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

1 Answer
Nov 15, 2015

Answer:

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)#

Explanation:

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.