Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.4 - Multiplying Polynomials - Exercise Set: 86

Answer

$a^2+7a+10$

Work Step by Step

Replacing $x$ with $a+5$ in the given function, $ f(x)=x^2-3x $, then $f(a+5)$ is equal to \begin{array}{l} (a+5)^2-3(a+5) \\\\= a^2+2(a)(5)+(5)^2-3a-15 \\\\= a^2+10a+25-3a-15 \\\\= a^2+(10a-3a)+(25-15) \\\\= a^2+7a+10 .\end{array}
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