Solve 4x^2+12x+6

$$4 x ^ { 2 } + 12 x + 6$$

Factor

$$4\left(x-\frac{-\sqrt{3}-3}{2}\right)\left(x-\frac{\sqrt{3}-3}{2}\right)$$

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.

$$4x^{2}+12x+6=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\times 4\times 6}}{2\times 4}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 4\times 6}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-12±\sqrt{144-16\times 6}}{2\times 4}$$

Multiply -16 times 6.

$$x=\frac{-12±\sqrt{144-96}}{2\times 4}$$

Add 144 to -96.

$$x=\frac{-12±\sqrt{48}}{2\times 4}$$

Take the square root of 48.

$$x=\frac{-12±4\sqrt{3}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-12±4\sqrt{3}}{8}$$

Now solve the equation x=\frac{-12±4\sqrt{3}}{8} when ± is plus. Add -12 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}-12}{8}$$

Divide -12+4\sqrt{3} by 8.

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

Now solve the equation x=\frac{-12±4\sqrt{3}}{8} when ± is minus. Subtract 4\sqrt{3} from -12.

$$x=\frac{-4\sqrt{3}-12}{8}$$

Divide -12-4\sqrt{3} by 8.

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

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

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

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 4

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

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

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

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

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

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{2}

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = \frac{3}{2}$$

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

$$\frac{9}{4} - u^2 = \frac{3}{2}$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = \frac{3}{2}-\frac{9}{4} = -\frac{3}{4}$$

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

$$u^2 = \frac{3}{4} u = \pm\sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$$

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

$$r =-\frac{3}{2} - \frac{\sqrt{3}}{2} = -2.366 s = -\frac{3}{2} + \frac{\sqrt{3}}{2} = -0.634$$

Evaluate

$$4x^{2}+12x+6$$