Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Appendix A - A.4 - Quadratic Equations - Exercises - Page 550: 22

Answer

$6a^2c^2+11abc^2-10b^2c^2 = (c^2)~(3a-2b)~(2a+5b)$

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: $~6a^2c^2+11abc^2-10b^2c^2$ First we can factor by using the GCF: $6a^2c^2+11abc^2-10b^2c^2 = (c^2)~(6a^2+11ab-10b^2)$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 11$ and $r\times s = -60$. We can see that $(15)+(-4) = 11~$ and $(15)\times (-4) = -60$ We can factor the trinomial as follows: $6a^2c^2+11abc^2-10b^2c^2 = (c^2)~(6a^2+11ab-10b^2)$ $6a^2c^2+11abc^2-10b^2c^2 = (c^2)~(6a^2+15ab-4ab-10b^2)$ $6a^2c^2+11abc^2-10b^2c^2 = (c^2)~[~(6a^2+15ab)+(-4ab-10b^2)~]$ $6a^2c^2+11abc^2-10b^2c^2 = (c^2)~[~(3a)(2a+5b)+(-2b)(2a+5b)~]$ $6a^2c^2+11abc^2-10b^2c^2 = (c^2)~(3a-2b)~(2a+5b)$
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.