How do you factor $x^6 - x^4 - 2x^2 + 2$ ?

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Answer 1 · 2015-06-12T09:44:18

Factor by grouping, then using difference of squares three times to find:

$x^6-x^4-2x^2+2$

$= (x^4-2)(x^2-1)$

$=(x-root(4)(2))(x+root(4)(2))(x^2+sqrt(2))(x-1)(x+1)$

In the following, we first factor by grouping then use the difference of squares identity:

$a^2-b^2 = (a-b)(a+b)$

up to three times...

$f(x) = x^6-x^4-2x^2+2$

$=(x^6-x^4)-(2x^2-2)$

$=x^4(x^2-1)-2(x^2-1)$

$=(x^4-2)(x^2-1)$

$=(x^4-2)(x-1)(x+1)$

If we allow irrational coefficients we can break this down some more using difference of squares a couple more times...

$=((x^2)^2-(sqrt(2))^2)(x-1)(x+1)$

$=(x^2-sqrt(2))(x^2+sqrt(2))(x-1)(x+1)$

$=(x^2-(root(4)(2))^2)(x^2+sqrt(2))(x-1)(x+1)$

$=(x-root(4)(2))(x+root(4)(2))(x^2+sqrt(2))(x-1)(x+1)$

There are no simpler linear factors with real coefficients since:

$x^2 + sqrt(2) >= sqrt(2) > 0$ for all $x in RR$

How do you factor x^6 - x^4 - 2x^2 + 2x^6 - x^4 - 2x^2 + 2?