Solve 9x^2+9x-4

$$9 { x }^{ 2 } +9x-4$$

Factor

$$\left(3x-1\right)\left(3x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=9 ab=9\left(-4\right)=-36$$

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 -36.

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

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 9.

$$a=-3 b=12$$

Rewrite 9x^{2}+9x-4 as \left(9x^{2}-3x\right)+\left(12x-4\right).

$$\left(9x^{2}-3x\right)+\left(12x-4\right)$$

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

$$3x\left(3x-1\right)+4\left(3x-1\right)$$

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

$$\left(3x-1\right)\left(3x+4\right)$$

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.

$$9x^{2}+9x-4=0$$

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.

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

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 9\left(-4\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-9±\sqrt{81-36\left(-4\right)}}{2\times 9}$$

Multiply -36 times -4.

$$x=\frac{-9±\sqrt{81+144}}{2\times 9}$$

Add 81 to 144.

$$x=\frac{-9±\sqrt{225}}{2\times 9}$$

Take the square root of 225.

$$x=\frac{-9±15}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-9±15}{18}$$

Now solve the equation x=\frac{-9±15}{18} when ± is plus. Add -9 to 15.

$$x=\frac{6}{18}$$

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

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

Now solve the equation x=\frac{-9±15}{18} when ± is minus. Subtract 15 from -9.

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

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{1}{3} for x_{1} and -\frac{4}{3} for x_{2}.

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

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

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

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

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

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

$$9x^{2}+9x-4=9\times \frac{3x-1}{3}\times \frac{3x+4}{3}$$

Multiply \frac{3x-1}{3} times \frac{3x+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

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

Multiply 3 times 3.

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

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

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

Evaluate

$$\left(3x-1\right)\left(3x+4\right)$$