Given x ^ 4-5x ^ 3 + 8x ^ 2-5x + 1 = 0, find x + 1 / X

Given x ^ 4-5x ^ 3 + 8x ^ 2-5x + 1 = 0, find x + 1 / X

x^4-5x^3+8x^2-5x+1=0
x^2(x^2-5x+8-5/x+1/x^2)=0
x^2[(x^2+1/x^2)-5(x+1/x)+8]=0
x^2[(x+1/x)^2-5(x+1/x)+6]=0
x^2(x+1/x-6)(x+1/x+1)=0
If x = 0, then the original formula does not hold
So x ≠ 0, x ^ 2 ≠ 0
∴(x+1/x-6)(x+1/x+1)=0
Ψ x + 1 / x = 6 or - 1

We know that x = 1 is a root of the equation x ^ 3-5X ^ 2 + 8x-4 = 0. Try factorization to find other roots of the equation. You are very smart!

I've got an answer for you. Then you can divide this formula by (x-1), and the result is x ^ 2-4x + 4, so the original formula can be decomposed into (x-1) (x ^ 2-4x + 4) = (x-1) (X-2) ^ 2 = 0, so the other root is x = 2