Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 9 - 9.4 - Partial Fractions - 9.4 Exercises - Page 668: 20

Answer

$[\frac{1}{5}][\frac{4}{x-3} + \frac{1}{x+2}]$

Work Step by Step

We must first find the Partial Fraction Decomposition: $\frac{A}{x-3} + \frac{B}{x+2} = \frac{x+1}{(x-3)(x+2)}$ We must then solve for the constants: $A(x+2) + B(x-3) = x+1$ $Ax + 2A + Bx - 3B = x +1$ $Ax + Bx = x$ $2A - 3B = 1$ This can be represented with the following matrix: $\begin{bmatrix} 1 & 1 & |1\\ 2 & -3 & |1\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 1 & |1\\ 0 & -5 & |-1\\ \end{bmatrix}$ $-5B = -1$ $B = \frac{1}{5}$ $A + B = 1$ $A + \frac{1}{5} = 1$ $A = \frac{4}{5}$ The Partial Fraction is: $[\frac{1}{5}][\frac{4}{x-3} + \frac{1}{x+2}]$ Checking the Result: $[\frac{1}{5}]\frac{4x+8+x-3}{(x-3)(x+2)} = [\frac{1}{5}]\frac{5x + 5}{(x-3)(x+2)} = \frac{x+1}{(x-3)(x+2)}$ Therefore, the answer is correct.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.