How do you factor #x^7 - 16x^5 + 5x^4 - 80x^2#?

1 Answer
Sep 17, 2016

Answer:

#x^7-16x^5+5x^4-80x^2#

#= x^2(x+root(3)(5))(x^2-root(3)(5)x+root(3)(25))(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)#

Given:

#x^7-16x^5+5x^4-80x^2#

All of the terms are divisible by #x^2#, so that can be separated out as a factor.

Note also that the ratio between the first and second terms is the same as that between the third and fourth terms. So this quadrinomial will factor by grouping:

#x^7-16x^5+5x^4-80x^2#

#= x^2((x^5-16x^3)+(5x^2-80))#

#= x^2(x^3(x^2-16)+5(x^2-16))#

#= x^2(x^3+5)(x^2-16)#

#= x^2(x^3+(root(3)(5))^3)(x^2-4^2)#

#= x^2(x+root(3)(5))(x^2-root(3)(5)x+root(3)(25))(x-4)(x+4)#