Solve 10q^2+12q

$$10 q ^ { 2 } + 12 q$$

Factor

$$2q\left(5q+6\right)$$

Solution Steps

Factor out 2.

$$2\left(5q^{2}+6q\right)$$

Consider 5q^{2}+6q. Factor out q.

$$q\left(5q+6\right)$$

Rewrite the complete factored expression.

$$2q\left(5q+6\right)$$

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.

$$10q^{2}+12q=0$$

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.

$$q=\frac{-12±\sqrt{12^{2}}}{2\times 10}$$

Take the square root of 12^{2}.

$$q=\frac{-12±12}{2\times 10}$$

Multiply 2 times 10.

$$q=\frac{-12±12}{20}$$

Now solve the equation q=\frac{-12±12}{20} when ± is plus. Add -12 to 12.

$$q=\frac{0}{20}$$

Divide 0 by 20.

$$q=0$$

Now solve the equation q=\frac{-12±12}{20} when ± is minus. Subtract 12 from -12.

$$q=-\frac{24}{20}$$

Reduce the fraction \frac{-24}{20} to lowest terms by extracting and canceling out 4.

$$q=-\frac{6}{5}$$

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{6}{5} for x_{2}.

$$10q^{2}+12q=10q\left(q-\left(-\frac{6}{5}\right)\right)$$

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

$$10q^{2}+12q=10q\left(q+\frac{6}{5}\right)$$

Add \frac{6}{5} to q by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10q^{2}+12q=10q\times \frac{5q+6}{5}$$

Cancel out 5, the greatest common factor in 10 and 5.

$$10q^{2}+12q=2q\left(5q+6\right)$$

Evaluate

$$2q\left(5q+6\right)$$