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

*print*Print*list*Cite

We have the equation:

x^4+x^3+x^2+x+1=5

x^4+x^3+x^2+x-4=0

According Viete's rule we know that:

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

x1*x2+x1*x3+x1*x4+x2*x3+x2*x4+x3*x4=1

x1*x2*x3+x1*x2*x4+x1*x3*x4+ x2*x3*x4=1

x1*x2*x3*x4=1

We need to prove that:

(1-x1)(1-x2)(1-x3)(1-x4)=5

Let us open the brackets:

(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+1+1+1+1=5

x^4+x^3+x^2+x+1 = 5.

Then to show that (1-x1)(1-x2)(1-x3)(1-x4) = 5, if x1,x2 ,x3 and x4 are the roots of x^4+x^3+x^2+x+1 = 5

Solution:

If x1, x2,x3 and x4 are the roots of x^4+x^3+x^2+x+1 = 5 Or x^4+x^3+x^2+x-4 =0, then we can write:

x^4+x^3+x^2+x-4 = (x-x1)(x-x2)(x-x3)(x-x4). Now since this should be identity. so putting x = 1, we get:

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

(1-x1)(1-x2)(1-x3)(1-x4) = 0 and not 5.

We'll write Viete expressions:

x1 + x2 + x3 + x4 =-1

x1*x2 + x1*x3 + x1*x4 + x2*x3 + x2*x4 + x3*x4=1

x1*x2*x3 + x1*x2*x4 + x1*x3*x4 + x2*x3*x4=1

x1*x2*x3*x4=1

We'll open the brackets and we'll get:

(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)

We'll substitute by the results from the Viete's relations:

(1-x1)(1-x2)(1-x3)(1-x4)= 1+1+1+1+1=5