# Demonstrate if (1-x1)(1-x2)(1-x3)(1-x4)=5 if x1,x2,x3,x4 are roots of the equation x^4 + x^3 + x^2 + x +1 =0

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If x^4+x^3+x^2+x+1 = 0.To show that (1-x1)(1-x2)(1-x3)1-x4) = 5,

Where x1,x2,x3 and x4 are the roots of the equation.

Proof.

If f(x) = ax^n+a1x^n-1 +a3x^-3 +....+an-*x+an = 0 has the roots x1,x2,x3,..and xn, then

f(x) = x^n +a1x^-1+a2x^(n-2)+....an-1*x+an = (x-x1)(x-x2)(x-x3)...(x-xn) is identity.

So in this case case, x^4+x^3+x^2+x+1 = (x-x1)(x-x2)(x-x3)(x-x4)...........(1).

Since above equation is an identity, we put x= 1 on both sides:

1^4+1^3+1^2+1^1+1 = (1-x1)(1-x2)(1-x3)(1-x4)

5 = (1-x1)(1-x2)(1-x3)(1-x4) which estabilses the proof for the enunciation.

We could write Viete expressions:

x1 + x2 + x3 + x4 = -b/a=-1

x1*x2 + x1*x3 + x1*x4 + x2*x3 + x2*x4 + x3*x4=c/a=1

x1*x2*x3 + x1*x2*x4 + x1*x3*x4 + x2*x3*x4 = -d/a=1

x1*x2*x3*x4= e/a=1

(1-x1)(1-x2)(1-x3)(1-x4)=1-(x1 + x2 + x3 + x4) + (x1*x2 + x1*x3 + x1*x4 + x2*x3 + x2*x4 + x3*x4) - (x1*x2*x3 + x1*x2*x4 + x1*x3*x4 + x2*x3*x4) + (x1*x2*x3*x4)=

= 1+b/a+c/a+d/a+e/a=1+1+1+1+1=5 true