Solve 6x^2+8x+1

$$6 x ^ { 2 } + 8 x + 1$$

Factor

$$6\left(x-\left(-\frac{\sqrt{10}}{6}-\frac{2}{3}\right)\right)\left(x-\left(\frac{\sqrt{10}}{6}-\frac{2}{3}\right)\right)$$

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.

$$6x^{2}+8x+1=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\times 6}}{2\times 6}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 6}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-8±\sqrt{64-24}}{2\times 6}$$

Add 64 to -24.

$$x=\frac{-8±\sqrt{40}}{2\times 6}$$

Take the square root of 40.

$$x=\frac{-8±2\sqrt{10}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-8±2\sqrt{10}}{12}$$

Now solve the equation x=\frac{-8±2\sqrt{10}}{12} when ± is plus. Add -8 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}-8}{12}$$

Divide -8+2\sqrt{10} by 12.

$$x=\frac{\sqrt{10}}{6}-\frac{2}{3}$$

Now solve the equation x=\frac{-8±2\sqrt{10}}{12} when ± is minus. Subtract 2\sqrt{10} from -8.

$$x=\frac{-2\sqrt{10}-8}{12}$$

Divide -8-2\sqrt{10} by 12.

$$x=-\frac{\sqrt{10}}{6}-\frac{2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{2}{3}+\frac{\sqrt{10}}{6} for x_{1} and -\frac{2}{3}-\frac{\sqrt{10}}{6} for x_{2}.

$$6x^{2}+8x+1=6\left(x-\left(\frac{\sqrt{10}}{6}-\frac{2}{3}\right)\right)\left(x-\left(-\frac{\sqrt{10}}{6}-\frac{2}{3}\right)\right)$$

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

$$x ^ 2 +\frac{4}{3}x +\frac{1}{6} = 0$$

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

$$r + s = -\frac{4}{3} rs = \frac{1}{6}$$

Two numbers r and s sum up to -\frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{4}{3} = -\frac{2}{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>

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

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{6}

$$(-\frac{2}{3} - u) (-\frac{2}{3} + u) = \frac{1}{6}$$

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

$$\frac{4}{9} - u^2 = \frac{1}{6}$$

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

$$-u^2 = \frac{1}{6}-\frac{4}{9} = -\frac{5}{18}$$

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

$$u^2 = \frac{5}{18} u = \pm\sqrt{\frac{5}{18}} = \pm \frac{\sqrt{5}}{\sqrt{18}}$$

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

$$r =-\frac{2}{3} - \frac{\sqrt{5}}{\sqrt{18}} = -1.194 s = -\frac{2}{3} + \frac{\sqrt{5}}{\sqrt{18}} = -0.140$$

Evaluate

$$6x^{2}+8x+1$$