Solve 18x^2+9x

$$18 x ^ { 2 } + 9 x$$

Factor

$$9x\left(2x+1\right)$$

Solution Steps

Factor out 9.

$$9\left(2x^{2}+x\right)$$

Consider 2x^{2}+x. Factor out x.

$$x\left(2x+1\right)$$

Rewrite the complete factored expression.

$$9x\left(2x+1\right)$$

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.

$$18x^{2}+9x=0$$

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.

$$x=\frac{-9±\sqrt{9^{2}}}{2\times 18}$$

Take the square root of 9^{2}.

$$x=\frac{-9±9}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-9±9}{36}$$

Now solve the equation x=\frac{-9±9}{36} when ± is plus. Add -9 to 9.

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

Divide 0 by 36.

$$x=0$$

Now solve the equation x=\frac{-9±9}{36} when ± is minus. Subtract 9 from -9.

$$x=-\frac{18}{36}$$

Reduce the fraction \frac{-18}{36} to lowest terms by extracting and canceling out 18.

$$x=-\frac{1}{2}$$

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}{2} for x_{2}.

$$18x^{2}+9x=18x\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$18x^{2}+9x=18x\left(x+\frac{1}{2}\right)$$

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}+9x=18x\times \frac{2x+1}{2}$$

Cancel out 2, the greatest common factor in 18 and 2.

$$18x^{2}+9x=9x\left(2x+1\right)$$

Evaluate

$$9x\left(2x+1\right)$$