Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.4 - Dividing Polynomials; Remainder and Factor Theorems - Exercise Set - Page 364: 52

Answer

a. See figure and explanations. b. $2\times4\times9\ in$.
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Work Step by Step

a. The coefficients of the dividend $f(x)=2h^3+14h^2-72$ can be identified in order as $\{2,14,0,-72\}$ and the divisor is $h-2$; using synthetic division as shown in the figure, we get the remainder as zero, indicating that $h=2$ is a solution of the equation. b. The volume of the box is given as $2h^2(h+7)=72$ which can be simplified to $2h^3+14h^2-72=0$ Based on the above result, we have $(2h^2+18h+36)(h-2)=0$ or $(h+3)(h+6)(h-2)=0$. Since $h\gt0$ the only positive solution to the equation is $h=2$ and we have the dimensions as $2\times4\times9\ in$.
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