Solve 8x^2+16x+3

$$8 x ^ { 2 } + 16 x + 3$$

Factor

$$8\left(x-\left(-\frac{\sqrt{10}}{4}-1\right)\right)\left(x-\left(\frac{\sqrt{10}}{4}-1\right)\right)$$

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.

$$8x^{2}+16x+3=0$$

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.

$$x=\frac{-16±\sqrt{16^{2}-4\times 8\times 3}}{2\times 8}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 8\times 3}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-16±\sqrt{256-32\times 3}}{2\times 8}$$

Multiply -32 times 3.

$$x=\frac{-16±\sqrt{256-96}}{2\times 8}$$

Add 256 to -96.

$$x=\frac{-16±\sqrt{160}}{2\times 8}$$

Take the square root of 160.

$$x=\frac{-16±4\sqrt{10}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-16±4\sqrt{10}}{16}$$

Now solve the equation x=\frac{-16±4\sqrt{10}}{16} when ± is plus. Add -16 to 4\sqrt{10}.

$$x=\frac{4\sqrt{10}-16}{16}$$

Divide -16+4\sqrt{10} by 16.

$$x=\frac{\sqrt{10}}{4}-1$$

Now solve the equation x=\frac{-16±4\sqrt{10}}{16} when ± is minus. Subtract 4\sqrt{10} from -16.

$$x=\frac{-4\sqrt{10}-16}{16}$$

Divide -16-4\sqrt{10} by 16.

$$x=-\frac{\sqrt{10}}{4}-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1+\frac{\sqrt{10}}{4} for x_{1} and -1-\frac{\sqrt{10}}{4} for x_{2}.

$$8x^{2}+16x+3=8\left(x-\left(\frac{\sqrt{10}}{4}-1\right)\right)\left(x-\left(-\frac{\sqrt{10}}{4}-1\right)\right)$$

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 8

$$x ^ 2 +2x +\frac{3}{8} = 0$$

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

$$r + s = -2 rs = \frac{3}{8}$$

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

$$r = -1 - u s = -1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{8}

$$(-1 - u) (-1 + u) = \frac{3}{8}$$

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

$$1 - u^2 = \frac{3}{8}$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = \frac{3}{8}-1 = -\frac{5}{8}$$

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

$$u^2 = \frac{5}{8} u = \pm\sqrt{\frac{5}{8}} = \pm \frac{\sqrt{5}}{\sqrt{8}}$$

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

$$r =-1 - \frac{\sqrt{5}}{\sqrt{8}} = -1.791 s = -1 + \frac{\sqrt{5}}{\sqrt{8}} = -0.209$$

Evaluate

$$8x^{2}+16x+3$$