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 - Page 496: 40

Answer

$(9k+2q)^{2}$=$81k^{2}$+36kq+$4q^{2}$

Work Step by Step

To simplify the product $(9k+2q)^{2}$ we'll apply the rule that states that $(a+b)^{2}$=$a^{2}$+2ab+$b^{2}$ and set a=9k and b=2q Therefore, $(9k+2q)^{2}$ =$(9k)^{2}$+2(9k)(2q)+$(2q)^{2}$ =$81k^{2}$+36kq+$4q^{2}$
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