Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to find an nth degree polynomial function that has real coefficients using the following conditions:

n=3; 3 and 4i are zeros; f(2)=40

I have no idea what I'm doing on this one. It's been too long. Also, there's no homework tag because this isn't something I have to do. I'm just brushing up in preparation for an upcoming math course.

share|cite|improve this question
Hint : Since $3$ and $4i$ are zeros, you can factorize your polynomial like this : $f(z)=(z-3)(z-4i)g(z)$. What is the degree of $g$ then ? – Philippe Malot Apr 26 '13 at 9:50
up vote 2 down vote accepted

We want $3$ and $4i$ to be roots. If a real polynomial has a complex root, the conjugate also is a root. So we can just write down a polynomial that has exactly these roots: \begin{align*} f(x)=(x-3)(x-4i)(x+4i)=(x-3)(x^2+16) \end{align*} which indeed has only real coefficients and is of degree $3$. But the value at $2$ is not $40$ but $-20$ so the polynomial you are looking for is $g(x)=-2f(x)$.

share|cite|improve this answer
I need to practice some more of these, but you've pointed me in the right direction. Thanks. – Dave Apr 26 '13 at 10:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.