How to turn into a product of $2$ factors the expression : $ x^4 -3x^2+1$? Source : Lebossé & Hémery, Algèbre et Analyse ( Classe de seconde , 1965).
The exercice requires  : turn into  a product of $2$ factors the expression $ x^4 -3 x^2 +1$.
The only way I see is to use the property : $P(x)$ being a polynomial , if $P(a) =0$ then $P(x)$ is divisible by $(x-a)$.
Using the  substitution $X = x^2$, the expression can be turned into : $X^2 -3X +1$
Setting $X^2 -3X +1= 0$ yields $ X = \frac {3+\sqrt5} {2}$ or $ X = \frac {3-\sqrt5} {2}$.
This implies that : $x= \pm\sqrt{ \frac {3+\sqrt5} {2}}$ or $x= \pm\sqrt{ \frac {3-\sqrt5} {2}}$.
Finally, using the property referred to above, one gets :
$ x^4 -3 x^2 +1 = ( x- \sqrt{\frac {3+\sqrt5} {2}}) ( x+ \sqrt{\frac {3+\sqrt5} {2} })( x - \sqrt{\frac {3-\sqrt5} {2}})(x+ \sqrt{\frac {3-\sqrt5} {2}})$.
But this method yields 4 factors instead of 2.
Is there a better way to meet the requirement of the exercise?
Is there a classical identity hidden below the original expression?

You can stop earlier:
$$
x^4 - 3x^2 + 1
= X^2 - 3X + 1
= (X - \frac{3+\sqrt5}{2})(X - \frac{3-\sqrt5}{2})
= (x^2 - \frac{3+\sqrt5}{2})(x^2 - \frac{3-\sqrt5}{2}).
$$
