# Demonstrate if (1+x1)(1+x2)(1+x3)(1+x4)=1 . x1,x2,x3,x4 roots of x^4+x^3+x^2+x^1=0

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### 2 Answers

f(x)=x^4+x^3+x^2+x+1 = 0.

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

To prove that (1+x1)(1+x2)(1+x3)(1+x4) = 1.

Since x1, x2 ,x3 and x4 are the roots of the equation, f(x) = 0,

we can write the polynomial x^4+x^3+x^2+x+1 = (x-x1)(x-x2)x-x3)(x-x4), by theory of equation, as the coefficient of x^4 agrees on both sides.

Now put x =-1 on both sides:

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

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

1 = (1+x1)(1+x2)(1+x3)(1+x4) which is established.

(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=1 true