## Factoring x^12 + x^6 - 2: where to start?

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
Luke53
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### Factoring x^12 + x^6 - 2: where to start?

How to factor the following expression: x^12 + x^6 - 2 ?
I really don't know how to start.
Luke.

maggiemagnet
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Joined: Mon Dec 08, 2008 12:32 am
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### Re: Factoring x^12 + x^6 - 2: where to start?

How to factor the following expression: x^12 + x^6 - 2 ?
I really don't know how to start.
How would you factor y^2 + y - 2? Now let x^6 = y.

Luke53
Posts: 54
Joined: Sun Mar 13, 2011 9:46 am
Contact:

### Re: Factoring x^12 + x^6 - 2: where to start?

How would you factor y^2 + y - 2? Now let x^6 = y.
I'll give it a try:
The factors of 2 are: 1 ; 2 and -1 ; -2 if I combine -1 and 2, the factors of the given quadratic are: (y - 1)(y + 2) expanding this gives: y² + y - 2 ; subst. y = x^6 gives: (x^6 - 1) (x^6 + 2).
To factor anything further out: ( x^6 + 2) won't factor into anything of the first degree, (just leaving it in this form).
Now : (x^6 - 1) ; I'll first check if 1 and -1 are roots of this expression: (x = 1) ; (1^6) - 1 = 0 so 1 is a root; ( x = -1) ; (-1^6) - 1 = 0, so -1 is also a root. Multiply these (x + 1) (x - 1) = (x^2 - 1).
I'll divide (x^6 -1 ) by (x^2 - 1) the result is: x^4+ x^2 + 1 this factors in to (x² + x + 1) (x² - x + 1).
So: x^12 + x^6 - 2 = (x + 1) (x - 1) (x² + x + 1) (x² - x + 1) (x^6 +2).
Thanks.
Luke.