List all solutions if x^4 + x^3 - 11x^2 + 9x + 20 = 0 and if one of the roots is x = (2+i)
here's how i'm thinking to do this. there may be an easier way:
if one root is 2 + i, then another is 2 - i, since complex roots always come in conjugates.
thus we can write: $\displaystyle x^4 + x^3 - 11x^2 + 9x + 20 = (x - a)(x - b)(x - (2 + i))(x - (2-i))$
where $\displaystyle a,b$ are the remaining 2 roots, they may be another complex conjugate pair or real (i think they are real though).
expand the right side and equate the coefficients to solve for $\displaystyle a$ and $\displaystyle b$
OR you can do it the long way and try to find real roots of the equation by plugging in factors of 20 and doing the necessary long divisions. that may work
there goes that method i saw Soroban use the other day, i still do not get how you got the answer for the sum of the roots though, i try to find that thread again to see if it was explained there
EDIT: Yup, looked at it, still don't get it
EDIT: Oh wait, i think i found something (in particular, see theorem 5): General theorems for polynomials
now to just figure out a way to memorize it
Another (slightly different) way:
We know that 2 + i is a solution so since the polynomial has real coefficients we know that 2 - i is also a solution. Thus
$\displaystyle (x - (2 + i))(x - (2 - i)) = x^2 - 4x + 5$
is a factor of $\displaystyle x^4 + x^3 - 11x^2 + 9x + 20$.
Doing the division:
$\displaystyle x^4 + x^3 - 11x^2 + 9x + 20 = (x^2 - 4x + 5)(x^2 + 5x + 4) = 0$
Thus you can solve $\displaystyle x^2 + 5x + 4 = 0$ to find the other two roots.
-Dan