How do you factor $x^4-8x$ ?

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Answer 1 · 2015-04-12T15:54:28

In this way:

$x^4-8x=x(x^3-8)=x(x-2)(x^2+2x+4)$ .

The last passage is because $(x^3-8)$ is a difference of two cubes, that can be factored with this rule:

$a^3-b^3=(a-b)(a^2+ab+b^2)$ .

The last quadratic polynomial can't be further factored.

How do you factor x^4-8xx^4-8x?