How do you factor completely x^5 - 16x^3 + 8x^2 - 128 x5−16x3+8x2−128?
1 Answer
Factor by grouping, sum of cubes identity and difference of squares identity to find:
x^5-16x^3+8x^2-128x5−16x3+8x2−128
=(x+2)(x^2-2x+4)(x-4)(x+4)=(x+2)(x2−2x+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)a3+b3=(a+b)(a2−ab+b2)
The difference of squares identity can be written:
a^2-b^2=(a-b)(a+b)a2−b2=(a−b)(a+b)
Hence:
x^5-16x^3+8x^2-128x5−16x3+8x2−128
=(x^5-16x^3)+(8x^2-128)=(x5−16x3)+(8x2−128)
=x^3(x^2-16)+8(x^2-16)=x3(x2−16)+8(x2−16)
=(x^3+8)(x^2-16)=(x3+8)(x2−16)
=(x^3+2^3)(x^2-4^2)=(x3+23)(x2−42)
=(x+2)(x^2-2x+4)(x-4)(x+4)=(x+2)(x2−2x+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)(x2−2x+4)=(x+2ω)(x+2ω2)
where
