College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 3 - Polynomial and Rational Functions - Exercise Set 3.3 - Page 375: 52

Answer

a. $(2h^{3}+14h^2-72)\div(h-2)=2h^2+18h+36$ b. $h=2, l=9, w=4$

Work Step by Step

a.$\begin{array}{lllll} \underline{2}| & 2 & 14 & 0 & -72\\ & & 4 & 36 & 72\\ \hline & & & & \\ & 2 & 18 & 36 & 0 \end{array}$ Since the last number on the bottom right of synthetic division known as the remainder is $0$, the equation divided perfectly. Therefore, $h=2$ is a solution to the equation. $(2h^{3}+14h^2-72)\div(h-2)=2h^2+18h+36$ b. For a rectangular prism, the volume formula is, $V=h\times l \times w$ whereas, $h$ is height, $l$ is length and $w$ is width. In our case, $h=h$, $l=h+7$ and $w=2h$, therefore the volume is, $V=2h^3+14h^2=72$, $V=2h^3+14h^2-72=0$. The equation for Volume when the volume equals $72$ is the same as equation from part(a). $V=2(h-2)(h^2+9h+18)=2(h-2)(h+3)(h+6)$ Therefore $h=2$ is the only positive solution to the equation. $h=2, l=9, w=4$
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