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: 1



Work Step by Step

The two numbers whose product is $ac= 1(9)=9 $ and whose sum is $b= 6 $ are $\{ 3,3 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ x^2+6x+9 $, is \begin{array}{l}\require{cancel} x^2+3x+3x+9 \\\\= (x^2+3x)+(3x+9) \\\\= x(x+3)+3(x+3) \\\\= (x+3)(x+3) \\\\= (x+3)^2 .\end{array}
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