algebra
posted by Alissa .
Use the given information about a polynomial whose coefficients are real numbers to find the remaining zeros.
degree: 6
Zeros: 8 + 11x(can't put this sign in looks like ii with slash over)i, 7 + 17i, 16  isqrt 2

You know that since the roots come in conjugate pairs, there will be
7 + 17i and 7 17i
16  isqrt2 and 16 + isqrt2
so the factors will be
(x^2  14x + 338) that's (x^2 2*7x + 17^2 + 7^2)
(x^2  32x + 258) that's (x^2  2*16x + 16^2 + sqrt2^2)
and the other pair you can figure out. I can't tell what you mean. 
Hmm. I think that's +14x in the 1st case (2*7)