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

$25x^{2}(9-\pi)$.
$A_{1}=\pi r^{2}\qquad$...area of a circle $A_{1}=\pi(5x)^{2}\qquad$...substitute 5x for r. $A_{1}=25\pi x^{2}\qquad...$simplify. $A_{2}=s^{2}\qquad...$area of a square. $A_{2}=(15x)^{2}\qquad...$substitute 15x for s. $A_{2}=225x^{2}\qquad...$simplify. The area of the metal frame is $A_{2}-A_{1}$, or $225x^{2}-25\pi x^{2}$. $225x^{2}-25\pi x^{2}\qquad...$find the GCF. $225x^{2}=3\cdot 3\cdot(5\cdot 5)\cdot(x\cdot x)$ $-25\pi x^{2}=-1\cdot(5\cdot 5)\cdot\pi\cdot(x\cdot x)$ GCF=$(5\cdot 5)\cdot(x\cdot x)$, or $25x^{2}$. $225x^{2}-25\pi x^{2}\qquad...$factor out the GCF $=25x^{2}(9)-25x^{2}(\pi)\qquad...$apply the Distributive Property. $=25x^{2}(9-\pi)$ The factored form of the area of the metal frame is $25x^{2}(9-\pi)$.