Solve 5a^2-4a

$$5 a ^ { 2 } - 4 a$$

Factor

$$a\left(5a-4\right)$$

Solution Steps

Factor out a.

$$a\left(5a-4\right)$$

Steps Using the Quadratic Formula

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.

$$5a^{2}-4a=0$$

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.

$$a=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2\times 5}$$

Take the square root of \left(-4\right)^{2}.

$$a=\frac{-\left(-4\right)±4}{2\times 5}$$

The opposite of -4 is 4.

$$a=\frac{4±4}{2\times 5}$$

Multiply 2 times 5.

$$a=\frac{4±4}{10}$$

Now solve the equation a=\frac{4±4}{10} when ± is plus. Add 4 to 4.

$$a=\frac{8}{10}$$

Reduce the fraction \frac{8}{10} to lowest terms by extracting and canceling out 2.

$$a=\frac{4}{5}$$

Now solve the equation a=\frac{4±4}{10} when ± is minus. Subtract 4 from 4.

$$a=\frac{0}{10}$$

Divide 0 by 10.

$$a=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{5} for x_{1} and 0 for x_{2}.

$$5a^{2}-4a=5\left(a-\frac{4}{5}\right)a$$

Subtract \frac{4}{5} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5a^{2}-4a=5\times \frac{5a-4}{5}a$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5a^{2}-4a=\left(5a-4\right)a$$

Evaluate

$$a\left(5a-4\right)$$