Solve 27x^2-12x-4

$$27 x ^ { 2 } - 12 x - 4$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 27x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-12 ab=27\left(-4\right)=-108$$

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 -108.

$$1,-108 2,-54 3,-36 4,-27 6,-18 9,-12$$

Calculate the sum for each pair.

$$1-108=-107 2-54=-52 3-36=-33 4-27=-23 6-18=-12 9-12=-3$$

The solution is the pair that gives sum -12.

$$a=-18 b=6$$

Rewrite 27x^{2}-12x-4 as \left(27x^{2}-18x\right)+\left(6x-4\right).

$$\left(27x^{2}-18x\right)+\left(6x-4\right)$$

Factor out 9x in the first and 2 in the second group.

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

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

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

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.

$$27x^{2}-12x-4=0$$

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.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 27\left(-4\right)}}{2\times 27}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 27\left(-4\right)}}{2\times 27}$$

Multiply -4 times 27.

$$x=\frac{-\left(-12\right)±\sqrt{144-108\left(-4\right)}}{2\times 27}$$

Multiply -108 times -4.

$$x=\frac{-\left(-12\right)±\sqrt{144+432}}{2\times 27}$$

Add 144 to 432.

$$x=\frac{-\left(-12\right)±\sqrt{576}}{2\times 27}$$

Take the square root of 576.

$$x=\frac{-\left(-12\right)±24}{2\times 27}$$

The opposite of -12 is 12.

$$x=\frac{12±24}{2\times 27}$$

Multiply 2 times 27.

$$x=\frac{12±24}{54}$$

Now solve the equation x=\frac{12±24}{54} when ± is plus. Add 12 to 24.

$$x=\frac{36}{54}$$

Reduce the fraction \frac{36}{54} to lowest terms by extracting and canceling out 18.

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

Now solve the equation x=\frac{12±24}{54} when ± is minus. Subtract 24 from 12.

$$x=-\frac{12}{54}$$

Reduce the fraction \frac{-12}{54} to lowest terms by extracting and canceling out 6.

$$x=-\frac{2}{9}$$

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{2}{9} for x_{2}.

$$27x^{2}-12x-4=27\left(x-\frac{2}{3}\right)\left(x-\left(-\frac{2}{9}\right)\right)$$

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

$$27x^{2}-12x-4=27\left(x-\frac{2}{3}\right)\left(x+\frac{2}{9}\right)$$

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

$$27x^{2}-12x-4=27\times \frac{3x-2}{3}\left(x+\frac{2}{9}\right)$$

Add \frac{2}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$27x^{2}-12x-4=27\times \frac{3x-2}{3}\times \frac{9x+2}{9}$$

Multiply \frac{3x-2}{3} times \frac{9x+2}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$27x^{2}-12x-4=27\times \frac{\left(3x-2\right)\left(9x+2\right)}{3\times 9}$$

Multiply 3 times 9.

$$27x^{2}-12x-4=27\times \frac{\left(3x-2\right)\left(9x+2\right)}{27}$$

Cancel out 27, the greatest common factor in 27 and 27.

$$27x^{2}-12x-4=\left(3x-2\right)\left(9x+2\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 27

$$x ^ 2 -\frac{4}{9}x -\frac{4}{27} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{4}{9} rs = -\frac{4}{27}$$

Two numbers r and s sum up to \frac{4}{9} exactly when the average of the two numbers is \frac{1}{2}*\frac{4}{9} = \frac{2}{9}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{2}{9} - u s = \frac{2}{9} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{4}{27}

$$(\frac{2}{9} - u) (\frac{2}{9} + u) = -\frac{4}{27}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{4}{81} - u^2 = -\frac{4}{27}$$

Simplify the expression by subtracting \frac{4}{81} on both sides

$$-u^2 = -\frac{4}{27}-\frac{4}{81} = -\frac{16}{81}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{16}{81} u = \pm\sqrt{\frac{16}{81}} = \pm \frac{4}{9}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{2}{9} - \frac{4}{9} = -0.222 s = \frac{2}{9} + \frac{4}{9} = 0.667$$

Evaluate

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