Question

Resolve the fraction $\frac{2x^2+3x+4}{(x-1)(x^2+2)}$ into partial fractions.

Answer 1

Since $x^2+2$ can not be factored

So, $\frac{2x^2+3x+4}{(x-1)(x^2+2)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+2}$

Now cross multiply, $2x^2+3x+4=A(x^2+2)+(Bx+C)(x-1)$

$\Rightarrow 2x^2+3x+4=A(x^2+2)+B{x}^2-Bx+Cx-C$

$\Rightarrow 2x^2+3x+4=(A+B)x^2+(C-B)x+(2A+C)$

Comparing the coefficients, $A+B=\mathrm{2.....}\left(1\right),C-B=\mathrm{3......}\left(2\right)$ and $2A-C=\mathrm{4.......}\left(3\right)$

Adding the three equations, we get

$3A=9\Rightarrow A=3$

From $\left(1\right)$, $A+B=2\Rightarrow 3+B=2\Rightarrow B=-1$

From $\left(3\right)$, $2A-C=4\Rightarrow 6-C=4\Rightarrow C=2$

$A=3,B=-1,C=2$

Hence,

$\frac{2x^2+3x+4}{(x-1)(x^2+2)}=\frac{3}{x-1}+\frac{-x+2}{x^2+2}$