Solve 2x^2-8x+4

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

Factor

$$2\left(x-\left(2-\sqrt{2}\right)\right)\left(x-\left(\sqrt{2}+2\right)\right)$$

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.

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

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.

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

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\times 4}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$x=\frac{-\left(-8\right)±\sqrt{64-32}}{2\times 2}$$

Add 64 to -32.

$$x=\frac{-\left(-8\right)±\sqrt{32}}{2\times 2}$$

Take the square root of 32.

$$x=\frac{-\left(-8\right)±4\sqrt{2}}{2\times 2}$$

The opposite of -8 is 8.

$$x=\frac{8±4\sqrt{2}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±4\sqrt{2}}{4}$$

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

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

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

$$x=\sqrt{2}+2$$

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

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

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

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

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

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

Evaluate

$$2\left(x^{2}-4x+2\right)$$