Can someone tell me if i have factorized correctly:

Equation is this:$\displaystyle x^4-6x^2+1$

First i completed the square:

$\displaystyle x^4-6x^2+1$

$\displaystyle x^4-6x^2+1-9x^2$

$\displaystyle (x^2-3)^2-(3x)^2$

$\displaystyle (x^2-3-3x)(x^2-3+3x)$

Now i use the difference of two squares where $\displaystyle a^2-b^2=(a-b)(a+b)$

$\displaystyle x^2-3x-3$

$\displaystyle (x-3)^2-(\sqrt3)^2$

$\displaystyle (x-3-\sqrt3)(x-3+\sqrt3)$

$\displaystyle x^2+3x-3$

$\displaystyle (x+3)^2-(\sqrt3)^2$

$\displaystyle (x+3-\sqrt3)(x+3+\sqrt3)$

Therefore answer is:

$\displaystyle (x-3-\sqrt3)(x-3+\sqrt3)(x+3-\sqrt3)(x+3+\sqrt3)$

P.S