Answer
$15x+10$.
Work Step by Step
First, find the total area of the outer rectangle.
$A=a\cdot b$ ...the area of a rectangle.
...substitute $(x+6)$ for $a$ and $(x+3)$ for $b$ in the formula.
$A=(x+6)(x+3)$
...use the FOIL method.
$A=(x)(x)+(x)(3)+(6)(x)+(6)(3)$
...simplify
$A=x^{2}+3x+6x+18$
...add like terms.
$\color{red}{A=x^{2}+9x+18}$ ... (area of the outer rectangle.)
Now find the area of the inner rectangle
$A=a\cdot b$
...the area of a rectangle.
...substitute $(x-2)$ for $a$ and $(x-4)$ for $b$ in the formula.
$A=(x-2)(x-4)$
...use the FOIL method.
$A=(x)(x)+(x)(-4)+(-2)(x)+(-2)(-4)$
...simplify
$A=x^{2}-4x-2x+8$
...add like terms.
$A=\color{red}{x^{2}-6x+8}$ ... (area of the inner rectangle.)
Finally, find the area of the shaded region.
Area of shaded region=Area of outer rectangle - Area of inner rectangle
$A=x^{2}+9x+18-(x^{2}-6x+8)$
$=x^{2}+9x+18-x^{2}+6x-8$
...add like terms.
$=\color{red}{15x+10}$
The area of the shaded region is $15x+10$.
