Give a polynomial with these roots

Give a polynomial, of smallest degree, with integer coefficients whose roots are:

1, -(2^.5), 3+i

this is a weird prob. bec. it has complex roots and square roots.
please show work and explain your procedure

Remember , 2^0.5 = √2

If one of the roots is 3+i, then................

The conjugate other root is 3 - i................so

two of the roots are :

(x-3-i)(x-3+i) = x^2 - 6x -(-1) = x^2 - 6x + 1

One of the factors is easy to find. If I have zero at x = 1 , then .............(x-1)


The other solution is messy, with the square root in it. Since they specified that the polynomial has "integer coefficients", you know you can't leave that square root by itself; have to find its pair that the Quadratic Formula created. In the Quadratic Formula, the "±" is right in front of the square root, so the pair for the solution "0 + √(2)" must be "0 – √(2)".

the other two roots are :

(x+√2)(x-√2) = x^2 - 2.....................so

y = (x^2 - 6x + 1)(x -1)(x^2 - 2)

y = (x^2 - 6x + 1)(x^3 - x^2 - 2x + 1)

y = x^5 - 7x^4 + 5x^3 .................etc, etc. ending with + 1