College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises: 48

Answer

$4x^{3}+6x^{2}h+4xh^{2}+h^{3}$

Work Step by Step

We apply the binomial theorem to expand $(x+h)^3$: $(x+h)^4=\left(\begin{array}{l}4\\0\end{array}\right)x^{4}+\left(\begin{array}{l} 4\\1\end{array}\right)x^{3}h^{1}+\left(\begin{array}{l} 4\\2\end{array}\right)x^{2}h^{2}+\left(\begin{array}{l} 4\\3\end{array}\right)x^{1}h^{3}+\left(\begin{array}{l} 4\\4\end{array}\right)h^{4}=1x^{4}+4x^{3}h^1+6x^{2}h^{2}+4x^1h^{3}+1h^{4}$ Now we simplify: $\displaystyle \frac{(x+h)^{4}-x^{4}}{h}=\frac{x^{4}+4x^{3}h+6x^{2}h^{2}+4xh^{3}+h^{4}-x^{4}}{h} =\frac{4x^{3}h+6x^{2}h^{2}+4xh^{3}+h^{4}}{h}=\frac{h(4x^{3}+6x^{2}h+4xh^{2}+h^{3})}{h}=4x^{3}+6x^{2}h+4xh^{2}+h^{3}$
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