Chapter 7.4, Problem 56E

To determine

To find:write the partial fraction of the given ration expression $\frac{x^3}{{\left(x+2\right)}^2{\left(x-2\right)}^2}$ .

Answer to Problem 56E

The final answer is $\frac{x^3}{{\left(x+2\right)}^2{\left(x-2\right)}^2}=\frac{1}{2\left(x+2\right)}-\frac{1}{2{\left(x+2\right)}^2}+\frac{1}{2\left(x-2\right)}+\frac{1}{2{\left(x-2\right)}^2}$ .

Explanation of Solution

Given information: we have the given rational fraction is $\frac{x^3}{{\left(x+2\right)}^2{\left(x-2\right)}^2}$

Let the partial fraction form of the rational expression is

  $\frac{x^3}{{\left(x+2\right)}^2{\left(x-2\right)}^2}=\frac{A}{\left(x+2\right)}+\frac{B}{{\left(x+2\right)}^2}+\frac{C}{\left(x-2\right)}+\frac{D}{{\left(x-2\right)}^2}$

Now we solve the equation for the constants A,B,C and D

Now, the above equation is equivalent to

  $A\left(x+2\right){\left(x-2\right)}^2+B{\left(x-2\right)}^2+C{\left(x+2\right)}^2\left(x-2\right)+D{\left(x+2\right)}^2=x^3$

We put $x=2$ we get

  $\begin{array}{c}D{\left(2+2\right)}^2={\left(2\right)}^3\\ 16D=8\\ D=\frac{1}{2}\end{array}$

We put $x=-2$ we get

  $\begin{array}{l}B{\left(-2-2\right)}^2={\left(-2\right)}^3\\ 16B=-8\\ B=\frac{-1}{2}\end{array}$

Equating the coefficient of $x^3$ on both sides

  $A+C=1$

Equating constant term on both sides we get

  $\begin{array}{c}8A+4B-8C+4D=0\\ 8A+4\left(\frac{-1}{2}\right)-8C+4\left(\frac{1}{2}\right)=0\\ 8A-8C=0\\ A-C=0\end{array}$

Solving for A and C we get

  $A=C=\frac{1}{2}$

Therefore, the required partial fraction is

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