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

$(8x^{2}+32x+32)\ \mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}^{2}.$ Drawing a diagram helps in visualizing which area needs to be subtracted (which is outer, which is inner) Area of red part=(Area of outer square) - (Area of inner square)
Find the total area of the inner square. $(x+2)^{2}$ ...square the binomial. $=x^{2}+2(x)(2)+2^{2}$ ...simplify. $=\color{red}{x^{2}+4x+4}$ The length of the side of the outer square is $3\cdot(x+2)$ Find the area of the outer square. $[3\cdot(x+2)]^{2}=3^{2}\cdot(x+2)^{2}$ ...square the binomial. $=9\cdot[x^{2}+2(x)(2)+2^{2}]$ ...simplify. $=9(x^{2}+4x+4)$ $=\color{red}{9x^{2}+36x+36}$ Find the area of the red part of the logo. Area of figure=(Area of outer square) - (Area of inner square) $A=9x^{2}+36x+36-(x^{2}+4x+4)$ ...remove parentheses. $=9x^{2}+36x+36-x^{2}-4x-4$ ...group like terms. $=9x^{2}-x^{2}+36x-4x+36-4$ ...add like terms. $=8x^{2}+32x+32$ The area of the red part of the logo is $(8x^{2}+32x+32)\ \mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}^{2}.$