Solve 2+8x+4x^2

$$2 + 8 x + 4 x ^ { 2 }$$

Factor

$$4\left(x-\left(-\frac{\sqrt{2}}{2}-1\right)\right)\left(x-\left(\frac{\sqrt{2}}{2}-1\right)\right)$$

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.

$$4x^{2}+8x+2=0$$

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.

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

Square 8.

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

Multiply -4 times 4.

$$x=\frac{-8±\sqrt{64-16\times 2}}{2\times 4}$$

Multiply -16 times 2.

$$x=\frac{-8±\sqrt{64-32}}{2\times 4}$$

Add 64 to -32.

$$x=\frac{-8±\sqrt{32}}{2\times 4}$$

Take the square root of 32.

$$x=\frac{-8±4\sqrt{2}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-8±4\sqrt{2}}{8}$$

Now solve the equation x=\frac{-8±4\sqrt{2}}{8} when ± is plus. Add -8 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}-8}{8}$$

Divide -8+4\sqrt{2} by 8.

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

Now solve the equation x=\frac{-8±4\sqrt{2}}{8} when ± is minus. Subtract 4\sqrt{2} from -8.

$$x=\frac{-4\sqrt{2}-8}{8}$$

Divide -8-4\sqrt{2} by 8.

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

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}.

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

Evaluate

$$4x^{2}+8x+2$$