Chapter 7.3, Problem 76E

Solution

To find: The partial fraction decomposition of $\frac{-7x-7}{x^2-3x-10}$ and support answer graphically.

The required partial fraction decomposition is $-\frac{6}{x-5}-\frac{1}{x+2}$ .

Given information:

The given rational function is $\frac{-7x-7}{x^2-3x-10}$ .

Explanation:

Consider the given rational function.

Factorize the denominator $x^2-3x-10$ .

  $x^2-3x-10=\left(x-5\right)\left(x+2\right)$

Thus,

  $\begin{array}{c}\frac{-7x-7}{x^2-3x-10}=\frac{A}{x-5}+\frac{B}{x+2}\\ -7x-7=A\left(x+2\right)+B\left(x-5\right)\\ -7x-7=Ax+Bx+2A-5B\end{array}$

Now compare the coefficient and constants.

  $\begin{array}{c}A+B=-7\\ 2A-5B=-7\end{array}$

Write the system as matrix equation $C=\left[\begin{array}{cc}1& 1\\ 2& -5\end{array}\right],X=\left[\begin{array}{c}A\\ B\end{array}\right]\text{ and }D=\left[\begin{array}{c}-7\\ -7\end{array}\right]$

Now solve the equation $CX=D$ .

  $\begin{array}{c}\left[\begin{array}{cc}1& 1\\ 2& -5\end{array}\right]\left[\begin{array}{c}A\\ B\end{array}\right]=\left[\begin{array}{c}-7\\ -7\end{array}\right]\\ \left[\begin{array}{c}A\\ B\end{array}\right]={\left[\begin{array}{cc}1& 1\\ 2& -5\end{array}\right]}^{-1}\left[\begin{array}{c}-7\\ -7\end{array}\right]\\ =\frac{1}{-5-2}\left[\begin{array}{cc}-5& -1\\ -2& 1\end{array}\right]\left[\begin{array}{c}-7\\ -7\end{array}\right]\\ =\frac{1}{-7}\left[\begin{array}{c}35+7\\ 14-7\end{array}\right]\\ =\left[\begin{array}{c}-6\\ -1\end{array}\right]\end{array}$

The required partial fraction decomposition is $-\frac{6}{x-5}-\frac{1}{x+2}$ .

Now draw the graph of both in the same window.