## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 4 - Polynomials - 4.6 Special Products - 4.6 Exercise Set - Page 277: 82

#### Answer

$9+3t^5+\dfrac{1}{4}t^{10}$

#### Work Step by Step

Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the given expression, $\left( 3+\dfrac{1}{2}t^5 \right)\left( 3+\dfrac{1}{2}t^5 \right)$, is equivalent to \begin{array}{l}\require{cancel} (3)^2+2(3)\left( \dfrac{1}{2}t^5 \right)+\left( \dfrac{1}{2}t^5 \right)^2 \\\\= 9+3t^5+\dfrac{1}{4}t^{10} .\end{array}

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