algerbra 2
posted by kaitlyn .
what are the other roots of 3x^412x^3+41x8x+26 that include 23i

Assume you mean:
3x^412x^3+41x^28x+26
Well then 2+3i must be one
multiply them
(x 2+3i)(x  23i) = x^2 4x +13
divide by that and get
3x^2+2
so the other two roots are
x=sqrt(2/3) and x = 2/3 
First of all, I will assume you have a typo and the third term should be 41x^2
complex roots always come in conjugate pairs, that is,
if 23i is a root, so is 2+3i
sum of those roots = 4
product of those roots = 39i^2 = 13
so the quadratic that yields those two roots is
x^2  4x + 13
I then did an algebraic long division by x^2  4x + 13 and that came out very nicely to an answer of 3x^2 + 2 with no remainder.
so ...
3x^412x^3+41x^28x+26= 0
(x^2  4x + 13)(3x^2 + 2) = 0
x = 23i, 2+3i from the first part and
x^2 = 2/3
x = ± √(2/3)
= ± i√(2/3) or ±(1/3)i √6