# Solve 5x^{3}-3x^{2}-20x+12=0

Solve $5{x}^{3}-3{x}^{2}-20x+12=0$
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scoollato7o
We have $5{x}^{3}-3{x}^{2}-20x+12=0Z$
${x}^{2}\left(5x-3\right)-4\left(5x-3\right)=0$
$\left({x}^{2}-4\right)\left(5x-3\right)=0$
$\left(x+2\right)\left(x-2\right)\left(5x-3\right)=0$
Thus, $x=±2,\frac{3}{5}$

redhotdevil13l3
$5{x}^{3}-3{x}^{2}-20x+12=0Z$
${x}^{2}\left(5x-3\right)-4\left(5x-3\right)=0$
$\left({x}^{2}-4\right)\left(5x-3\right)=0$
$\left(x+2\right)\left(x-2\right)\left(5x-3\right)=0$
Using the Zero Factor Principle
$x=-2$, $x=2$, $x=\frac{3}{5}$