Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - 6-3 Binomial Radical Expressions - Practice and Problem-Solving Exercises - Page 378: 24

Answer

$23+7\sqrt{7}$

Work Step by Step

Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression, $ (2+\sqrt{7})(1+3\sqrt{7}) ,$ is equivalent to \begin{align*} & 2(1)+2(3\sqrt{7})+\sqrt{7})(1)+\sqrt{7}(3\sqrt{7}) \\&= 2+6\sqrt{7}+\sqrt{7}+3(7) \\&= 2+6\sqrt{7}+\sqrt{7}+21 \\&= (2+21)+(6\sqrt{7}+\sqrt{7}) &\left( \text{combine like terms} \right) \\&= 23+7\sqrt{7} .\end{align*} Hence, the product of the given expression is $ 23+7\sqrt{7} $.
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