Talon Mcbride

2022-07-28

Find all the zeros ofthe function and write the polynomial as a product of linear factors:

$f(x)=5{x}^{3}-9{x}^{2}+28x=6$

$f(x)=5{x}^{3}-9{x}^{2}+28x=6$

Caylee Davenport

Beginner2022-07-29Added 14 answers

$f(x)=5{x}^{3}-9{x}^{2}+28x+6$

if x=-1/5 is a root of f(x) then f(-1/5) =0.

$f(-1/5)=5\ast (\frac{-1}{5}{)}^{3}-9\ast (\frac{-1}{5}{)}^{2}+28\ast (\frac{-1}{5})+6=0$

$(5{x}^{3}-9{x}^{2}+28x+6)\xf7(x+1/5)=30-10x+5{x}^{2}$

$30-10x+5{x}^{2}=0\Rightarrow (x-(1-i\sqrt{5}))\ast (x-(1+i\sqrt{5}))=0$

$5{x}^{3}-9{x}^{2}+28x+6=(x+1/5)(x-(1-i\sqrt{5}))\ast (x-(1+i\sqrt{5}))=0$

Hence

if x=-1/5 is a root of f(x) then f(-1/5) =0.

$f(-1/5)=5\ast (\frac{-1}{5}{)}^{3}-9\ast (\frac{-1}{5}{)}^{2}+28\ast (\frac{-1}{5})+6=0$

$(5{x}^{3}-9{x}^{2}+28x+6)\xf7(x+1/5)=30-10x+5{x}^{2}$

$30-10x+5{x}^{2}=0\Rightarrow (x-(1-i\sqrt{5}))\ast (x-(1+i\sqrt{5}))=0$

$5{x}^{3}-9{x}^{2}+28x+6=(x+1/5)(x-(1-i\sqrt{5}))\ast (x-(1+i\sqrt{5}))=0$

Hence

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