Solve 18x^2+6x-4

$$18 { x }^{ 2 } +6x-4$$

Factor

$$2\left(3x-1\right)\left(3x+2\right)$$

Solution Steps

Factor out 2.

$$2\left(9x^{2}+3x-2\right)$$

Consider 9x^{2}+3x-2. Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=3 ab=9\left(-2\right)=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 3.

$$a=-3 b=6$$

Rewrite 9x^{2}+3x-2 as \left(9x^{2}-3x\right)+\left(6x-2\right).

$$\left(9x^{2}-3x\right)+\left(6x-2\right)$$

Factor out 3x in the first and 2 in the second group.

$$3x\left(3x-1\right)+2\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(3x+2\right)$$

Rewrite the complete factored expression.

$$2\left(3x-1\right)\left(3x+2\right)$$

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.

$$18x^{2}+6x-4=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\times 18\left(-4\right)}}{2\times 18}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 18\left(-4\right)}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-6±\sqrt{36-72\left(-4\right)}}{2\times 18}$$

Multiply -72 times -4.

$$x=\frac{-6±\sqrt{36+288}}{2\times 18}$$

Add 36 to 288.

$$x=\frac{-6±\sqrt{324}}{2\times 18}$$

Take the square root of 324.

$$x=\frac{-6±18}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-6±18}{36}$$

Now solve the equation x=\frac{-6±18}{36} when ± is plus. Add -6 to 18.

$$x=\frac{12}{36}$$

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

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{-6±18}{36} when ± is minus. Subtract 18 from -6.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{3} for x_{1} and -\frac{2}{3} for x_{2}.

$$18x^{2}+6x-4=18\left(x-\frac{1}{3}\right)\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$18x^{2}+6x-4=18\left(x-\frac{1}{3}\right)\left(x+\frac{2}{3}\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}+6x-4=18\times \frac{3x-1}{3}\left(x+\frac{2}{3}\right)$$

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

$$18x^{2}+6x-4=18\times \frac{3x-1}{3}\times \frac{3x+2}{3}$$

Multiply \frac{3x-1}{3} times \frac{3x+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18x^{2}+6x-4=18\times \frac{\left(3x-1\right)\left(3x+2\right)}{3\times 3}$$

Multiply 3 times 3.

$$18x^{2}+6x-4=18\times \frac{\left(3x-1\right)\left(3x+2\right)}{9}$$

Cancel out 9, the greatest common factor in 18 and 9.

$$18x^{2}+6x-4=2\left(3x-1\right)\left(3x+2\right)$$

Evaluate

$$18x^{2}+6x-4$$