Solve 12x^2+18x

$$12 x ^ { 2 } + 18 x$$

Factor

$$6x\left(2x+3\right)$$

Solution Steps

Factor out 6.

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

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

$$x\left(2x+3\right)$$

Rewrite the complete factored expression.

$$6x\left(2x+3\right)$$

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.

$$12x^{2}+18x=0$$

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.

$$x=\frac{-18±\sqrt{18^{2}}}{2\times 12}$$

Take the square root of 18^{2}.

$$x=\frac{-18±18}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-18±18}{24}$$

Now solve the equation x=\frac{-18±18}{24} when ± is plus. Add -18 to 18.

$$x=\frac{0}{24}$$

Divide 0 by 24.

$$x=0$$

Now solve the equation x=\frac{-18±18}{24} when ± is minus. Subtract 18 from -18.

$$x=-\frac{36}{24}$$

Reduce the fraction \frac{-36}{24} to lowest terms by extracting and canceling out 12.

$$x=-\frac{3}{2}$$

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

$$12x^{2}+18x=12x\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$12x^{2}+18x=12x\left(x+\frac{3}{2}\right)$$

Add \frac{3}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+18x=12x\times \frac{2x+3}{2}$$

Cancel out 2, the greatest common factor in 12 and 2.

$$12x^{2}+18x=6x\left(2x+3\right)$$

Evaluate

$$6x\left(2x+3\right)$$