College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.3 - Polynomials - R.3 Exercises - Page 30: 95

Answer

$9,999$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ 99\times101 ,$ get the middle number between $99$ and $101$. Then use the special product for the sum and difference of like terms. $\bf{\text{Solution Details:}}$ The middle number between $99$ and $101$ is $100.$ Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent\begin{array}{l}\require{cancel} (100-1)(100+1) \\\\= (100)^2-(1)^2 \\\\= 10000-1 \\\\= 9,999 .\end{array}
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