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: 20

Answer

$30x^2-35x+10 = (5)~(2x-1)~(3x-2)$

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: $~30x^2-35x+10$ First we can factor by using the GCF: $30x^2-35x+10 = (5)~(6x^2-7x+2)$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = -7$ and $r\times s = 12$. We can see that $(-4)+(-3) = -7~$ and $(-4)\times (-3) = 12$ We can factor the trinomial as follows: $30x^2-35x+10 = (5)~(6x^2-7x+2)$ $30x^2-35x+10 = (5)~(6x^2-4x-3x+2)$ $30x^2-35x+10 = (5)~[~(6x^2-4x)+(-3x+2)~]$ $30x^2-35x+10 = (5)~[(2x)(3x-2)+(-1)(3x-2)]$ $30x^2-35x+10 = (5)~(2x-1)~(3x-2)$
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