Is factoring just random trial and error?

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SUMMARY

The discussion centers on the factorization of the polynomial expression x^4+x^2+1 into (x^2-x+1)(x^2+x+1). The key insight is the substitution of x^2 with y, transforming the expression into a quadratic y^2+y+1. By rewriting this quadratic as a perfect square and applying the difference of squares technique, the factorization is achieved without prior knowledge of the factors. This method highlights the importance of recognizing patterns and employing algebraic identities in polynomial factorization.

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Homework Statement


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

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