# How can I solve this: x^4+x^3+x^2+x+1=0 ? thx

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To solve x^4+x^3+x^2+x+1 = 0.

We know that 1+x+x^2+x^4 = (x^5-1)/(x-1).

So the given equation is therefore rewritten as:

(x^5-1) /(x-1) = 0

Multiply by (x-1). So ,** mind x-1 is our factor**.

(x^5 -1) = 0

x^5 = 1.

We know 1 = cos2npi +isin2npi.

Therefore ,

x^5 = cos2npi+isin2npi

Take the 5 th root.

x = (co2npi+isin2npi) ^(1/5)

x = (cos(2npi)/5 +isin(2npi)/5 , for n = 0,1,2,3,4....by D'Moivres theorem.Actually after n=4, for the next integral values the roots repeat.

x0 = 1 is not the root as we have multiplied by **our factor (x-1)** to the given expression (x^4+x^2+x^3+1 ).

x1 = cos72+isin72

x2 = cos144 + isin 144

x3 = cos216 +i sin 216

x4 = cos288 +isin288.

So x1,x2,x3 and x4 are the solutions.