1
$\begingroup$

I have been struggling to create a polynomial that adheres to the following guidelines and I am hoping someone can help me out. I have a series of questions regarding the polynomial and I am struggling to get past the first step, creating the polynomial itself.


  1. Create a polynomial in standard form of least degree with integer coefficients that has 5 – 2i, √3, 0, and -1 as zeros. Show your work.

  2. Check that your answer is correct by using division. Show your work, and make it clear and organized.

  3. Show that 1 is not a zero of the polynomial

  4. What is the end behavior of your polynomial?

  5. How many real zeros, irrational zeros, and imaginary zeros does it have, respectively?

  6. What is the degree of your polynomial?

  7. Use your calculator to determine the

a. Relative maxima

b. Relative minima

c. Intervals over which the polynomial is increasing

d. Intervals over which the polynomial is decreasing

  1. Does your polynomial have an absolute maximum? If so, what is it?

  2. Does your polynomial have an absolute minimum? If so, what is it?

$\endgroup$
  • $\begingroup$ If you have multiple questions, please refrain from asking them all at once. It would be better if there was only one question. $\endgroup$ – S.C.B. Jan 26 '17 at 3:57
  • $\begingroup$ The question is imprecisely stated. Almost surely, it was asking for a rational polynomial, i.e. with coefficients in the rational numbers. If it had been asking only for a polynomial with complex coefficients, then $X(X+1)(X-\sqrt3\,)(X-5+2i)$ would have sufficed. $\endgroup$ – Lubin Jan 31 '17 at 23:48
1
$\begingroup$

Conjugate Theorem: Given a root of the polynomial $f(x)$ is $x=a-bi$, another root is its conjugate: $\overline{a-bi}=a+bi$.

Irrational Root Theorem: Given an irrational root say, $x=a+b\sqrt{c}$, another root will be $x=a-b\sqrt{c}$.


With that in mind, let's construct our polynomial. We can build our polynomial by starting with its roots in expanded form, and multiplying everything out at the end.

For $5-2i$:

By the conjugate theorem, $5+2i$ is another root. Hence, we have a quadratic$$(x-(5-2i))(x-(5+2i))=\color{red}{x^2-10x+29}\tag1$$

For $\sqrt3$:

And by the irrational root theorem, we have $x=-\sqrt3$ as a root. Hence, its minimal polynomial is $\color{blue}{x^2-3}$.


Thus, our polynomial is$$\begin{align*}f(x) & =x(x+1)(\color{red}{x^2-10x+29})(\color{blue}{x^2-3})\tag2\\ & =x^6-9x^5+16x^4+56x^3-57x^2-87x\tag3\end{align*}$$

I'll let you solve the rest of the problems. If you need help with any of them, ask me and I'll reply with a hint.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you very much. I understand how you got that answer very clearly now. I think I should be able to work my way through the rest of them; although I am just curious if dividing that polynomial by one of the zeros (5-2i) would be a sufficient answer for number two. $\endgroup$ – Kyle Papili Jan 26 '17 at 4:11
  • $\begingroup$ @KylePapili Yes, dividing $(3)$ by any of its zeroes will leave a remainder of $0$. Although, I believe the problem is saying to do the division for all of the roots. I may have misinterpreted the problem (I'm not the best at English) $\endgroup$ – Frank Jan 26 '17 at 4:13
  • $\begingroup$ Nvm, I understand your numbering of the steps. Thank you very much, I think that you may be correct. $\endgroup$ – Kyle Papili Jan 26 '17 at 4:18
  • $\begingroup$ @KylePapili To take this off unanswered, I suggest you accept and upvote this answer. $\endgroup$ – S.C.B. Jan 26 '17 at 4:25
  • $\begingroup$ I have been trying although I do not have a high enough reputation. I marked the answer with the green check, although that is the best I am able to do at this time. My apoligies $\endgroup$ – Kyle Papili Jan 26 '17 at 4:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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