Solve 6x^2+34x-12

$$6 x ^ { 2 } + 34 x - 12$$

Factor

$$2\left(3x-1\right)\left(x+6\right)$$

Solution Steps

Factor out 2.

$$2\left(3x^{2}+17x-6\right)$$

Consider 3x^{2}+17x-6. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=17 ab=3\left(-6\right)=-18$$

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 -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 17.

$$a=-1 b=18$$

Rewrite 3x^{2}+17x-6 as \left(3x^{2}-x\right)+\left(18x-6\right).

$$\left(3x^{2}-x\right)+\left(18x-6\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(3x-1\right)+6\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(x+6\right)$$

Rewrite the complete factored expression.

$$2\left(3x-1\right)\left(x+6\right)$$

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.

$$6x^{2}+34x-12=0$$

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.

$$x=\frac{-34±\sqrt{34^{2}-4\times 6\left(-12\right)}}{2\times 6}$$

Square 34.

$$x=\frac{-34±\sqrt{1156-4\times 6\left(-12\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-34±\sqrt{1156-24\left(-12\right)}}{2\times 6}$$

Multiply -24 times -12.

$$x=\frac{-34±\sqrt{1156+288}}{2\times 6}$$

Add 1156 to 288.

$$x=\frac{-34±\sqrt{1444}}{2\times 6}$$

Take the square root of 1444.

$$x=\frac{-34±38}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-34±38}{12}$$

Now solve the equation x=\frac{-34±38}{12} when ± is plus. Add -34 to 38.

$$x=\frac{4}{12}$$

Reduce the fraction \frac{4}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{-34±38}{12} when ± is minus. Subtract 38 from -34.

$$x=-\frac{72}{12}$$

Divide -72 by 12.

$$x=-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{3} for x_{1} and -6 for x_{2}.

$$6x^{2}+34x-12=6\left(x-\frac{1}{3}\right)\left(x-\left(-6\right)\right)$$

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

$$6x^{2}+34x-12=6\left(x-\frac{1}{3}\right)\left(x+6\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+34x-12=6\times \frac{3x-1}{3}\left(x+6\right)$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6x^{2}+34x-12=2\left(3x-1\right)\left(x+6\right)$$

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 6

$$x ^ 2 +\frac{17}{3}x -2 = 0$$

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

$$r + s = -\frac{17}{3} rs = -2$$

Two numbers r and s sum up to -\frac{17}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{17}{3} = -\frac{17}{6}. 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>

$$r = -\frac{17}{6} - u s = -\frac{17}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -2

$$(-\frac{17}{6} - u) (-\frac{17}{6} + u) = -2$$

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

$$\frac{289}{36} - u^2 = -2$$

Simplify the expression by subtracting \frac{289}{36} on both sides

$$-u^2 = -2-\frac{289}{36} = -\frac{361}{36}$$

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

$$u^2 = \frac{361}{36} u = \pm\sqrt{\frac{361}{36}} = \pm \frac{19}{6}$$

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

$$r =-\frac{17}{6} - \frac{19}{6} = -6 s = -\frac{17}{6} + \frac{19}{6} = 0.333$$

Evaluate

$$2\left(3x-1\right)\left(x+6\right)$$