Factoring Polynomials (x^4) - using completing squares 

Exercise 6. By viewing the polynomials as a difference of two squares, factorise the following polymomials.
  
  
*
  
*$x^4+x^2+1$,
  
*$x^4+3x^2+4$,
  
*$x^4+4$.
  

To solve difference of two squares isnt it necessary to have negatives? im confused about how to start solving these questions. I cant seem to factorise by grouping either /:
Information on how to factorise these types of questions would be appreciated. thank You!

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*$x^4+x^2+1= (x^2)^2+(x^2)\cdot 1+1^2=(x^2)^2+2\cdot (x^2)\cdot 1+1^2- (x^2)\cdot 1=(x^2+1)^2-x^2=(x^2+1-x)(x^2+1+x)=(x^2+x+1)(x^2-x+1).$

*$x^4+3x^2+4= (x^2)^2+3\cdot (x^2)+2^2=(x^2)^2+4\cdot (x^2)+2^2-x^2=(x^2)^2+2\cdot 2\cdot x^2+2^2-x^2=(x^2+2)^2-x^2=(x^2+2-x)(x^2+2+x)=(x^2+x+2)(x^2-x+2).$

*$x^4+4=(x^2)^2+2^2 =(x^2)^2+2\cdot 2\cdot x^2+2^2-4x^2=(x^2+2)^2-(2x)^2=(x^2-2x+2)(x^2+2x+2).$
The idea is that if you have something like: $a^2+b^2+kab$ you can add and subtract something which in your case gives you a square (the exercise is of such a nature). $a^2+b^2+2ab+(k-2)ab=a^2+b^2+2ab-(2-k)ab=(a+b)^2-(2-k)ab$.
In your examples: 


*

*$a=x^2, b=1, k=1$

*$a=x^2, b=2, k=1$

*$a=x^2, b=2, k=0$.

