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 - Page 289: 88

Answer

$f(a-b)=a^2-2ab+b^2-3a+3b$

Work Step by Step

Substituting $x=a-b$ in $f(x)=x^2-3x$, then, \begin{align*} f(a-b)=(a-b)^2-3(a-b) = a^2-2ab+b^2-3a+3b .\end{align*}
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