Solve 15a-3a^2

$$15 a - 3 a ^ { 2 }$$

Factor

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

Solution Steps

Factor out 3.

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

Consider 5a-a^{2}. Factor out a.

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

Rewrite the complete factored expression.

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

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.

$$-3a^{2}+15a=0$$

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.

$$a=\frac{-15±\sqrt{15^{2}}}{2\left(-3\right)}$$

Take the square root of 15^{2}.

$$a=\frac{-15±15}{2\left(-3\right)}$$

Multiply 2 times -3.

$$a=\frac{-15±15}{-6}$$

Now solve the equation a=\frac{-15±15}{-6} when ± is plus. Add -15 to 15.

$$a=\frac{0}{-6}$$

Divide 0 by -6.

$$a=0$$

Now solve the equation a=\frac{-15±15}{-6} when ± is minus. Subtract 15 from -15.

$$a=-\frac{30}{-6}$$

Divide -30 by -6.

$$a=5$$

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 5 for x_{2}.

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

Evaluate

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