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:
$V\left( x \right)=2{{x}^{3}}+12{{x}^{2}}+12x+10$.
