# Polynomial Question

#### Taran

##### New member
Hi, this question was in a year 11 extension maths textbook in the enrichment section. I have the answer as k>17 and k<-11 because I graphed it on GeoGebra. The Graph can be found here: https://ggbm.at/xpegwwtq. While I know the answers I would like to know how to work it out using algebra.

Here is the Question:
Consider the cubic equation x^3 - kx + (k + 11) = 0, find all the integer values of k for which the equation has at least one positive integer solution for x

Thanks, Taran

#### Jomo

##### Elite Member
Taran,
Welcome to the forum.
According to the forum policy you need to show us your work so that we know where you are making a mistake. It also helps us have an idea what level you are at. Some 11th grader know Calculus. Are you one of them? This would be helpful to know.
Please get back to us.
Thanks!

#### Taran

##### New member
Hi, I don't know Calculus. My class has just started on polynomials and we are up to dividing polynomials with long division. Also, I don't know any working out for this, that is what I was wanting to find out. I found the answers by just putting the equation into GeoGebra (which shows the graph but not the working).

#### pka

##### Elite Member
Here is the Question:
Consider the cubic equation x^3 - kx + (k + 11) = 0, find all the integer values of k for which the equation has at least one positive integer solution for x
Look HERE.

• topsquark

#### Taran

##### New member
Thank you for that but that doesn't show me the integer values of k for which the equation has at least one positive integer solution for x

#### Otis

##### Senior Member
… I have the answer as k > 17 OR k < -11 because I graphed it on GeoGebra …

… x^3 - kx + (k + 11) = 0, find all the integer values of k for which the equation has at least one positive integer solution for x
Hello Taran. Please note the phrase highlighted in red above and explain why you think that is the answer shown by the graph. (Note that I changed 'and' to 'OR' because no number can be both positive and negative. Let us know, if you don't understand this difference.)

Let's start with your solution k = 18. Why do you think it works?

… that [site] doesn't show me the [answer]
You can see all the relevant values of k, if you keep clicking the 'More solutions' button until it disappears. The answer to the exercise (as posted) is a set of five positive Integers.

By the way, is this homework?

EDITS: Changed 'and' to OR (and explained why). Reworded, to simplify my first question. Last edited by a moderator:

#### Taran

##### New member
Hi, I think I know one thing I did wrong. The graphing program that I was using showed it as one line, rather than two seperate equations. Using the wolfram alpha graph, it show five values for k that give positive integers (see attached image). What I am wanting to know is how it got these values.

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#### Denis

##### Senior Member
Just confirmed those results through a looper program;
though I'm still wondering what the "purpose" of this is #### Otis

##### Senior Member
… The graphing program that I was using showed it as one line, rather than two separate equations …

… What I am wanting to know is how [wolfram] got these values.
I'm not sure what it means, above.

I don't know what wolfram did exactly. You'll have to pay wolfram some money, to see their steps. Also, they don't explain their steps (they just list them), so after you pay to see the steps it will be up to you to figure them out.

You didn't answer either of my two questions, so I have nothing more to post for you right now.

Cheers #### MarkFL

##### Super Moderator
Staff member
This question was posted on another site, and one of our scholars there posted this insightful reply:

If $$x=n$$ is a positive integer solution of the equation, then $$n^3 - kn + k + 11 = 0$$, so that $k = \frac{n^3+11}{n-1} = \frac{(n-1)(n^2+n+1) + 12}{n-1} = n^2+n+1 + \frac{12}{n-1}.$ For that to be an integer, $$n-1$$ must be a factor of $$12$$. You can then tabulate the possible values of $$n$$ and $$k = \frac{n^3+11}{n-1}$$, as follows: $\begin{array}{c|cccccc} n-1&1&2&3&4&6&12 \\ n&2&3&4&5&7&13 \\ k&19&19&25&34&59&184 \end{array}.$ So the only possible values for $$k$$ are $$19,\ 25,\ 34,\ 59,\ 184$$ (which all agree with your condition that $$k>17$$).

• topsquark and Otis

#### Otis

##### Senior Member
… [the solutions for k] all agree with your condition that $$k>17$$).
Good copy from that other site, but I note that k>17 is not Taran's condition. It is Taran's answer (in addition to k<-11).

I'd like to know what Taran was thinking because something major went wrong with their process. #### Taran

##### New member
Hi, Thank you so much!!! This question had my class stumped. That answer makes so much sense. It's been bugging me for a while and I'm very thankful for your help. Thanks Otis for the spell checks . Yes, this was in my homework but was as an enrichment challenge question. When you look at the original link in my first post it comes up with a graph. This graph removed the brackets I added around (k+11), causing it to graph it incorrectly. Using this graph, I moved the slider for k to find the values of k that gave a positive x intercept. These values were: k>17 and k<-11. But seeing that answer shows me that the graph actually isn't important and I was overthinking it.

Thanks again, Taran

#### Taran

##### New member
Hang on, Couldn't the negative factors of 12 work too. If I put in x=-3, x=-5 or x=-11, it still works out as a positive integer.

#### Otis

##### Senior Member
… brackets I added around (k+11) [were removed] …
Thank you for the explanation.

… I moved the slider for k to find the values of k that gave a positive x intercept …
Okay, but the intercepts must be positive Integers.

Don't rely on a graph, for distinguishing exact values. A plotted intercept may appear as x=2, but it might actually be 1.9905 or 2.1073. Graphs are not precise enough for us to "see" Integers because machines can display only to the nearest pixel. #### Otis

##### Senior Member
Hang on, Couldn't the negative factors of 12 work too. If I put in x=-3, x=-5 or x=-11, it still works out as a positive integer.
Why are you considering negative values of x? All solutions for k must lead to positive Integer solutions for x. #### Taran

##### New member
Yeah, you're right. I was thinking of positive solutions for k.

Thanks again, Taran