Solve 4c^2-15c

$$4 c ^ { 2 } - 15 c$$

Factor

$$c\left(4c-15\right)$$

Solution Steps

Factor out c.

$$c\left(4c-15\right)$$

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.

$$4c^{2}-15c=0$$

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.

$$c=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}}}{2\times 4}$$

Take the square root of \left(-15\right)^{2}.

$$c=\frac{-\left(-15\right)±15}{2\times 4}$$

The opposite of -15 is 15.

$$c=\frac{15±15}{2\times 4}$$

Multiply 2 times 4.

$$c=\frac{15±15}{8}$$

Now solve the equation c=\frac{15±15}{8} when ± is plus. Add 15 to 15.

$$c=\frac{30}{8}$$

Reduce the fraction \frac{30}{8} to lowest terms by extracting and canceling out 2.

$$c=\frac{15}{4}$$

Now solve the equation c=\frac{15±15}{8} when ± is minus. Subtract 15 from 15.

$$c=\frac{0}{8}$$

Divide 0 by 8.

$$c=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{15}{4} for x_{1} and 0 for x_{2}.

$$4c^{2}-15c=4\left(c-\frac{15}{4}\right)c$$

Subtract \frac{15}{4} from c by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4c^{2}-15c=4\times \frac{4c-15}{4}c$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4c^{2}-15c=\left(4c-15\right)c$$

Evaluate

$$c\left(4c-15\right)$$