Chapter 16, Problem 16.6EAA

To determine

Partial fraction decomposition of the given rational expression.

Answer to Problem 16.6EAA

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

Therefore, option (a) is the correct answer.

Explanation of Solution

Given information:

A rational expression: $\frac{2}{x\left(x^2-3x+2\right)}$.

Calculation:

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

$\begin{array}{l}{x}^2-3x+2=x^2-2x-x+2\left(\text{Splitting middle term}\right)\\ x^2-3x+2=x\left(x-2\right)-1\left(x-2\right)\left(\text{Pulling GCF from each pair}\right)\\ x^2-3x+2=\left(x-2\right)\left(x-1\right)\end{array}$.

The given rational expression becomes: $\frac{2}{x\left(x-1\right)\left(x-2\right)}$.

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

$\frac{2}{x\left(x-1\right)\left(x-2\right)}=\frac{A}{x}+\frac{B}{\left(x-1\right)}+\frac{C}{\left(x-2\right)}$.

Upon multiplying by $x\left(x-1\right)\left(x-2\right)$ throughout this equation, we get:

$2=\left(x-1\right)\left(x-2\right)A+x\left(x-2\right)B+x\left(x-1\right)C$.

Now we need to find the values of A, B and C by substituting some values of x. The easiest way to find the values of A, B and C would be to substitute 0, 1 and 2 for x.

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

Upon substituting $A=1,B=-2,C=1$ in our set up for partial fractions, we get:

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

Conclusion:

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