What are the roots of $3x^4-28x^3-3x^2+112x-36=0$ ?

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Answer 1 · 2017-08-03T20:38:26

$3x^4-28x^3-3x^2+112x-36=0$

$3x^4-12x^2-28x^3+112x+9x^2-36=0$

$3x^2*(x^2-4)-28x*(x^2-4)+9*(x^2-4)=0$

$(x^2-4)*(3x^2-28x+9)=0$

$(x+2)* (x-2)* (3x^2-27x+x+9)=0$

$(x+2)* (x-2)* (3x-1)* (x-9)=0$

From these multipliers, roots of it $x=-2$ , $x=2$ , $x=1/3$ and $x=9$ .

1) I regrouped terms.

2) I found multipliers

3) I found roots.

What are the roots of 3x^4-28x^3-3x^2+112x-36=03x^4-28x^3-3x^2+112x-36=0?