How do you factor #x^7 - 16x^5 + 5x^4 - 80x^2#?
1 Answer
Sep 17, 2016
#= 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
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)#
