Solve 12x^2+x-6

$$12 x ^ { 2 } + x - 6$$

Factor

$$\left(3x-2\right)\left(4x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=1 ab=12\left(-6\right)=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 1.

$$a=-8 b=9$$

Rewrite 12x^{2}+x-6 as \left(12x^{2}-8x\right)+\left(9x-6\right).

$$\left(12x^{2}-8x\right)+\left(9x-6\right)$$

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

$$4x\left(3x-2\right)+3\left(3x-2\right)$$

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

$$\left(3x-2\right)\left(4x+3\right)$$

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.

$$12x^{2}+x-6=0$$

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.

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

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 12\left(-6\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-1±\sqrt{1-48\left(-6\right)}}{2\times 12}$$

Multiply -48 times -6.

$$x=\frac{-1±\sqrt{1+288}}{2\times 12}$$

Add 1 to 288.

$$x=\frac{-1±\sqrt{289}}{2\times 12}$$

Take the square root of 289.

$$x=\frac{-1±17}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-1±17}{24}$$

Now solve the equation x=\frac{-1±17}{24} when ± is plus. Add -1 to 17.

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

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

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

Now solve the equation x=\frac{-1±17}{24} when ± is minus. Subtract 17 from -1.

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

Reduce the fraction \frac{-18}{24} to lowest terms by extracting and canceling out 6.

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

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

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

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

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

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

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

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

$$12x^{2}+x-6=12\times \frac{3x-2}{3}\times \frac{4x+3}{4}$$

Multiply \frac{3x-2}{3} times \frac{4x+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

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

Multiply 3 times 4.

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

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

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

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

$$x ^ 2 +\frac{1}{12}x -\frac{1}{2} = 0$$

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

$$r + s = -\frac{1}{12} rs = -\frac{1}{2}$$

Two numbers r and s sum up to -\frac{1}{12} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{12} = -\frac{1}{24}. 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>

$$r = -\frac{1}{24} - u s = -\frac{1}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{2}

$$(-\frac{1}{24} - u) (-\frac{1}{24} + u) = -\frac{1}{2}$$

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

$$\frac{1}{576} - u^2 = -\frac{1}{2}$$

Simplify the expression by subtracting \frac{1}{576} on both sides

$$-u^2 = -\frac{1}{2}-\frac{1}{576} = -\frac{289}{576}$$

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

$$u^2 = \frac{289}{576} u = \pm\sqrt{\frac{289}{576}} = \pm \frac{17}{24}$$

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

$$r =-\frac{1}{24} - \frac{17}{24} = -0.750 s = -\frac{1}{24} + \frac{17}{24} = 0.667$$

Evaluate

$$12x^{2}+x-6$$