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 $ax^{2}+bx+c=0$.
$$-3x^{2}+x=0$$
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.
$$x=\frac{-1±\sqrt{1^{2}}}{2\left(-3\right)}$$
Take the square root of $1^{2}$.
$$x=\frac{-1±1}{2\left(-3\right)}$$
Multiply $2$ times $-3$.
$$x=\frac{-1±1}{-6}$$
Now solve the equation $x=\frac{-1±1}{-6}$ when $±$ is plus. Add $-1$ to $1$.
$$x=\frac{0}{-6}$$
Divide $0$ by $-6$.
$$x=0$$
Now solve the equation $x=\frac{-1±1}{-6}$ when $±$ is minus. Subtract $1$ from $-1$.
$$x=-\frac{2}{-6}$$
Reduce the fraction $\frac{-2}{-6}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{1}{3}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $0$ for $x_{1}$ and $\frac{1}{3}$ for $x_{2}$.
$$-3x^{2}+x=-3x\left(x-\frac{1}{3}\right)$$
Subtract $\frac{1}{3}$ from $x$ by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
$$-3x^{2}+x=-3x\times \frac{-3x+1}{-3}$$
Cancel out $3$, the greatest common factor in $-3$ and $-3$.
