Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$6x^{2}-4x=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
Now solve the equation $x=\frac{4±4}{12}$ when $±$ is plus. Add $4$ to $4$.
$$x=\frac{8}{12}$$
Reduce the fraction $\frac{8}{12}$ to lowest terms by extracting and canceling out $4$.
$$x=\frac{2}{3}$$
Now solve the equation $x=\frac{4±4}{12}$ when $±$ is minus. Subtract $4$ from $4$.
$$x=\frac{0}{12}$$
Divide $0$ by $12$.
$$x=0$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{2}{3}$ for $x_{1}$ and $0$ for $x_{2}$.
$$6x^{2}-4x=6\left(x-\frac{2}{3}\right)x$$
Subtract $\frac{2}{3}$ from $x$ by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
$$6x^{2}-4x=6\times \frac{3x-2}{3}x$$
Cancel out $3$, the greatest common factor in $6$ and $3$.
