Intermediate Algebra (6th Edition)

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

Chapter 11 - Section 11.5 - The Binomial Theorem - Exercise Set - Page 664: 28

Answer

$-243 x^5 + 1620 x^4 - 4320 x^3 + 5760 x^2 - 3840 x + 1024$

Work Step by Step

Using the Binomial Formula, the expression $ (4-3x)^5 $ expands to \begin{array}{l} 4^5(-3x)^0 \dfrac{5}{1!}4^4(-3x)^1+ \dfrac{5\cdot4}{2!}4^3(-3x)^2+ \dfrac{5\cdot4\cdot3}{3!}4^2(-3x)^3+ \dfrac{5\cdot4\cdot3\cdot2}{4!}4^1(-3x)^4+ \dfrac{5\cdot4\cdot3\cdot2\cdot1}{5!}4^0(-3x)^5+ \\\\= -243 x^5 + 1620 x^4 - 4320 x^3 + 5760 x^2 - 3840 x + 1024 \end{array}
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