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?

  • 0
    If you have multiple questions, please refrain from asking them all at once. It would be better if there was only one question. – 2017-01-26
  • 0
    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. – 2017-01-31

1 Answers 1

1

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.

  • 0
    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. – 2017-01-26
  • 0
    @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) – 2017-01-26
  • 0
    Nvm, I understand your numbering of the steps. Thank you very much, I think that you may be correct. – 2017-01-26
  • 0
    @KylePapili To take this off unanswered, I suggest you accept and upvote this answer. – 2017-01-26
  • 0
    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 – 2017-01-26
  • 0
    @Frank I have been able to get answers for all of the other questions EXCEPT for #5. Could you assist me in finding the number of real, irrational, and imaginary zeros. I believe it contains 4 real zeros but that is as far as I have gotten. – 2017-01-26
  • 0
    @KylePapili Sure! We're given the roots of $f(x)$. Now, we just have to determine whether they are real or imaginary. The given roots are $5-2i,\sqrt{3},0,-1$ and recall that $a+bi$ has a real part of $a$, and an imaginary part of $b$. For example, the real part of $3+i$ is $3$ because here, $a=3$. While the imaginary part is $1$ because $b=1$. – 2017-01-26