Big Ideas Math - Algebra 1, A Common Core Curriculum

Published by Big Ideas Learning LLC
ISBN 10: 978-1-60840-838-2
ISBN 13: 978-1-60840-838-2

Chapter 7 - Polynomial Equations and Factoring - 7.3 - Special Products of Polynomials - Exercises - Page 375: 21

Answer

$p^2-100q^2$.

Work Step by Step

The given expression is $=(p-10q)(p+10q)$ Use sum and difference pattern $(a-b)(a+b)=a^2-b^2$. We have $a=p$ and $b=10q$. $=(p)^2-(10q)^2$ Simplify. $=p^2-100q^2$ Hence, the product is $p^2-100q^2$.
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