Intermediate Algebra (6th Edition)

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

Chapter 11 - Review - Page 668: 70

Answer

$81y^4-108y^3+54y^2z^2-12yz^3+z^4$

Work Step by Step

Using the coefficients in the $r=5$ row of the Pascal's triangle, $\left\{ 1,4,6,4,1 \right\}$, and the pattern in expanding binomials, then $ (3y-z)^4 $ expands to \begin{array}{l} (3y)^4(-z)^0+4(3y)^3(-z)^1+6(3y)^2(-z)^2+4(3y)^1(-z)^3+(3y)^0(-z)^4 \\\\= 81y^4-108y^3+54y^2z^2-12yz^3+z^4 .\end{array}
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