List all solutions if x^4 + x^3 - 11x^2 + 9x + 20 = 0 and if one of the roots is x = (2+i)

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- Sep 16th 2007, 09:20 AMMr_Greenimaginary number problem
List all solutions if x^4 + x^3 - 11x^2 + 9x + 20 = 0 and if one of the roots is x = (2+i)

- Sep 16th 2007, 09:26 AMJhevon
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 - Sep 16th 2007, 09:39 AMThePerfectHacker
- Sep 16th 2007, 09:41 AMJhevon
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 - Sep 16th 2007, 10:09 AMKrizalid
- Sep 16th 2007, 10:42 AMtopsquark
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 - Sep 16th 2007, 10:45 AMJhevon