Solve 5r^2-3r

$$5 r ^ { 2 } - 3 r$$

Factor

$$r\left(5r-3\right)$$

Solution Steps

Factor out r.

$$r\left(5r-3\right)$$

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.

$$5r^{2}-3r=0$$

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.

$$r=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\times 5}$$

Take the square root of \left(-3\right)^{2}.

$$r=\frac{-\left(-3\right)±3}{2\times 5}$$

The opposite of -3 is 3.

$$r=\frac{3±3}{2\times 5}$$

Multiply 2 times 5.

$$r=\frac{3±3}{10}$$

Now solve the equation r=\frac{3±3}{10} when ± is plus. Add 3 to 3.

$$r=\frac{6}{10}$$

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

$$r=\frac{3}{5}$$

Now solve the equation r=\frac{3±3}{10} when ± is minus. Subtract 3 from 3.

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

Divide 0 by 10.

$$r=0$$

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}.

$$5r^{2}-3r=5\left(r-\frac{3}{5}\right)r$$

Subtract \frac{3}{5} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5r^{2}-3r=5\times \frac{5r-3}{5}r$$

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

$$5r^{2}-3r=\left(5r-3\right)r$$

Evaluate

$$r\left(5r-3\right)$$