Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.3 Rational Expressions - R.3 Exercises - Page R-10: 10



Work Step by Step

Step 1: First you have to factor both the numerator and the denominator to get them completely simplified. To factor the numerator, you have to use the process of factoring a trinomial. Since the coefficient on the first term $(z^{2})$ is 1, you only have to look to see which pair of numbers is the product of the third term (positive 6), but the sum of the middle coefficient (-5). You can first try: -6 and 1. They add up to -5, but they are not the product of a positive 6. Next you can try: -2 and -3. They both add up to -5, and they are the product of a positive 6 so these work to use in your factoring. You can then factor the trinomial completely to result in: (z-2)(z-3) for your numerator. Next you have to factor the denominator. Since both $z^{2}$ and 4 are both perfect squares and being subtracted, you are allowed to use the factoring method of difference of squares. To do this you separately take the sum and difference of the simplified squares of the terms and multiply them together in parenthesis to get: (z-2)(z+2) Step 2: Once everything is factored, you will get this fraction: $\frac{(z-2)(z-3)}{(z-2)(z+2)}$. You can now cancel out any like terms that are on the numerator and denominator because anything divided by itself equals 1. Therefore you can cancel out the two (z-2) terms leaving you with the final answer of: $\frac{z-3}{z+2}$.
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