Solve 9c^2-18c

$$9 c ^ { 2 } - 18 c$$

Factor

$$9c\left(c-2\right)$$

Solution Steps

Factor out 9.

$$9\left(c^{2}-2c\right)$$

Consider c^{2}-2c. Factor out c.

$$c\left(c-2\right)$$

Rewrite the complete factored expression.

$$9c\left(c-2\right)$$

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.

$$9c^{2}-18c=0$$

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.

$$c=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}}}{2\times 9}$$

Take the square root of \left(-18\right)^{2}.

$$c=\frac{-\left(-18\right)±18}{2\times 9}$$

The opposite of -18 is 18.

$$c=\frac{18±18}{2\times 9}$$

Multiply 2 times 9.

$$c=\frac{18±18}{18}$$

Now solve the equation c=\frac{18±18}{18} when ± is plus. Add 18 to 18.

$$c=\frac{36}{18}$$

Divide 36 by 18.

$$c=2$$

Now solve the equation c=\frac{18±18}{18} when ± is minus. Subtract 18 from 18.

$$c=\frac{0}{18}$$

Divide 0 by 18.

$$c=0$$

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

$$9c^{2}-18c=9\left(c-2\right)c$$

Evaluate

$$9c\left(c-2\right)$$