## Intermediate Algebra (12th Edition)

$x^{3}-3x^{2}-9x+27$
We are given that $f(x)=x^{2}-9$, $g(x)=2x$, and $h(x)=x-3$. We know that $(fh)(x)=f(x)\times h(x)$. Therefore, $(fg)(x)=(x^{2}-9)\times(x-3)$. We can use the FOIL method to further simplify this function. First terms: $x^{2}\times x=x^{2+1}=x^{3}$ Outer terms: $x^{2}\times-3=-3x^{2}$ Inner terms: $-9\times x=-9x$ Last terms: $-9\times-3$ Add together: $x^{3}+(-3x^{2})+(-9x)+27=x^{3}-3x^{2}-9x+27$