If f(x)=x^4 + 1 calculate x1+x2+x3+x4 and x1^2+x2^2+x3^2+x4^2.

f(x) = x^4+1.

We pressume
x1,x2,x3 and x4 are the roots of x^4 +1 = 0.

To calculate
x1+x2+x3+x4.

By the relation  between the roots and
coeficints, x1+x2+x3+x4 = - (coefficientnt x^3/ coeficient of
x^4).

So x1 +x2+x3++x4 = 0

and
x1^2+x^2+x3^2+x4^2.

x1^2+x2^2+x3^2+x4^2 = (x1+x2+x3+x4)^2
-  2( sum of xixj) .....(1)

By the relation  between roots
and coefficients of the equation x^4+1=o,

sum of (xixj) i
not equal to j  = (coefficient of x^2)/coefficient of x^4) =
0.

Therefore x1^2+x2^2+x3^2+x4^2 = 0-2*0 =
0.