2X^4-9X^3+3X^2-X-5=0

Given that one root is 2+i, solve the equation.

So far I have it set up like this:

2X^4-9X^3+3X^2-X-5/X^2-4X+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

To solve the given equation, you need to find the other roots of the equation using the given root, 2+i. Here's how you can do it:

Step 1: Express 2+i as a complex conjugate to simplify the calculation. The complex conjugate of 2+i is 2-i. Since complex conjugates have the same real part but opposite signs for the imaginary part, 2+i and 2-i are a conjugate pair.

Step 2: The equation states that 2+i is a root. Therefore, the equation (x - root)(x - conjugate of root) = 0 must be a factor of the given equation.

(x - (2+i))(x - (2-i)) = 0
Expanding the equation:
(x - 2 - i)(x - 2 + i) = 0
(x - 2)^2 - (i^2) = 0

Step 3: Simplify the equation.

(x - 2)^2 - i^2 = 0
(x - 2)^2 + 1 = 0
(x - 2)^2 = -1

Step 4: Take the square root of both sides.

x - 2 = ±√(-1)
x - 2 = ± i
x = 2 ± i

So, the two roots of the equation are 2 + i and 2 - i.

Note: The expression you attempted, 2X^4-9X^3+3X^2-X-5/X^2-4X+5, seems to be an attempt at long division. However, it is not directly applicable in this case as the given root is complex.