## Intermediate Algebra (6th Edition)

$x^{2}-8x+16$
We are given the expression $(x-4)^{2}$, which is equivalent to $(x-4)(x-4)$. We can use the FOIL method to simplify. F=$x\times x=x^{2}$ O= $x\times-4=-4x$ I=$-4\times x=-4x$ L=$-4\times-4=16$ Now, we can add these terms together. $(x-4)^{2}=(x^{2})+(-4x)+(-4x)+(16)=x^{2}-8x+16$