Solve 16x^2-24x-27

$$16 { x }^{ 2 } -24x-27$$

Factor

$$\left(4x-9\right)\left(4x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 16x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=-24 ab=16\left(-27\right)=-432$$

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 -432.

$$1,-432 2,-216 3,-144 4,-108 6,-72 8,-54 9,-48 12,-36 16,-27 18,-24$$

Calculate the sum for each pair.

$$1-432=-431 2-216=-214 3-144=-141 4-108=-104 6-72=-66 8-54=-46 9-48=-39 12-36=-24 16-27=-11 18-24=-6$$

The solution is the pair that gives sum -24.

$$a=-36 b=12$$

Rewrite 16x^{2}-24x-27 as \left(16x^{2}-36x\right)+\left(12x-27\right).

$$\left(16x^{2}-36x\right)+\left(12x-27\right)$$

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

$$4x\left(4x-9\right)+3\left(4x-9\right)$$

Factor out common term 4x-9 by using distributive property.

$$\left(4x-9\right)\left(4x+3\right)$$

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.

$$16x^{2}-24x-27=0$$

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.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 16\left(-27\right)}}{2\times 16}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 16\left(-27\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-24\right)±\sqrt{576-64\left(-27\right)}}{2\times 16}$$

Multiply -64 times -27.

$$x=\frac{-\left(-24\right)±\sqrt{576+1728}}{2\times 16}$$

Add 576 to 1728.

$$x=\frac{-\left(-24\right)±\sqrt{2304}}{2\times 16}$$

Take the square root of 2304.

$$x=\frac{-\left(-24\right)±48}{2\times 16}$$

The opposite of -24 is 24.

$$x=\frac{24±48}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{24±48}{32}$$

Now solve the equation x=\frac{24±48}{32} when ± is plus. Add 24 to 48.

$$x=\frac{72}{32}$$

Reduce the fraction \frac{72}{32} to lowest terms by extracting and canceling out 8.

$$x=\frac{9}{4}$$

Now solve the equation x=\frac{24±48}{32} when ± is minus. Subtract 48 from 24.

$$x=-\frac{24}{32}$$

Reduce the fraction \frac{-24}{32} to lowest terms by extracting and canceling out 8.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9}{4} for x_{1} and -\frac{3}{4} for x_{2}.

$$16x^{2}-24x-27=16\left(x-\frac{9}{4}\right)\left(x-\left(-\frac{3}{4}\right)\right)$$

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

$$16x^{2}-24x-27=16\left(x-\frac{9}{4}\right)\left(x+\frac{3}{4}\right)$$

Subtract \frac{9}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$16x^{2}-24x-27=16\times \frac{4x-9}{4}\left(x+\frac{3}{4}\right)$$

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

$$16x^{2}-24x-27=16\times \frac{4x-9}{4}\times \frac{4x+3}{4}$$

Multiply \frac{4x-9}{4} times \frac{4x+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$16x^{2}-24x-27=16\times \frac{\left(4x-9\right)\left(4x+3\right)}{4\times 4}$$

Multiply 4 times 4.

$$16x^{2}-24x-27=16\times \frac{\left(4x-9\right)\left(4x+3\right)}{16}$$

Cancel out 16, the greatest common factor in 16 and 16.

$$16x^{2}-24x-27=\left(4x-9\right)\left(4x+3\right)$$

Evaluate

$$\left(4x-9\right)\left(4x+3\right)$$