How do you factor completely $x^5 - 16x^3 + 8x^2 - 128$ ?

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Answer 1 · 2015-11-15T20:48:20

Factor by grouping, sum of cubes identity and difference of squares identity to find:

$x^5-16x^3+8x^2-128$

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

The sum of cubes identity can be written:

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

The difference of squares identity can be written:

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

Hence:

$x^5-16x^3+8x^2-128$

$=(x^5-16x^3)+(8x^2-128)$

$=x^3(x^2-16)+8(x^2-16)$

$=(x^3+8)(x^2-16)$

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

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

This has no simpler factors with Real coefficients.

If you allow Complex coefficients then:

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

where $omega = -1/2+sqrt(3)/2i = cos((2pi)/3) + i sin((2pi)/3)$ is the primitive Complex cube root of unity.

How do you factor completely x^5 - 16x^3 + 8x^2 - 128 x^5 - 16x^3 + 8x^2 - 128 ?