Solve 18q^2+6q

$$18 q ^ { 2 } + 6 q$$

Factor

$$6q\left(3q+1\right)$$

Solution Steps

Factor out 6.

$$6\left(3q^{2}+q\right)$$

Consider 3q^{2}+q. Factor out q.

$$q\left(3q+1\right)$$

Rewrite the complete factored expression.

$$6q\left(3q+1\right)$$

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.

$$18q^{2}+6q=0$$

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.

$$q=\frac{-6±\sqrt{6^{2}}}{2\times 18}$$

Take the square root of 6^{2}.

$$q=\frac{-6±6}{2\times 18}$$

Multiply 2 times 18.

$$q=\frac{-6±6}{36}$$

Now solve the equation q=\frac{-6±6}{36} when ± is plus. Add -6 to 6.

$$q=\frac{0}{36}$$

Divide 0 by 36.

$$q=0$$

Now solve the equation q=\frac{-6±6}{36} when ± is minus. Subtract 6 from -6.

$$q=-\frac{12}{36}$$

Reduce the fraction \frac{-12}{36} to lowest terms by extracting and canceling out 12.

$$q=-\frac{1}{3}$$

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{1}{3} for x_{2}.

$$18q^{2}+6q=18q\left(q-\left(-\frac{1}{3}\right)\right)$$

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

$$18q^{2}+6q=18q\left(q+\frac{1}{3}\right)$$

Add \frac{1}{3} to q by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$18q^{2}+6q=18q\times \frac{3q+1}{3}$$

Cancel out 3, the greatest common factor in 18 and 3.

$$18q^{2}+6q=6q\left(3q+1\right)$$

Evaluate

$$6q\left(3q+1\right)$$