Chapter 16, Problem 16.7EAA

To determine

Partial fraction decomposition of the given rational expression.

Answer to Problem 16.7EAA

The required partial fraction decomposition of the given rational expression is $\frac{4}{x^2{\left(x-2\right)}^2}=\frac{1}{x}+\frac{1}{x^2}-\frac{1}{\left(x-2\right)}+\frac{1}{{\left(x-2\right)}^2}$.

Therefore, option (b) is the correct answer.

Explanation of Solution

Given information:

A rational expression: $\frac{4}{x^2\left(x^2-4x+4\right)}$.

Calculation:

First of all, we factor the quadratic in the denominator:

$\begin{array}{l}{x}^2-4x+4=x^2-2x-2x+4\left(\text{Splitting middle term}\right)\\ x^2-4x+4=x\left(x-2\right)-2\left(x-2\right)\left(\text{Pulling GCF from each pair}\right)\\ x^2-4x+4=\left(x-2\right)\left(x-2\right)\\ x^2-4x+4={\left(x-2\right)}^2\end{array}$.

The given rational expression becomes: $\frac{4}{x^2{\left(x-2\right)}^2}$.

Now we set up the partial fraction composition as shown below:

$\frac{4}{x^2{\left(x-2\right)}^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{\left(x-2\right)}+\frac{D}{{\left(x-2\right)}^2}$.

Upon multiplying by $x^2{\left(x-2\right)}^2$ throughout this equation, we get:

$4=Ax{\left(x-2\right)}^2+B{\left(x-2\right)}^2+C{x}^2\left(x-2\right)+D{x}^2$.

Now we need to find the values of A, B, C and D by substituting some values of x.

$\begin{array}{l}\text{At }x=0\\ \Rightarrow 4=A\left(0\right){\left(0-2\right)}^2+B{\left(0-2\right)}^2+C{\left(0\right)}^2\left(0-2\right)+D{\left(0\right)}^2\\ \Rightarrow 4B=4\\ \Rightarrow B=1\\ \text{At }x=2\\ \Rightarrow 4=A\left(2\right){\left(2-2\right)}^2+B{\left(2-2\right)}^2+C{\left(2\right)}^2\left(2-2\right)+D{\left(2\right)}^2\\ \Rightarrow 4D=4\\ \Rightarrow D=1\\ \text{At }x=1\\ \Rightarrow 4=A\left(1\right){\left(1-2\right)}^2+B{\left(1-2\right)}^2+C{\left(1\right)}^2\left(1-2\right)+D{\left(1\right)}^2\\ \Rightarrow 4=A+B-C+D\\ \Rightarrow 4=A+1-C+1\\ \Rightarrow A-C=2\\ \text{At }x=3\\ \Rightarrow 4=A\left(3\right){\left(3-2\right)}^2+B{\left(3-2\right)}^2+C{\left(3\right)}^2\left(3-2\right)+D{\left(3\right)}^2\\ \Rightarrow 4=3A+B+9C+9D\\ \Rightarrow 3A+1+9C+9\left(1\right)=4\\ \Rightarrow 3A+9C=-6\\ \Rightarrow A+3C=-2\end{array}$.

We have the values of B and D. In order to get the values of A and C, we solve the system of linear equations: $A-C=2,A+3C=-2$ . Upon solving this system, we get: $A=1,C=-1$.

Upon substituting $A=1,B=1,C=-1,D=1$ in our set up for partial fractions, we get:

$\begin{array}{l}\frac{4}{x^2{\left(x-2\right)}^2}=\frac{1}{x}+\frac{1}{x^2}+\frac{-1}{\left(x-2\right)}+\frac{1}{{\left(x-2\right)}^2}\\ \frac{4}{x^2{\left(x-2\right)}^2}=\frac{1}{x}+\frac{1}{x^2}-\frac{1}{\left(x-2\right)}+\frac{1}{{\left(x-2\right)}^2}\end{array}$.

Conclusion:

The required partial fraction decomposition of the given rational expression is $\frac{4}{x^2{\left(x-2\right)}^2}=\frac{1}{x}+\frac{1}{x^2}-\frac{1}{\left(x-2\right)}+\frac{1}{{\left(x-2\right)}^2}$.