Algebra 1

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

Chapter 8 - Polynomials and Factoring - 8-7 Factoring Special Cases - Practice and Problem-Solving Exercises - Page 514: 12

Answer

$(m+9)^{2}$

Work Step by Step

In order to factor $m^{2}$+18m+81, we must apply the rule that states that $(a+b)^{2}$=$a^{2}$+2ab+$b^{2}$, and if we set $m^{2}$=$a^{2}$, 2ab=18m then we can solve for a and b $m^{2}$=$a^{2}$, then to solve for a, we square root both sides $\sqrt m^{2}$=$\sqrt a^{2}$ a=m Then, 2ab=18m, and since we know that a=m, we can substitute in a for , then solve for b. 2ab=18a, then, to solve for b, we'll divide by 2a on both sides of the equation $\frac{2ab}{2a}$=$\frac{18a}{2a}$ b=9 Then we sub in a and b into $(a+b)^{2}$ and get $m^{2}$+18m+81=$(m+9)^{2}$
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