Answer
$\dfrac{10x^2+9x-7}{(x+2)(x^2-1)}=\dfrac{5}{x+2}+\dfrac{3}{x-1}+\dfrac{2}{x-1}$
Work Step by Step
We are given the fraction:
$\dfrac{10x^2+9x-7}{(x+2)(x^2-1)}$
We will decompose the fraction: first we factor the denominator:
$(x+2)(x^2-1)=(x+2)(x-1)(x+1)$
Decompose the fraction:
$\dfrac{10x^2+9x-7}{(x+2)(x^2-1)}=\dfrac{A}{x+2}+\dfrac{B}{x-1}+\dfrac{C}{x-1}$
Bring the fractions at the same denominator:
$\dfrac{10x^2+9x-7}{(x+2)(x^2-1)}=\dfrac{A(x+1)(x-1)}{(x+2)(x+1)(x-1)}+\dfrac{B(x+2)(x-1)}{(x+2)(x+1)(x-1)}+\dfrac{C(x+2)(x+1)}{(x+2)(x+1)(x-1)}$
$10x^2+9x-7=A(x+1)(x-1)+B(x+2)(x-1)+C(x+2)(x+1)$
$10x^2+9x-7=A(x^2-1)+B(x^2-x+2x-2)+C(x^2+x+2x+2)$
$10x^2+9x-7=A(x^2-1)+B(x^2+x-2)+C(x^2+3x+2)$
$10x^2+9x-7=Ax^2-A+Bx^2+Bx-2B+Cx^2+3Cx+2C$
$10x^2+9x-7=(A+B+C)x^2+(B+3C)x+(-A-2B+2C)$
Identify the coefficients and build the system:
$\begin{cases}
A+B+C=10\\
B+3C=9\\
-A-2B+2C=-7
\end{cases}$
Add Equation 1 and Equation 3:
$\begin{cases}
A+B+C-A-2B+2C=10-7\\
B+3C=9
\end{cases}$
$\begin{cases}
-B+3C=3\\
B+3C=9
\end{cases}$
Add the two equations:
$-B+3C+B+3C=3+9$
$6C=12$
$C=2$
Determine $B$:
$B+3C=9$
$B+3(2)=9$
$B=9-6$
$B=3$
Determine $A$:
$A+B+C=10$
$A+3+2=10$
$A+5=10$
$A=5$
Write the decomposed fraction:
$\dfrac{10x^2+9x-7}{(x+2)(x^2-1)}=\dfrac{5}{x+2}+\dfrac{3}{x-1}+\dfrac{2}{x-1}$
