Solve 12x^2+2x-4

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

Factor

$$2\left(2x-1\right)\left(3x+2\right)$$

Solution Steps

Factor out 2.

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

Consider 6x^{2}+x-2. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=1 ab=6\left(-2\right)=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite 6x^{2}+x-2 as \left(6x^{2}-3x\right)+\left(4x-2\right).

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

Factor out 3x in the first and 2 in the second group.

$$3x\left(2x-1\right)+2\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(3x+2\right)$$

Rewrite the complete factored expression.

$$2\left(2x-1\right)\left(3x+2\right)$$

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.

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

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.

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

Square 2.

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

Multiply -4 times 12.

$$x=\frac{-2±\sqrt{4-48\left(-4\right)}}{2\times 12}$$

Multiply -48 times -4.

$$x=\frac{-2±\sqrt{4+192}}{2\times 12}$$

Add 4 to 192.

$$x=\frac{-2±\sqrt{196}}{2\times 12}$$

Take the square root of 196.

$$x=\frac{-2±14}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-2±14}{24}$$

Now solve the equation x=\frac{-2±14}{24} when ± is plus. Add -2 to 14.

$$x=\frac{12}{24}$$

Reduce the fraction \frac{12}{24} to lowest terms by extracting and canceling out 12.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-2±14}{24} when ± is minus. Subtract 14 from -2.

$$x=-\frac{16}{24}$$

Reduce the fraction \frac{-16}{24} to lowest terms by extracting and canceling out 8.

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

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

$$12x^{2}+2x-4=12\left(x-\frac{1}{2}\right)\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$12x^{2}+2x-4=12\left(x-\frac{1}{2}\right)\left(x+\frac{2}{3}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+2x-4=12\times \frac{2x-1}{2}\left(x+\frac{2}{3}\right)$$

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

$$12x^{2}+2x-4=12\times \frac{2x-1}{2}\times \frac{3x+2}{3}$$

Multiply \frac{2x-1}{2} times \frac{3x+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+2x-4=12\times \frac{\left(2x-1\right)\left(3x+2\right)}{2\times 3}$$

Multiply 2 times 3.

$$12x^{2}+2x-4=12\times \frac{\left(2x-1\right)\left(3x+2\right)}{6}$$

Cancel out 6, the greatest common factor in 12 and 6.

$$12x^{2}+2x-4=2\left(2x-1\right)\left(3x+2\right)$$

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 12

$$x ^ 2 +\frac{1}{6}x -\frac{1}{3} = 0$$

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

$$r + s = -\frac{1}{6} rs = -\frac{1}{3}$$

Two numbers r and s sum up to -\frac{1}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{6} = -\frac{1}{12}. 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>

$$r = -\frac{1}{12} - u s = -\frac{1}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{3}

$$(-\frac{1}{12} - u) (-\frac{1}{12} + u) = -\frac{1}{3}$$

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

$$\frac{1}{144} - u^2 = -\frac{1}{3}$$

Simplify the expression by subtracting \frac{1}{144} on both sides

$$-u^2 = -\frac{1}{3}-\frac{1}{144} = -\frac{49}{144}$$

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

$$u^2 = \frac{49}{144} u = \pm\sqrt{\frac{49}{144}} = \pm \frac{7}{12}$$

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

$$r =-\frac{1}{12} - \frac{7}{12} = -0.667 s = -\frac{1}{12} + \frac{7}{12} = 0.500$$

Evaluate

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