Create a fifth degree polynomial

[frogfriend]

Hello, everyone. I'm a little confused by a math problem so I thought I would ask here. The question states, create a fifth degree polynomial that could have the possible roots +/- 1,2,3,6,1/3,4/3. The thing that is confusing me is that I don't remember running across this in any of the homework problems. Maybe if someone could explain what it is asking, or maybe give the first step it would jog something in my memory. Thank you for any help, I really appreciate it. Also if anything needs clarification let me know.

 

[soroban]

Hello, frogfriend!

There are few issues with this problem.

Create a fifth-degree polynomial that could have the possible roots +/- 1, 2, 3, 6, 1/3, 4/3.

You have given six distinct roots for a fifth-degree polynomial.
. . This is clearly impossible.

A fifth-degree polynomial can have a graph like this:

                       *
          *     *     *
         * *   * *   *
        *   * *   * *
       *     *     *
      *

It cannot cross the x-axis more than five times.


Does the \(\displaystyle \pm\) apply to just the 1, or to all the roots?**

 

[lookagain]

The question states, create a fifth degree polynomial that could have the possible roots +/- 1,2,3,6,1/3,4/3.

No, soroban, "could have" combined with "possible" means they are potential roots. Think of the Rational Root Theorem. The plus/minus would apply to all of them when using the Rational Root Theorem.

 

[Deleted member 4993]

Why not go as "simple" as possible: (x+1)(x-1)(x+1)(x-1)(x+1) ?

Then only possible roots are ±1 - to the corner - on the double.....

This polynomial should have a leading coefficient of 3 - or leading term 3x⁵ - and constant term of 12 (assuming the given list of possible roots was not exhaustive)

 

[frogfriend]

thanks for the help so far guys. I'm sorry say though I have no idea how to solve this equation. I'm not even sure where to begin still or where this problem leads.

 

[DrPhil]

thanks for the help so far guys. I'm sorry say though I have no idea how to solve this equation. I'm not even sure where to begin still or where this problem leads.

What I would do is choose five from the set of 12 possible roots, then multiply together five binomials of the form (x - root).

For another interpretation of the question, what lookagain suggested (I think) is to make a polynomial that would appear to have potential roots of all the values - when using the Rational Root Theorem to test for possible roots. That would be more fun... What set of 5 roots would mimic all the others...

 

[HallsofIvy]

thanks for the help so far guys. I'm sorry say though I have no idea how to solve this equation. I'm not even sure where to begin still or where this problem leads.

First, you understand, don't you, that this problem does not ask you to "solve this equation"- you are asked to set up a function that has certain properties.

The problem asked you to "create a fifth degree polynomial that could have the possible roots +/- 1,2,3,6,1/3,4/3". As you were told before, you need to use the rational root theorem": the constant term must have 1, 2, 3, and 4 as factors and the coefficient of [itex]x^5[/itex] must haver 3 as a factor.

 

[lookagain]

Then only possible roots are ±1 - to the corner - on the double.....

This polynomial should have a leading coefficient of 3 - or leading term 3x⁵ -
and constant term of 12 (assuming the given list of possible roots was not exhaustive)

Or it could have a leading term \(\displaystyle of \ \ -3x^5\) and/or a constant term of -12. (Four cases.)

And if those are the possible roots in the given list, then there should also be these: \(\displaystyle \ \pm \frac{2}{3}.\)

 

[Deleted member 4993]

Or it could have a leading term \(\displaystyle of \ \ -3x^5\) and/or a constant term of -12. (Four cases.)

And if those are the possible roots in the given list, then there should also be these: \(\displaystyle \ \pm \frac{2}{3}.\)

Actually, 12 has 6 factors (1,2,3,4,6 & 12) → 12 is also missing from list of possible roots.