Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 4 - Polynomials - Study Summary - Practice Exercises: 33

Answer

$y^3-2y+4$

Work Step by Step

Divide $4y^2$ to each term of the numerator to obtain: $=\dfrac{4y^5}{4y^2}-\dfrac{8y^3}{4y^2}+\dfrac{16y^2}{4y^2}$ Simplify by dividing the coefficients and using the Quotient Rule of Exponents $(\frac{a^m}{a^n} = a^{m-n})$ to obtain: $=1y^{5-2} - 2y^{3-2} + 4y^{2-2} \\=y^3-2y+4y^{0}$ Since $a^0=1, a\ne0$, then the expression above simplifies to: $=y^3-2y+4(1) \\=y^3-2y+4$
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