# Is factoring just random trial and error?

1. Apr 27, 2007

### pivoxa15

1. The problem statement, all variables and given/known data
How does one know x^4+x^2+1=(x^2-x+1)(x^2+x+1)?

How does one get to that step?

I can find the roots of x but that doesn't seem to help. Is it be random trial and error?

2. Apr 27, 2007

### HallsofIvy

Staff Emeritus
One knows that x^4+x^2+1=(x^2-x+1)(x^2+x+1) by multiplying out the right side! But I suspect that was not your question. Your question really is "how does one factor x^4+ x^2+ 1 if you do NOT know that factorization in advance?"

Notice that there are no ODD powers of x in the expression. Replace "x^2" by "y" so that you have the quadratic y^2+ y+ 1. Rewrite that as (y^2+ 2y+ 1)- y= (y+1)^2- y so that the first is a "perfect square". Think of it as the "difference of two squares" so that it can be factored as $[(y+1)- \sqrt{y}][(y+ 1)+ \sqrt{y}]$. One wouldn't normally do that because of the $\sqrt{y}$ but since y= x^2, $\sqrt{y}= x$ and we have [(x^2+1)-x][(x^2+1)+x]= (x^2-x+1)(x^2+x+1).

3. Apr 27, 2007

### pivoxa15

Nice trick. I wonder how many of those tricks are there and which book contains most or all of them?