Question

Resolve into partial fractions:
$\frac{2x^2-11x+5}{(x-3)(x^2+2x-5)}$

Answer 1

Consider $\frac{2x^2-11x+5}{(x-3)(x^2+2x-5)}=\frac{Ax+B}{x^2+2x-5}+\frac{C}{x-3}$

$\Rightarrow 2x^2-11x+5=(Ax+B)(x-3)+C(x^2+2x-5)$

Comparing the coefficients of $x^2$, $x$ and constant on both sides, we get

$2=A+C$ ....... $\left(1\right)$

$-11=-3A+B+2C$ ........... $\left(2\right)$

$5=-3B-5C$ .......... $\left(3\right)$

Solving $\left(1\right),\left(2\right)$ and $\left(3\right)$ we get $A=3,B=0,C=-1$

Therefore, $\frac{2x^2-11x+5}{(x-3)(x^2+2x-5)}=\frac{3x}{x^2+2x-5}-\frac{1}{x-3}$