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: 87

Answer

$b^2-7b+10$

Work Step by Step

Replacing $x$ with $b-2$ in the given function, $ f(x)=x^2-3x $, then $f(b-2)$ is equal to \begin{array}{l} (b-2)^2-3(b-2) \\\\= (b)^2+2(b)(-2)+(-2)^2-3b+6 \\\\= b^2-4b+4-3b+6 \\\\= b^2+(-4b-3b)+(4+6) \\\\= b^2-7b+10 .\end{array}
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