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 $ax^{2}+bx+c=0$.
$$5x^{2}-3x=0$$
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 $x=\frac{3±3}{10}$ when $±$ is plus. Add $3$ to $3$.
$$x=\frac{6}{10}$$
Reduce the fraction $\frac{6}{10}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{3}{5}$$
Now solve the equation $x=\frac{3±3}{10}$ when $±$ is minus. Subtract $3$ from $3$.
$$x=\frac{0}{10}$$
Divide $0$ by $10$.
$$x=0$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{3}{5}$ for $x_{1}$ and $0$ for $x_{2}$.
$$5x^{2}-3x=5\left(x-\frac{3}{5}\right)x$$
Subtract $\frac{3}{5}$ from $x$ by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
$$5x^{2}-3x=5\times \frac{5x-3}{5}x$$
Cancel out $5$, the greatest common factor in $5$ and $5$.
