# How do you factor completely: 2x^2 + 4x − 2?

Jul 29, 2015

Either: $2 \left({x}^{2} + 2 x - 1\right)$

or 2(x+1-sqrt2))(x+1-sqrt2)).

#### Explanation:

The phrase "factor completely" is not unambiguous as one might think.

When we factor poynomials, we usually have some perticula set of numbers we are willing to use as coefficients. In early aalgebra classes, we usually restrict ourselves to integer coefficinents.

Using Integers:

2x^2 + 4x − 2 = 2(x^2+2x-1) which cannot be factored further.

(The only possibility to get ${x}^{2}$ first ans $- 1$ last might be $\left(x + 1\right) \left(x - 1\right)$. But the product of these is not ${x}^{2} + 2 x - 1$)

Using Rational Numbers

We can get the factors for the plolynomial ${x}^{2} + 2 x - 1$

by solving the equation ${x}^{2} + 2 x - 1 = 0$.

Use either completing the square or the quadratic formula to get:

$x = - 1 + \sqrt{2} \text{ }$ or $\text{ } x = - 1 - \sqrt{2}$

So the factors are:

$\left(x - \left(- 1 + \sqrt{2}\right)\right) \text{ }$ and $\text{ } \left(x - \left(- 1 - \sqrt{2}\right)\right)$

We get:

2x^2 + 4x − 2 = 2(x^2+2x-1)

 = 2(x+1-sqrt2))(x+1-sqrt2)).