How do you factor x^5-y^5?
(using only real coefficients)
(using only real coefficients)
2 Answers
Explanation:
We know that :
So let's use this :
Explanation:
Given:
x^5-y^5
First note that if
x^5-y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)
We can factor the remaining quartic by making use of its symmetry, expressing it in terms of a quadratic in
Note that:
(x/y+y/x)^2 = x^2/y^2+2+y^2/x^2
So we find:
x^4+x^3y+x^2y^2+xy^3+y^4
= x^2y^2(x^2/y^2+x/y+1+y/x+y^2/x^2)
= x^2y^2((x/y+y/x)^2+(x/y+y/x)-1)
= x^2y^2((x/y+y/x)^2+(x/y+y/x)+1/4-5/4)
= x^2y^2(((x/y+y/x)+1/2)^2-(sqrt(5)/2)^2)
= x^2y^2((x/y+y/x)+1/2)-sqrt(5)/2)((x/y+y/x)+1/2)+sqrt(5)/2)
= x^2y^2(x/y+(1/2-sqrt(5)/2)+y/x)(x/y+(1/2+sqrt(5)/2)+y/x)
= (x^2+(1/2-sqrt(5)/2)xy+y^2)(x^2+(1/2+sqrt(5)/2)xy+y^2)
So putting it all together, we have: