If x^4 + x^3 + x^2 + x + 1 = 0 what is the value of the product

(x1 + 1)(x2 + 1)(x3 +1)(x4 + 1) ?

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The roots of the given equation are:x1, x2, x3, x4.

We'll use Viete's relations:

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

We'll calculate the product:

P = (1+x1)(1+x2)(1+x3)(1+x4)

P = 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)

P = 1 - b/a + c/a - d/a + e/a = 1-1+1-1+1 = 1

**P = 1**

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

To find (x1+1)(x2+1)(x3+1)(x4+1).

Since x1, x2 ,x3 and x4 are the roots of f(x), we make a transformation, x+1 = y , Or x= y-1 in the equation to get an equation whose roots are increased by 1.

So g (y) = (y-1)^4+(y-1)^3+(y-1)^2+(y-1)+1......(1)

By the relation of the roots and coefficients,

y1y2y3y4 = ((x1+1)(x2+1)(x3+1)(x4+1) = constant term in eq (1)

divided by coefficient of y^4 = (1-1+1-1+1)/1 = 3-2 = 1.

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

Usind Viete's rule:

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

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

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

x1*x2*x3*x4 = 1

let P = (x1+1)(x2+1)(x3+1)(x4+1) = x1+x2+x3+x4 +(x1x2+x1x3+x2x4+x2x3+x3x4) +x1x2x3x4 +x1x2x3+x2x3x4+x1x3x4+1

= -1 + 1 -1 + 1 + 1 = 1

Then: P = 1

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