Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.3 Polynomials - R.3 Exercises: 31

Answer

$ \frac{256n^8}{t^4p^8}$.

Work Step by Step

The rule $(\frac{a}{b})^n = \frac{a^n}{b^n}$. $(\frac{-4n^2}{tp^2})^4 = \frac{(-4)^4n^{2\times4}}{t^4p^{2\times4}} = \frac{256n^8}{t^4p^8}$.
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