# x4+x3+x2+x+1=0 (4,3,2-->exponents) Could anyone please solve this without D'Moivres theorem???

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### 6 Answers

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

Divide by x^2.

x^2+x+1+1/x+1/x^2 = 0.

(x^2+1/x^2) +(x+1/x) +1 = 0

(x+1/x)^2 -2 + (x+1/x)+1 =0

(x+1/x)^2 +(x+1/x) -1 = 0. This is aquadratic equation in (x+1/x). Or

y^2 +y -1 = 0 , where y = x+1/x.

y1 = (-1+sqrt5)/2 , or y2 = (-1-sqrt5)/2

y1 = x+1/x = (-1+sqrt5)/2

x^2 +(1-sqrt5)/2 *x +1 = 0

x1 = {(-1+sqrt5)/2 + sqrt[(1-sqrt5)^2/4 -4}/2

x1 = {-(1-sqrt5)/4 + sqrt(-5-sqrt5)/2}/2

x1 = {-(1-sqrt5)/4 +sqrt(-1)sqrt(5+sqrt5)/4}

x2 = {-(1-sqrt5)/4 -sqrt(-1) sqrt(5+sqrt5)/4

Similarly we can solve for y2 = x+1/x = (-1-srqt5)/2

x^2 +(1+sqrt5)x+1 = 0

x3 = -(1+sqrt5) /4+ [sqrt(sqrt5 -5)]/4 or

x3 = -(1+sqrt5)/4 +{sqrt(-1)sqrt(5-sqrt5)}/4

x4 = -(1+sqrt5)- {sqrt(-1) sqrt(5-sqrt5)}/4

S0 there are as above 4 complex roots x1,x,2 ,x3 and x4..

The polynomial x^4+x^3+x^2+x+1 is cyclotomic.

Then the roots are self evident in the complex field.

Since (x^4+x^3+x^2+x+1) * (x-1) = x^5 - 1

the roots are easy to find now ;)

what about x4+x3+x2+x+1=m2 where m is a positive integer?

X^4+X^3+X^2+X+1= a(r^n-1)/(r-1)=0

(**Geometrical progression**)

where r=x , n=5 , a=1

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

, **X** not equal (**1**)

x^5-1=0

x^5=1

X=(1)^(1/5)

X=(cos(0) + i sin(0))^(1/5)

X=(cos((0+180k)/5) + i sin((0+180k)/5)

k={0,1,2,3,4}

**k=0**

X=(cos(0) + i sin(0))=1refused

**k=1**

X1=(cos(36) + i sin(36))

**k=2**

X2=(cos(72) + i sin(72))

**k=3**

X3=(cos(108)+ i sin(108))

**k=4**

X4=(cos(144)+ i sin(144))

very good answer neela@ :)

Thank u very much!!!