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: 13



Work Step by Step

In order to factor $q^{2}$+2q+1, we must apply the rule that states that $(a+b)^{2}$=$a^{2}$+2ab+$b^{2}$, and if we set $q^{2}$=$a^{2}$, 2ab=2q then we can solve for a and b $q^{2}$=$a^{2}$, then to solve for a, we square root both sides $\sqrt q^{2}$=$\sqrt a^{2}$ a=q Then, 2ab=2q, and since we know that a=q, we can substitute in a for q, then solve for b. 2ab=2a, then, to solve for b, we'll divide by 2a on both sides of the equation $\frac{2ab}{2a}$=$\frac{2a}{2a}$ b=1 Then we sub in a and b into $(a+b)^{2}$ and get $q^{2}$+2q+1=$(q+1)^{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.