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: 23

Answer

$x^3+5x^2+4x = (x)~(x+4)~(x+1)$

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: $~x^3+5x^2+4x$ First we can factor by using the GCF: $x^3+5x^2+4x = (x)~(x^2+5x+4)$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 5$ and $r\times s = 4$. We can see that $4+1 = 5~$ and $4\times 1 = 4$ We can factor the trinomial as follows: $x^3+5x^2+4x = (x)~(x^2+5x+4)$ $x^3+5x^2+4x = (x)~(x+4)~(x+1)$
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