## Algebra 1: Common Core (15th Edition)

$15x+10$.
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 $15x+10$.
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