Answer
The area of the provided figure as a polynomial function in standard form is \[A\left( x \right)=6{{x}^{2}}+26x\].
Work Step by Step
Consider the provided figure:
There are two triangles and two rectangles.
The length of the first triangle is $l=x$ and the base is $b=2x$.
The length of the second triangle is $l=x+2$ and the base is $b=6x-4x$.
The length of the first rectangle is $l=x+2$ and the width is $b=4x$.
The length of the second rectangle is $l=\left( x+10 \right)-\left( x+2 \right)$ and the width is $b=2x$.
Now, the area of the provided figure is the sum of the area of both triangles and both rectangles.
$\begin{align}
& A\left( x \right)=\left[ \frac{1}{2}x\left( 2x \right) \right]+\left[ \frac{1}{2}\left( 6x-4x \right)\left( x+2 \right) \right]+\left[ \left( 4x \right)\left( x+2 \right) \right]+\left[ 2x\left( x+10-x-2 \right) \right] \\
& ={{x}^{2}}+x\left( x+2 \right)+4{{x}^{2}}+8x+16x \\
& ={{x}^{2}}+{{x}^{2}}+2x+4{{x}^{2}}+8x+16x \\
& =6{{x}^{2}}+26x
\end{align}$
Hence, the area of the provided figure as a polynomial function in standard form is $A\left( x \right)=6{{x}^{2}}+26x$.
