Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.2 - More Work with Exponents and Scientific Notation - Exercise Set: 6

Answer

$16x^{6}y^{2}z^{2}$

Work Step by Step

We are given the expression $(4x^{3}yz)^{2}=(4)^{2}\times(x^{3})^{2}\times(y)^{2}\times (z)^{2}$. To simplify further, we can use the power rule, which holds that $(a^{m})^{n}=a^{m\times n}$ (where a is a real number, and m and n are integers). $(4)^{2}\times(x^{3})^{2}\times(y)^{2}\times (z)^{2}=16\times x^{3\times2}\times y^{2}\times z^{2}=16x^{6}y^{2}z^{2}$
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