Help with Polynomials (2 Viewers)

[goobi]

This question is from Ex. 3.2 of Terry Lee's textbook.

Factorise as real factors, given that it has a repeated zero.

The suggested solution is:





As f'''(x) has no real linear factors, we must find the multiple root from f''(x).

(^This is the part that I don't understand. When we factorise f'''(x), we get which are actually real and linear factors despite not being rational.)





Therefore, (x+1) is the triple root.





The solution is undoubtedly correct, but I still don't understand the bold part.
Any help would be greatly appreciated

 

[bleakarcher]

It wants the polynomial factored over the rational field of numbers although it is not being clear.

 

[goobi]

It wants the polynomial factored over the rational field of numbers although it is not being clear.

Thanks but why do we assume that rational factors are only allowed when the question requires to get real factors (which include irrational ones)?

 

[goobi]

BTW in Ex 3.1 in the same textbook, questions 2 and 3 specifically ask for factorisations over the rational field, the real field and the complex field respectively. Therefore, as the question above actually requires factorisation over the real field, I reckon it is reasonable and logical to get rational and/or irrational factors.

 

[bleakarcher]

True, I agree. The question isn't being specific enough. You may as well have factored over the irrational field and gotten full marks in an exam if the exact question was to come up.

 

[goobi]

True, I agree. The question isn't being specific enough. You may as well have factored over the irrational field and gotten full marks in an exam if the exact question was to come up.

I am still not convinced though Terry Lee must have done it that way for some reason?

 

[lolcakes52]

Arguably it could be said that because of the surd part of the root, the root is non-linear as it has a part with a degree not equal to 1? I think the whole thing is just a misunderstanding of semantics.

 

[bleakarcher]

Its nothing but a constant so it doesn't matter.

 

[lolcakes52]

Exactly, its only semantics.

 

[goobi]

Arguably it could be said that because of the surd part of the root, the root is non-linear as it has a part with a degree not equal to 1? I think the whole thing is just a misunderstanding of semantics.

IMO the adjective "linear" in this context is used to describe the noun "factor" but NOT the "root". Moreover, a linear factor is defined as being of the form (ax+b)... So, I wonder if I really misunderstand anything...

 

[goobi]

^ That's why I'm confused

 

[lolcakes52]

IMO the adjective "linear" in this context is used to describe the noun "factor" but NOT the "root". Moreover, a linear factor is defined as being of the form (ax+b)... So, I wonder if I really misunderstand anything...

I know, I'm scraping the barrel for answers here. I think that it's probably just an error in the textbook and it doesn't matter as the answer is still the same.

 

[goobi]

I know, I'm scraping the barrel for answers here. I think that it's probably just an error in the textbook and it doesn't matter as the answer is still the same.

Thank you so much anyways Is it just me or did anyone interpret the question that way?

 

[goobi]

I know, I'm scraping the barrel for answers here. I think that it's probably just an error in the textbook and it doesn't matter as the answer is still the same.

So what do you think the question should have been worded to avoid confusion?

 

[bleakarcher]

What I think is that what you did is fine and what Terry Lee did was fine to. It answers the question which asked to factorise the polynomial over the real field since the real field of numbers includes both rational and irrational numbers. Don't worry about it dude. Looks like you know the definition of a polynomial pretty well anyways.

 

[SpiralFlex]

We cannot take the irrational solutions to be a multiple root.

For example,

IF



This will imply that the root has a multiplicity of 5.

Then your polynomial will end up being,

(If you expand it, then your polynomial is obviously not the one in your question.)

 

[goobi]

We cannot take the irrational solutions to be a multiple root.

For example,

IF



This will imply that the root has a multiplicity of 5.

Then your polynomial will end up being,

(If you expand it, then your polynomial is obviously not the one in your question.)

But I think what you said may only be true for this question as to the best of my knowledge there is no theorem to show that an irrational solution can/cannot be a multiple root. But then again, prove me wrong

 

[SpiralFlex]

To convince you more.

Suppose the solution above was a multiple root with multiplicity of 5.

Our polynomial was

By applying Descartes' Rule of Signs,

Let

4 sign changes. There can not be more than 4 negative real roots. We have just contradicted our statement. Therefore in this case, there cannot be irrational multiple roots here.

In the general case, I am sure you will never come across a question with an obscure "irrational multiple root". It would be likely that your coefficients of your polynomial will not be integer values.

 

[goobi]

Thanks Spiral. That makes more sense now

 

[Carrotsticks]

Goobi, I can assure you that your exam will be much better worded. If for some freakish reason they copy pasted this question into your exam and you lost marks, just contest. You won't be the only one.

Though I do agree that T. Lee could have worded it more clearly.