Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Appendix A - A.4 - Quadratic Equations - Exercises - Page 550: 19

Answer

$3y^2+24y+45 = (3)~(y+5)~(y+3)$

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: $~3y^2+24y+45$ First we can factor by using the GCF: $3y^2+24y+45 = (3)~(y^2+8y+15)$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 8$ and $r\times s = 15$. We can see that $3+5 = 8~$ and $3\times 5 = 15$ We can factor the trinomial as follows: $3y^2+24y+45 = (3)~(y^2+8y+15)$ $3y^2+24y+45 = (3)~(y+5)~(y+3)$
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