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: 12

Answer

$y^2-4y-96 = (y-12)~(y+8)$

Work Step by Step

Let's consider a trinomial in this form: $~x^2+bx+c$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$ Then we can factor the trinomial as follows: $~x^2+bx+c = (x+r)~(x+s)$ Let's consider this trinomial: $~y^2-4y-96$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = -4$ and $r\times s = -96$. We can see that $(-12)+(8) = -4~$ and $(-12)\times (8) = -96$ We can factor the trinomial as follows: $y^2-4y-96 = (y-12)~(y+8)$
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