Expand and simplify [2x - (root 2)x - 1][2x + (root 2)x - 1].

Step-by-step solution

1. 1

   Collect the x terms inside each bracket. Both $2x$ and $\sqrt2\,x$ are multiples of $x$, so factor $x$ out of each pair:

   $\left[(2 - \sqrt2)x - 1\right]\left[(2 + \sqrt2)x - 1\right]$

   Written this way the two brackets are visibly the same shape, with conjugate coefficients $2 - \sqrt2$ and $2 + \sqrt2$.

2. 2

   Expand with the distributive law. Multiplying out term by term:

   $= (2-\sqrt2)(2+\sqrt2)x^2 - (2-\sqrt2)x - (2+\sqrt2)x + 1$

   Keep the two middle terms separate for now; they are where the surds will cancel.

3. 3

   Use the conjugate product for the x² coefficient. Because $(a-b)(a+b) = a^2 - b^2$,

   $(2-\sqrt2)(2+\sqrt2) = 2^2 - \left(\sqrt2\right)^2 = 4 - 2 = 2$

   So the quadratic term is $2x^2$ — rational, even though each factor was irrational. That is exactly what conjugates buy you.

4. 4

   Combine the two linear terms. The surds appear with opposite signs and cancel:

   $-(2-\sqrt2)x - (2+\sqrt2)x = \left(-2 + \sqrt2 - 2 - \sqrt2\right)x = -4x$

5. 5

   Assemble the result and sanity-check it.

   $\left[(2-\sqrt2)x - 1\right]\left[(2+\sqrt2)x - 1\right] = 2x^2 - 4x + 1$

   Testing $x = 1$: the left side is $(1-\sqrt2)(1+\sqrt2) = 1 - 2 = -1$, and the right side is $2 - 4 + 1 = -1$. The roots of $2x^2 - 4x + 1$ are $x = \frac{2 \pm \sqrt2}{2}$, the reciprocals of $2 \mp \sqrt2$, which is what the factored form predicts.