}$
Work Step by Step
We are given the fraction:
$\dfrac{x^4+2x^3-4x^2+x-3}{x^2-x-2}$
Because the degree of the numerator is greater than the degree of the denominator, we divide the numerator by the denominator:
$\dfrac{x^4+2x^3-4x^2+x-3}{x^2-x-2}=\dfrac{x^2(x^2-x-2)+3x^3-2x^2+x-3}{x^2-x-2}$
$=\dfrac{x^2(x^2-x-2)+3x(x^2-x-2)+x^2+7x-3}{x^2-x-2}$
$=\dfrac{x^2(x^2-x-2)+3x(x^2-x-2)+(x^2-x-2)+8x-1}{x^2-x-2}$
$=x^2+3x+1+\dfrac{8x-1}{x^2-x-2}$
We will write the partial fraction decomposition for the fraction $\dfrac{8x-1}{x^2-x-2}$. Factor the denominator and write the partial fraction decomposition:
$\dfrac{8x-1}{x^2-x-2}=\dfrac{8x-1}{x^2+x-2x-2}=\dfrac{8x-1}{x(x+1)-2(x+1)}=\dfrac{8x-1}{(x+1)(x-2)}$
Write the partial fraction decomposition:
$\dfrac{8x-1}{(x+1)(x-2)}=\dfrac{A}{x+1}+\dfrac{B}{x-2}$
Multiply all terms by the least common denominator $(x+1)(x-2)$:
$(x+1)(x-2)\cdot\dfrac{8x-1}{(x+1)(x-2)}=(x+1)(x-2)\cdot\dfrac{A}{x+1}+(x+1)(x-2)\cdot \dfrac{B}{x-2}$
$8x-1=A(x-2)+B(x+1)$
$8x-1=Ax-2A+Bx+B$
$8x-1=(A+B)x^2+(-2A+B)$
Identify the coefficients:
$\begin{cases}
A+B=8\\
-2A+B=-1\\
-B=2
\end{cases}$
$\begin{cases}
A+B=8\\
2A-B=1\\
-B=2
\end{cases}$
$A+B+2A-B=8+1$
$3A=9$
$A=3$
$A+B=8$
$3+B=8$
$B=5$
The partial fraction decomposition is:
$\dfrac{8x-1}{(x+1)(x-2)}=\dfrac{3}{x+1}+\dfrac{5}{x-2}$
The final result is:
$\dfrac{x^4+2x^3-4x^2+x-3}{x^2-x-2}=x^2+3x+1+\dfrac{3}{x+1}+\dfrac{5}{x-2}$
