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

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Answer 1 · 2015-04-12T15:54:28

In this way:

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

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

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

The last quadratic polynomial can't be further factored.

How do you factor x4−8xx^4-8x?