## Precalculus (6th Edition) Blitzer

The area of the provided figure as a polynomial function in standard form is: $A\left( x \right)=3{{x}^{2}}+x-4$
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: $A\left( x \right)=3{{x}^{2}}+x-4$.