Chapter 7.4, Problem 57E

To determine

To find: write the partial fraction of the given rational expression $\frac{2x^3-4x^2-15x+5}{x^2-2x-8}$

Answer to Problem 57E

The final answer is $\frac{x^2-4x+6}{{\left(x^2+4\right)}^3}=\frac{1}{{\left(x^2+4\right)}^2}+\frac{-4x+2}{{\left(x^2+4\right)}^3}$

Explanation of Solution

Given information: we have the given rational fraction is $\frac{2x^3-4x^2-15x+5}{x^2-2x-8}$

Since the degree of numerator is greater than denominator therefore using long division method we get

  $\frac{2x^3-4x^2-15x+5}{x^2-2x-8}=2x+\frac{x+5}{x^2-2x-8}$

Now we find partial fraction on remainder.

  $\frac{x+5}{x^2-2x-8}=\frac{x+5}{\left(x+2\right)\left(x-4\right)}=\frac{A}{x+2}+\frac{B}{x-4}$

Further we can write

  $A\left(x-4\right)+B\left(x+2\right)=x+5$

Expanding the L.H.S and group element’s according to powers of x we get

  $\left(A+B\right)x+\left(-4A+2B\right)=x+5$

Now, equating the coefficients of similar term on both sides to make a list of equations

  $\begin{array}{l}A+B=1\\ -4A+2B=5\end{array}$

Solving the above system we get

  $A=\frac{-1}{2}andB=\frac{3}{2}$

Therefore, we get

  $\frac{x+5}{x^2-2x-8}=\frac{-1}{2\left(x+2\right)}+\frac{3}{2\left(x-4\right)}$

Finally, the required partial fraction is

  $\frac{2x^3-4x^2-15x+5}{x^2-2x-8}=2x+\frac{-1}{2\left(x+2\right)}+\frac{3}{2\left(x-4\right)}$