Intermediate Algebra (6th Edition)

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

Chapter 5 - Review - Page 330: 69



Work Step by Step

Using $(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$ or the square of a multinomial, then, \begin{array}{l} (x^2+9x+1)^2 \\= (x^2)^2+(9x)^2+(1)^2+2(x^2)(9x)+2(x^2)(1)+2(9x)(1) \\= x^4+81x^2+1+18x^3+2x^2+18x \\= x^4+18x^3+(81x^2+2x^2)+18x+1 \\= x^4+18x^3+83x^2+18x+1 .\end{array}
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