Question

Resolve $\frac{5x^2-4x^2+3}{(x-2{)}^2(x+1)}$ into partial fractions.

Answer 1

We have

$\begin{array}{c}\frac{5x^2-4x^2+3}{{\left(x-2\right)}^2\left(x+1\right)}=\frac{x^2+3}{{\left(x-2\right)}^2\left(x+1\right)}\\ \frac{x^2+3}{{\left(x-2\right)}^2\left(x+1\right)}=\frac{A}{\left(x+1\right)}+\frac{B}{\left(x-2\right)}+\frac{C}{{\left(x-2\right)}^2}\\ \frac{x^2+3}{{\left(x-2\right)}^2\left(x+1\right)}=\frac{A{\left(x-2\right)}^2+B\left(x+1\right)\left(x-2\right)+C\left(x+1\right)}{\left(x+1\right){\left(x-2\right)}^2}\\ x^2+3=A{\left(x-2\right)}^2+B\left(x+1\right)\left(x-2\right)+C\left(x+1\right)\end{array}$

Equating the coefficients

$A+B=1$ $4A-2B+C=3$

$-4A-B+C=0$

Solving we get

$\begin{array}{c}A=\frac{4}{9},B=\frac{5}{9},C=\frac{7}{3}\\ Now,\\ \frac{5x^2-4x^2+3}{{\left(x-2\right)}^2\left(x+1\right)}=\frac{4}{9\left(x+1\right)}+\frac{5}{9\left(x-2\right)}+\frac{7}{3{\left(x-2\right)}^2}\end{array}$

Hence, this is the answer.