Solve 9x^2+6x-8

$$9 { x }^{ 2 } +6x-8$$

Factor

$$\left(3x-2\right)\left(3x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=6 ab=9\left(-8\right)=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 6.

$$a=-6 b=12$$

Rewrite 9x^{2}+6x-8 as \left(9x^{2}-6x\right)+\left(12x-8\right).

$$\left(9x^{2}-6x\right)+\left(12x-8\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(3x-2\right)+4\left(3x-2\right)$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(3x+4\right)$$

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.

$$9x^{2}+6x-8=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9\left(-8\right)}}{2\times 9}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 9\left(-8\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-6±\sqrt{36-36\left(-8\right)}}{2\times 9}$$

Multiply -36 times -8.

$$x=\frac{-6±\sqrt{36+288}}{2\times 9}$$

Add 36 to 288.

$$x=\frac{-6±\sqrt{324}}{2\times 9}$$

Take the square root of 324.

$$x=\frac{-6±18}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-6±18}{18}$$

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

$$x=\frac{12}{18}$$

Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

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

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

$$x=-\frac{24}{18}$$

Reduce the fraction \frac{-24}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{4}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{3} for x_{1} and -\frac{4}{3} for x_{2}.

$$9x^{2}+6x-8=9\left(x-\frac{2}{3}\right)\left(x-\left(-\frac{4}{3}\right)\right)$$

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

$$9x^{2}+6x-8=9\left(x-\frac{2}{3}\right)\left(x+\frac{4}{3}\right)$$

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

$$9x^{2}+6x-8=9\times \frac{3x-2}{3}\left(x+\frac{4}{3}\right)$$

Add \frac{4}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}+6x-8=9\times \frac{3x-2}{3}\times \frac{3x+4}{3}$$

Multiply \frac{3x-2}{3} times \frac{3x+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$9x^{2}+6x-8=9\times \frac{\left(3x-2\right)\left(3x+4\right)}{3\times 3}$$

Multiply 3 times 3.

$$9x^{2}+6x-8=9\times \frac{\left(3x-2\right)\left(3x+4\right)}{9}$$

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

$$9x^{2}+6x-8=\left(3x-2\right)\left(3x+4\right)$$

Evaluate

$$\left(3x-2\right)\left(3x+4\right)$$