Solve 2n^2+4n+1

$$2 n ^ { 2 } + 4 n + 1$$

Factor

$$2\left(n-\left(-\frac{\sqrt{2}}{2}-1\right)\right)\left(n-\left(\frac{\sqrt{2}}{2}-1\right)\right)$$

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.

$$2n^{2}+4n+1=0$$

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.

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

Square 4.

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

Multiply -4 times 2.

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

Add 16 to -8.

$$n=\frac{-4±\sqrt{8}}{2\times 2}$$

Take the square root of 8.

$$n=\frac{-4±2\sqrt{2}}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{-4±2\sqrt{2}}{4}$$

Now solve the equation n=\frac{-4±2\sqrt{2}}{4} when ± is plus. Add -4 to 2\sqrt{2}.

$$n=\frac{2\sqrt{2}-4}{4}$$

Divide -4+2\sqrt{2} by 4.

$$n=\frac{\sqrt{2}}{2}-1$$

Now solve the equation n=\frac{-4±2\sqrt{2}}{4} when ± is minus. Subtract 2\sqrt{2} from -4.

$$n=\frac{-2\sqrt{2}-4}{4}$$

Divide -4-2\sqrt{2} by 4.

$$n=-\frac{\sqrt{2}}{2}-1$$

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

$$2n^{2}+4n+1=2\left(n-\left(\frac{\sqrt{2}}{2}-1\right)\right)\left(n-\left(-\frac{\sqrt{2}}{2}-1\right)\right)$$

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 2

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

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

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

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{1}{2}

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

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

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

Simplify the expression by subtracting 1 on both sides

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

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

$$u^2 = \frac{1}{2} u = \pm\sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}$$

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

$$r =-1 - \frac{1}{\sqrt{2}} = -1.707 s = -1 + \frac{1}{\sqrt{2}} = -0.293$$

Evaluate

$$2n^{2}+4n+1$$