Chapter 7, Problem 14CT

To determine

To find: Partial expression of the given rational expression.

Answer to Problem 14CT

The partial fraction is given by

  $\frac{x^2-4}{x^3-2x}=\frac{2}{x}+\frac{3x}{x^2+2}$

Explanation of Solution

Given:

Rational Expression:

  $\frac{x^2-4}{x^3+2x}$

Consider the rational expression

  $\frac{x^2-4}{x^3+2x}$

Since the degree of denominator is greater than numerator, the expression is in proper form.

Therefore, the denominator can be written as

  $x³+2x=x\left(x²+2\right)$

Therefore,

  $\begin{array}{l}\frac{x^2-4}{x^3+2x}=\frac{x^2-4}{x(x^2+2)}\\ =\frac{A}{x}+\frac{Bx+C}{x^2+2}\end{array}$

Simplify as follows

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

Thus

  $x^2–4=A\left(x²+2\right)+\left(Bx+C\right)x\mathrm{......}\left(1\right)$

Substitute x=0 in equation (1) to get

  $\begin{array}{l}{\left(0\right)}^2–4=A\left(0^2+2\right)+\left[B\left(0\right)+C\right]0\hfill \\ –4=2A\hfill \\ –2=A\hfill \end{array}$

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

  $\begin{array}{l}1²–4=–2\left(1²+2\right)+\left[B\left(1\right)+C\right]1\hfill \\ –3=–6+B+C\hfill \\ 3=B+C\mathrm{........}\left(2\right)\hfill \end{array}$

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

  $\begin{array}{l}2²–4=–2\left(2²+2\right)+\left[B\left(2\right)+C\right]\left(2\right)\hfill \\ 0=–12+4B+2C\hfill \\ 12=4B+2C\hfill \\ 6=2B+C\mathrm{......}\left(3\right)\hfill \end{array}$

Subtract equation (2) from (3) to get

  $\begin{array}{l}6–3=2B+\cancel{C}–B–\cancel{C}\hfill \\ 3=B\hfill \end{array}$

Put B=3 in equation $3=B+C$

  $\begin{array}{l}3=3+C\\ 0=C\end{array}$

Hence, the partial fraction is given by

  $\frac{x^2-4}{x^3-2x}=\frac{2}{x}+\frac{3x}{x^2+2}$