Solve 6x^2+4x+6

$$6 x ^ { 2 } + 4 x + 6$$

Factor

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

Solution Steps

Factor out 2. Polynomial 3x^{2}+2x+3 is not factored since it does not have any rational roots.

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

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.

$$6x^{2}+4x+6=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\times 6\times 6}}{2\times 6}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 6\times 6}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-4±\sqrt{16-24\times 6}}{2\times 6}$$

Multiply -24 times 6.

$$x=\frac{-4±\sqrt{16-144}}{2\times 6}$$

Add 16 to -144.

$$x=\frac{-4±\sqrt{-128}}{2\times 6}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$6x^{2}+4x+6$$

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

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

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

$$r + s = -\frac{2}{3} rs = 1$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = 1

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

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

$$\frac{1}{9} - u^2 = 1$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = 1-\frac{1}{9} = \frac{8}{9}$$

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

$$u^2 = -\frac{8}{9} u = \pm\sqrt{-\frac{8}{9}} = \pm \frac{\sqrt{8}}{3}i$$

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

$$r =-\frac{1}{3} - \frac{\sqrt{8}}{3}i = -0.333 - 0.943i s = -\frac{1}{3} + \frac{\sqrt{8}}{3}i = -0.333 + 0.943i$$

Evaluate

$$6x^{2}+4x+6$$