Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 4 - Quadratic Functions and Equations - 4-4 Factoring Quadratic Expressions - Practice and Problem-Solving Exercises - Page 222: 75

Answer

$10(x+3)(x-3)$

Work Step by Step

Factoring the $GCF= 10 ,$ the given expression, $ 10x^2-90 ,$ is equivalent to \begin{align*} 10(x^2-9) .\end{align*} Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{align*} & 10[(x)^2-(3)^2] \\&= 10[(x)^2-(3)^2] \\&= 10[(x+3)(x-3)] \\&= 10(x+3)(x-3) .\end{align*}
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