This is a question motivated from GRE subgect test exam.

if f(x) over the real number has the complex numbers $2+i$ and $1-i$ as roots,then f(x) could be:

a) $x^4+6x^3+10$
b) $x^4+7x^2+10$
c) $x^3-x^2+4x+1$
d) $x^3+5x^2+4x+1$
e) $x^4-6x^3+15x^2-18x+10$
What I thought at first was to calculate $(x-2-i)(x-1+i)$ and find the polynomial that is divisible by it.Unfortunately it turns out that $(x-2-i)(x-1+i)$ is complex polynomial which makes thing harder to calculate and since this is a multiple choice question with very limited time there must be an easier way. Then I thought maybe it will be easy if I write the complex numbers in polar form and check explicitly if they are the roots.But I don't think that's a very efficient way as well. Then I noticed that the question ends with "$f(x)$ could be" which may suggest that there is a way eliminate the wrong choices, however I have no idea what to eliminate or not.
Does anyone have any ideas?
Thanks in advance


3 Answers 3


I think we should nortice that- in these cases- if $z$ is a root so is $\bar{z}$. And secemdly, I saw a nice practical problem in a book saying that:

If we have $a_0z^n+\cdots+a_n=0$ where $a_0\neq0$ then the sum of all roots and the product of all roots are $$-a_1/a_0$$ and $$(-1)^na_n/a_0$$ respectively. I hope you can solve the problem by yourself.

  • 1
    $\begingroup$ Thanks for your answer, but how do we know that if $z$ is a root then it's conjugate is also a root.? $\endgroup$ Oct 7, 2016 at 13:48
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    $\begingroup$ Y r Welcome! You can show it easily or search this simple fact on the web. It is easy. $\endgroup$
    – Mikasa
    Oct 7, 2016 at 13:56

By the complex conjugate root theorem,


if 2+i is one root, the other root is its conjugate 2-i, the quadratic would be x^2-4x+5, the other quadratic is x^2-2x+2, multiply them both to get f(x) as option e

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    $\begingroup$ Thanks a lot! I had no idea about that theorem $\endgroup$ Oct 7, 2016 at 13:50
















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