algebra 2
posted by Jessica .
2X^49X^3+3X^2X5=0
Given that one root is 2+i, solve the equation.
So far I have it set up like this:
2X^49X^3+3X^2X5/X^24X+5
I don't know how to get the answer, I know that I have to use long division but when I started to try it I kept getting weird answers. Please Help

I think there is something wrong here,
either you have a typo in the opening equation, or the question is faulty.
If one root is 2+i, then 2i must be another root
You correctly determined that
x^2  4x + 5 must then be a factor
When I divided that into the original, I got an answer of 2x^2  x 11 with a remainder of 40x + 50 
my bad the original equation is:
2X^4  9X^3 + 13X^2  X  5 = 0 
Ok, this time my division was exact, as expected,
and I got 2x^2  x  1, which factors to
(2x + 1)(x  1)
so the roots are
2+i, 2i, 1/2, and 1 
oh I get it thanks