## Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

# Appendix A - A.4 - Quadratic Equations - Exercises - Page 550: 16

#### Answer

$4a^2+12ab+9b^2 = (2a+3b)~(2a+3b)$

#### Work Step by Step

Let's consider a trinomial in this form: $~ax^2+bxy+cy^2$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = a\times c$ Then the next step is to write the trinomial as follows: $~x^2+bxy+cy^2 = ax^2+rxy+sxy+cy^2$ Let's consider this trinomial: $~4a^2+12ab+9b^2$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 12$ and $r\times s = 36$. We can see that $6+6 = 12~$ and $6\times 6 = 36$ We can factor the trinomial as follows: $4a^2+12ab+9b^2 = 4a^2+6ab+6ab+9b^2$ $4a^2+12ab+9b^2 = (4a^2+6ab)+(6ab+9b^2)$ $4a^2+12ab+9b^2 = (2a)(2a+3b)+(3b)(2a+3b)$ $4a^2+12ab+9b^2 = (2a+3b)~(2a+3b)$

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