## Algebra: A Combined Approach (4th Edition)

Published by Pearson

# Chapter 5 - Section 5.6 - Special Products - Practice: 5

#### Answer

$y^{2}+6y+9$

#### Work Step by Step

Step 1: $(y+3)^{2}$ Step 2: Using the special product, $(y+3)^{2}=(y)^{2}+2(y)(3)+(3)^{2}.$ Step 3: $(y)^{2}+2(y)(3)+(3)^{2}=y^{2}+6y+9.$ Step 4: Therefore, the required answer is $y^{2}+6y+9.$

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