Solve 15a^2+12a

$$15 a ^ { 2 } + 12 a$$

Factor

$$3a\left(5a+4\right)$$

Solution Steps

Factor out 3.

$$3\left(5a^{2}+4a\right)$$

Consider 5a^{2}+4a. Factor out a.

$$a\left(5a+4\right)$$

Rewrite the complete factored expression.

$$3a\left(5a+4\right)$$

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.

$$15a^{2}+12a=0$$

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.

$$a=\frac{-12±\sqrt{12^{2}}}{2\times 15}$$

Take the square root of 12^{2}.

$$a=\frac{-12±12}{2\times 15}$$

Multiply 2 times 15.

$$a=\frac{-12±12}{30}$$

Now solve the equation a=\frac{-12±12}{30} when ± is plus. Add -12 to 12.

$$a=\frac{0}{30}$$

Divide 0 by 30.

$$a=0$$

Now solve the equation a=\frac{-12±12}{30} when ± is minus. Subtract 12 from -12.

$$a=-\frac{24}{30}$$

Reduce the fraction \frac{-24}{30} to lowest terms by extracting and canceling out 6.

$$a=-\frac{4}{5}$$

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 -\frac{4}{5} for x_{2}.

$$15a^{2}+12a=15a\left(a-\left(-\frac{4}{5}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$15a^{2}+12a=15a\left(a+\frac{4}{5}\right)$$

Add \frac{4}{5} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15a^{2}+12a=15a\times \frac{5a+4}{5}$$

Cancel out 5, the greatest common factor in 15 and 5.

$$15a^{2}+12a=3a\left(5a+4\right)$$

Evaluate

$$3a\left(5a+4\right)$$