Solve 15c-9c^2

$$15c-9 { c }^{ 2 }$$

Factor

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

Solution Steps

Factor out 3.

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

Consider 5c-3c^{2}. Factor out c.

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

Rewrite the complete factored expression.

$$3c\left(-3c+5\right)$$

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.

$$-9c^{2}+15c=0$$

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.

$$c=\frac{-15±\sqrt{15^{2}}}{2\left(-9\right)}$$

Take the square root of 15^{2}.

$$c=\frac{-15±15}{2\left(-9\right)}$$

Multiply 2 times -9.

$$c=\frac{-15±15}{-18}$$

Now solve the equation c=\frac{-15±15}{-18} when ± is plus. Add -15 to 15.

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

Divide 0 by -18.

$$c=0$$

Now solve the equation c=\frac{-15±15}{-18} when ± is minus. Subtract 15 from -15.

$$c=-\frac{30}{-18}$$

Reduce the fraction \frac{-30}{-18} to lowest terms by extracting and canceling out 6.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and \frac{5}{3} for x_{2}.

$$-9c^{2}+15c=-9c\left(c-\frac{5}{3}\right)$$

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

$$-9c^{2}+15c=-9c\times \frac{-3c+5}{-3}$$

Cancel out 3, the greatest common factor in -9 and -3.

$$-9c^{2}+15c=3c\left(-3c+5\right)$$

Evaluate

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