Solve 4x^2-9x-9

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

Factor

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

Solution Steps

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

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

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. 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.

$$a=-12 b=3$$

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

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

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

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

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

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

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.

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

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.

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

Square -9.

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

Multiply -4 times 4.

$$x=\frac{-\left(-9\right)±\sqrt{81-16\left(-9\right)}}{2\times 4}$$

Multiply -16 times -9.

$$x=\frac{-\left(-9\right)±\sqrt{81+144}}{2\times 4}$$

Add 81 to 144.

$$x=\frac{-\left(-9\right)±\sqrt{225}}{2\times 4}$$

Take the square root of 225.

$$x=\frac{-\left(-9\right)±15}{2\times 4}$$

The opposite of -9 is 9.

$$x=\frac{9±15}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{9±15}{8}$$

Now solve the equation x=\frac{9±15}{8} when ± is plus. Add 9 to 15.

$$x=\frac{24}{8}$$

Divide 24 by 8.

$$x=3$$

Now solve the equation x=\frac{9±15}{8} when ± is minus. Subtract 15 from 9.

$$x=-\frac{6}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

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

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

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

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

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

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

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

Cancel out 4, the greatest common factor in 4 and 4.

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

Evaluate

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