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 $ax^{2}+bx+c=0$.
$$8a^{2}-2a=0$$
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 $a=\frac{2±2}{16}$ when $±$ is plus. Add $2$ to $2$.
$$a=\frac{4}{16}$$
Reduce the fraction $\frac{4}{16}$ to lowest terms by extracting and canceling out $4$.
$$a=\frac{1}{4}$$
Now solve the equation $a=\frac{2±2}{16}$ when $±$ is minus. Subtract $2$ from $2$.
$$a=\frac{0}{16}$$
Divide $0$ by $16$.
$$a=0$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{1}{4}$ for $x_{1}$ and $0$ for $x_{2}$.
$$8a^{2}-2a=8\left(a-\frac{1}{4}\right)a$$
Subtract $\frac{1}{4}$ from $a$ by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
$$8a^{2}-2a=8\times \frac{4a-1}{4}a$$
Cancel out $4$, the greatest common factor in $8$ and $4$.
