Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set - Page 309: 6

Answer

$(2a+3)^2$

Work Step by Step

The two numbers whose product is $ac= 4(9)=36 $ and whose sum is $b= 12 $ are $\{ 6,6 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ 4a^2+12a+9 $, is \begin{array}{l}\require{cancel} 4a^2+6a+6a+9 \\\\= (4a^2+6a)+(6a+9) \\\\= 2a(2a+3)+3(2a+3) \\\\= (2a+3)(2a+3) \\\\= (2a+3)^2 .\end{array}
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.