Answer
The area of the provided figure as a polynomial function in standard form is:
\[A\left( x \right)=3{{x}^{2}}+x-4\]
Work Step by Step
Consider the provided figure: there are two triangle and one rectangle.
The length of the first triangle is $l=x$ and the base length $b=x-5$.
The length of the second triangle is $l=x$ and the base length $b=x+3$.
The length of the rectangle is $l=x+2$ and width is $b=\left( x-5 \right)+\left( x+3 \right)$.
Now, the area of the provided figure is the sum of the area of both the triangle and rectangle.
$\begin{align}
& A\left( x \right)=\left[ \frac{1}{2}x\left( x-5 \right) \right]+\left[ \frac{1}{2}x\left( x+3 \right) \right]+\left[ \left( x+2 \right)\left[ \left( x-5 \right)+\left( x+3 \right) \right] \right] \\
& =\frac{1}{2}{{x}^{2}}-\frac{5}{2}x+\frac{1}{2}{{x}^{2}}+\frac{3}{2}x+2{{x}^{2}}+2x-4 \\
& ={{x}^{2}}-x+2{{x}^{2}}+2x-4 \\
& =3{{x}^{2}}+x-4
\end{align}$
Hence, the area of the provided figure as a polynomial function in standard form is:
$A\left( x \right)=3{{x}^{2}}+x-4$.
