# Find all the zeros ofthe function and write the polynomial as a product of linear factors: f(x)=5x^3-9x^2+28x=6

Find all the zeros ofthe function and write the polynomial as a product of linear factors:
$f\left(x\right)=5{x}^{3}-9{x}^{2}+28x=6$
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Caylee Davenport
$f\left(x\right)=5{x}^{3}-9{x}^{2}+28x+6$
if x=-1/5 is a root of f(x) then f(-1/5) =0.
$f\left(-1/5\right)=5\ast \left(\frac{-1}{5}{\right)}^{3}-9\ast \left(\frac{-1}{5}{\right)}^{2}+28\ast \left(\frac{-1}{5}\right)+6=0$
$\left(5{x}^{3}-9{x}^{2}+28x+6\right)÷\left(x+1/5\right)=30-10x+5{x}^{2}$
$30-10x+5{x}^{2}=0⇒\left(x-\left(1-i\sqrt{5}\right)\right)\ast \left(x-\left(1+i\sqrt{5}\right)\right)=0$
$5{x}^{3}-9{x}^{2}+28x+6=\left(x+1/5\right)\left(x-\left(1-i\sqrt{5}\right)\right)\ast \left(x-\left(1+i\sqrt{5}\right)\right)=0$
Hence