College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.3 - Polynomials - R.3 Exercises: 71

Answer

$9a^2+6ab+b^2-6a-2b+1$

Work Step by Step

Using $(a\pm b)^2=a^2\pm2ab+b^2$ or the square of a binomial, the expression, $ [(3a+b)-1]^2 ,$ is equivalent to \begin{array}{l}\require{cancel} (3a+b)^2-2(3a+b)(1)+(1)^2 \\\\= (3a+b)^2-6a-2b+1 \\\\= (3a)^2+2(3a)(b)+(b)^2-6a-2b+1 \\\\= 9a^2+6ab+b^2-6a-2b+1 .\end{array}
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