How do you factor completely #x^5 - 16x^3 + 8x^2 - 128 #?
1 Answer
Nov 15, 2015
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
