Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Appendix A - A.4 - Quadratic Equations - Exercises - Page 556: 14

Answer

$12a^2+31a+20 = (3a+4)~(4a+5)$

Work Step by Step

Let's consider a trinomial in this form: $~ax^2+bx+c$ 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: $~x^2+bx+c = ax^2+rx+sx+c$ Let's consider this trinomial: $~12a^2+31a+20$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 31$ and $r\times s = 240$. We can see that $15+16 = 31~$ and $15\times 16 = 240$ We can factor the trinomial as follows: $12a^2+31a+20 = 12a^2+15a+16a+20$ $12a^2+31a+20 = (12a^2+15a)+(16a+20)$ $12a^2+31a+20 = (3a)(4a+5)+(4)(4a+5)$ $12a^2+31a+20 = (3a+4)~(4a+5)$
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