Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 8 - Polynomials and Factoring - 8-4 Multiplying Special Cases - Practice and Problem-Solving Exercises: 50

Answer

Their mistake is that they did not multiply the middle term by two according to the rule. $(a-b)^{2}$=$a^{2}$-2ab+$b^{2}$ Therefore, $(3a-7)^{2}$=$9a^{2}$-42a+49

Work Step by Step

Their mistake is that they did not multiply the middle term by two according to the rule. $(a-b)^{2}$=$a^{2}$-2ab+$b^{2}$ They did. $(3a-7)^{2}$=$9a^{2}$-21a+49 However according to the rule a= 3a and b=7. Therefore, $(3a-7)^{2}$= $9a^{2}$-2(3a)(7)+$7^{2}$ =$9a^{2}$-42a+49
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