Chapter 7, Problem 12CT

To determine

To write: the partial decomposition of the given expression and check the result algebraically.

Answer to Problem 12CT

The partial fraction is given by $\frac{3x^2-2x+4}{x^2\left(2-x\right)}=\frac{2}{x^2}+\frac{3}{2-x}$ .

Explanation of Solution

Given:

The given rational expression is

  $\frac{3x^2-2x+4}{x^2\left(2-x\right)}$

Calculation:

Consider the rational expression

  $\frac{3x^2-2x+4}{x^2\left(2-x\right)}$

Since the degree of denominator is equal to numerator, the expression is in proper form.

Take one partial fraction with constant numerator for linear factor and for each power of $x$ in denominator.

Therefore

  $\frac{3x^2-2x+4}{x^2\left(2-x\right)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{2-x}$

Simplify as follows

  $\frac{3x^2-2x+4}{x^2\left(2-x\right)}=\frac{Ax\left(2-x\right)+B\left(2-x\right)+C{x}^2}{x^2\left(2-x\right)}$

Thus

  $3x^2-2x+4=Ax\left(2-x\right)+B\left(2-x\right)+C{x}^2\mathrm{............}(1)$

Substitute $x=2$ in equation (1) to get

  $\begin{array}{c}3{\left(2\right)}^2-2\left(2\right)+4=A\left(2\right)\left(2-2\right)+B\left(2-2\right)+C{\left(2\right)}^2\\ 12=4C\\ 3=C\end{array}$

Substitute $x=0$ in equation (1) to get

  $\begin{array}{c}3{\left(0\right)}^2-2\left(0\right)+4=A\left(0\right)\left(2-0\right)+B\left(2-0\right)+C{\left(0\right)}^2\\ 4=2B\\ 2=B\end{array}$

Substitute $x=1,B=2$ and $C=3$ in equation (1) to get

  $\begin{array}{c}3{\left(1\right)}^2-2\left(1\right)+4=A\left(1\right)\left(2-1\right)+2\left(2-1\right)+3{\left(1\right)}^2\\ 5=A+2+3\\ 5=A+5\\ 0=A\end{array}$

Hence the partial fraction is given by

  $\frac{3x^2-2x+4}{x^2\left(2-x\right)}=\frac{2}{x^2}+\frac{3}{2-x}$ .