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

Published by Brooks Cole

# Appendix A - A.4 - Quadratic Equations - Exercises - Page 550: 15

#### Answer

$3x^2+11xy-4y^2 = (3x-y)(x+4y)$

#### 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: $~3x^2+11xy-4y^2$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 11$ and $r\times s = -12$. We can see that $(12)+(-1) = 11~$ and $(12)\times (-1) = -12$ We can factor the trinomial as follows: $3x^2+11xy-4y^2 = 3x^2+12xy-xy-4y^2$ $3x^2+11xy-4y^2 = (3x^2+12xy)+(-xy-4y^2)$ $3x^2+11xy-4y^2 = (3x)(x+4y)+(-y)(x+4y)$ $3x^2+11xy-4y^2 = (3x-y)(x+4y)$

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