## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 1 - Section 1.10 - Modeling with Functions - Exercise Set - Page 295: 45

#### Answer

The volume of the provided figure as a polynomial function in standard form is: $V\left( x \right)=2{{x}^{3}}+12{{x}^{2}}+12x+10$.

#### Work Step by Step

Consider the provided figure: there are two rectangles -- one is the outer rectangle and the second is the inner rectangle which is removed from the outer rectangle. The length, width and height of the outer rectangle is $l=x+5,w=2x+1$ and $h=x+2$. The length, width and height of the inner rectangle is $l=x+5,w=x$ and $h=3$. Now, the volume of the provided figure is the difference of the volume of the both outer and inner rectangles because the inner rectangle is removed from the outer rectangle. \begin{align} & V\left( x \right)=\left[ \left( x+5 \right)\left( 2x+1 \right)\left( x+2 \right) \right]-\left[ \left( x+5 \right)\left( 3 \right)\left( x \right) \right] \\ & =\left[ \left( x+5 \right)\left( 2{{x}^{2}}+4x+x+2 \right) \right]-\left[ 3x\left( x+5 \right) \right] \\ & =2{{x}^{3}}+15{{x}^{2}}+27x+10-3{{x}^{2}}-15x \\ & =2{{x}^{3}}+12{{x}^{2}}+12x+10 \end{align} Hence, the volume of the provided figure as a polynomial function in standard form is: $V\left( x \right)=2{{x}^{3}}+12{{x}^{2}}+12x+10$.

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