Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 1 - Test - Page 111: 18



Work Step by Step

Substitute $-10$ to $x$ to obtain: $=(-10)^2-5(-10) \\=(-10)(-10)-(-50) \\=(-10)(-10)+50$ The product of two negative integers is positive. Simplify to obtain: $=100+50 \\=150$
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