Question

Resolve $\frac{x^2+5x+7}{{(x-3)}^3}$ into partial fraction

Answer 1

Given :

$\frac{x^2+5x+7}{{(x-3)}^3}=\frac{A}{x-3}+\frac{B}{{(x-3)}^2}+\frac{C}{{(x-3)}^3}$

$x^2+5x+7=A{(x-3)}^2+B(x-3)+C$

$x^2+5x+7=A(x^2+9-6x)+B(x-3)+C$

$x^2+5x+7=A{x}^2+(B-6A)x+(9A-3B+C)$

Comparing the coefficient of same powers of $x$ on both sides, we have

$A=1$

$B-6A=5$

$\Rightarrow B=6+5=11$

$9A-3B+C=7$

$\Rightarrow 9-33+C=7$

$\Rightarrow C=31$

Substituting the value of $A,B$ and $C$ in equation $\left(1\right)$,

We get

$\frac{x^2+5x+7}{{(x-3)}^3}=\frac{1}{x-3}+\frac{11}{{(x-3)}^2}+\frac{31}{{(x-3)}^3}$