[tex]6x^3-9x^2-12x=x(6x^2-9x-12)=\\\\Finding\ roots\ of\ equation:\\6x^2-9x-12=0\\\\
\Delta=b^2-4ac\\\\a=6,\ b=-9,\ c=-12 \\\\\Delta=(-9)^2-4*6*(-12)=81+288=369\\\\ \sqrt{\Delta}=3\sqrt{41}\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a}\ \ x_1=\frac{9-3\sqrt{41}}{2*6}=\frac{3-\sqrt{41}}{4}\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\ \ x_2=\frac{9+3\sqrt{41}}{2*6}=\frac{3+\sqrt{41}}{4}\\
6x^2-9x-12=(x-\frac{3+\sqrt{41}}{4})(x-\frac{3-\sqrt{41}}{4})[/tex][tex]so \\\\ \textbf{Answer} \\ 6x^3-9x^2-12x=x(6x^2-9x-12)=\\
\underline{x(x-\frac{3+\sqrt{41}}{4})(x-\frac{3-\sqrt{41}}{4})}[/tex]
