Factoring Polynomials (x^4) - using completing squares

Question

Exercise 6. By viewing the polynomials as a difference of two squares, factorise the following polymomials.

1. $x^4+x^2+1$,
2. $x^4+3x^2+4$,
3. $x^4+4$.

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!

Answer 1

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

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

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

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:

1. $a=x^2, b=1, k=1$
2. $a=x^2, b=2, k=1$
3. $a=x^2, b=2, k=0$.