Polynomial problem about division of a remainder

[Ognjen]

I just joined the forum so I attached an image instead of typing the problem out due to lack of knowledge about the math functions and such ( which I will acquire with time ).

The problem is concerned with finding the remainder of a polynomial division with the dividend being P(x), given the divisor.
Two other relations of P(x) are given ( divisors and remainders of 2 distinct dividing calculations ).

I read the solution of the problem and came across a point that is assumed to be comprehensible to the reader, but just struck me as inexplicable by any of the polynomial division rules that I know of.
Basically, P(x) and R(x), with R(x) being the remainder of the initial problematic equation, when divided by the given divisor(s) respectively, supposedly give out the SAME remainder. From this point on the problem solving is clear to me.

Can anybody explain to me how this happens to be ? I would greatly appreciate it.

 

[Dr.Peterson]

I just joined the forum so I attached an image instead of typing the problem out due to lack of knowledge about the math functions and such ( which I will acquire with time ).

The problem is concerned with finding the remainder of a polynomial division with the dividend being P(x), given the divisor.
Two other relations of P(x) are given ( divisors and remainders of 2 distinct dividing calculations ).

I read the solution of the problem and came across a point that is assumed to be comprehensible to the reader, but just struck me as inexplicable by any of the polynomial division rules that I know of.
Basically, P(x) and R(x), with R(x) being the remainder of the initial problematic equation, when divided by the given divisor(s) respectively, supposedly give out the SAME remainder. From this point on the problem solving is clear to me.

Can anybody explain to me how this happens to be ? I would greatly appreciate it.

It would be helpful if you actually showed us the problem as given to you, and also stated in words what each line of your work means, though I can guess. Also, if you're saying that the work here is not your own, but something you were given, or found, can you show that in its original form? Paraphrases by someone who is just learning are often not quite right, and I'd rather explain what was really said than what you think it said.

But if you are primarily asking how two polynomials can leave the same remainder, just think about the same question using numbers. If I divide 45 and 97 by 13, I get the same remainder: 45/13 = 3 R 6; 97/13 = 7 R 6. How did that happen (and how did I know they would have the same remainder without having done the divisions yet)?

By the way, images are fine, and so is typing without using any fancy formatting (e.g. "P(x) = (x^2 + x + 1)(x^2 + x + 2)Q(x) + R(x)".

You can also insert your image so it takes less work to read; here's a cleaned-up version:

 

[Ognjen]

It would be helpful if you actually showed us the problem as given to you, and also stated in words what each line of your work means, though I can guess. Also, if you're saying that the work here is not your own, but something you were given, or found, can you show that in its original form? Paraphrases by someone who is just learning are often not quite right, and I'd rather explain what was really said than what you think it said.

But if you are primarily asking how two polynomials can leave the same remainder, just think about the same question using numbers. If I divide 45 and 97 by 13, I get the same remainder: 45/13 = 3 R 6; 97/13 = 7 R 6. How did that happen (and how did I know they would have the same remainder without having done the divisions yet)?

By the way, images are fine, and so is typing without using any fancy formatting (e.g. "P(x) = (x^2 + x + 1)(x^2 + x + 2)Q(x) + R(x)".

You can also insert your image so it takes less work to read; here's a cleaned-up version:
View attachment 30470

Dividing a polynomial P(x) by the multiple of polynomials M(x) and N(x) gives S(x) and a remainder R(x).
When we multiply by M(x) * N(x) and move R(x) on the left hand side ( P(x) - R(x) = M(x) * N(x) * S(x) ), it says that from this we can SEE that dividing P(x) OR R(x) by either M(x) or N(x) inevitably gives us the same remainder. So, basically:

P(x) / M(x) = S'(x) + R'(x)
R(x) / M(x) = S''(x) + R'(x)
( M(x) and N(x) are totally exchangeable )

So, the remainders are the same, even though the quotient is not. That's the point. But why is this ? Why do the last 2 lines hold true ? I don't get it.

Thank you for your feedback by the way, I appreciate it and will incorporate your advice in future submissions

 

[Dr.Peterson]

Dividing a polynomial P(x) by the multiple of polynomials M(x) and N(x) gives S(x) and a remainder R(x).
When we multiply by M(x) * N(x) and move R(x) on the left hand side ( P(x) - R(x) = M(x) * N(x) * S(x) ), it says that from this we can SEE that dividing P(x) OR R(x) by either M(x) or N(x) inevitably gives us the same remainder. So, basically:

P(x) / M(x) = S'(x) + R'(x)
R(x) / M(x) = S''(x) + R'(x)
( M(x) and N(x) are totally exchangeable )

So, the remainders are the same, even though the quotient is not. That's the point. But why is this ? Why do the last 2 lines hold true ? I don't get it.

Thank you for your feedback by the way, I appreciate it and will incorporate your advice in future submissions

I really wish you would do as I asked and show the actual original; an image from book or screen would be good. I could help you far more easily if you showed exactly what you are asking about, rather than describe it in your own words.

In particular, it is nonsense to write "P(x) / M(x) = S'(x) + R'(x)" when you mean that S'(x) is the quotient and R'(x) is the remainder. You simply are not saying what you mean, which would be "P(x) = S'(x)*M(x) + R'(x)". Possibly if you wrote what you meant, you would see it. But presumably the author did, and you are just not understanding what you see! I want to point out what you should be seeing.

But I'll try to extract from this what is actually being said.

Can you see that when you write P(x) - R(x) = M(x) * N(x) * S(x), the right-hand side is a multiple of M(x), and also of N(x)? That is the key idea.

I would say it a little differently.

You are told that when you divide P(x) by [M(x)*N(x)], you get quotient S(x) and remainder R(x). This can be written as P(x) = [M(x)*N(x)] * S(x) + R(x). It also tells us that the degree of R(x) is less than that of the divisor, M(x)*N(x).

But this can also be written as P(x) = M(x)*[N(x)*S(x)] + R(x), which tells us that when you divide P(x) by M(x), you get quotient N(x)*S(x) and remainder R(x). There's a problem, however: In order to say that R(x) is still the remainder, we would need to know that its degree is less than that of M(x), our new divisor.

Now, in the image you gave, you had specific polynomials, and the remainder did in fact have a lower degree than either divisor, so the conclusion is valid in that case. Was the claim made in general, or just about this specific case? That's another reason we really need you to do as we ask, and show the exact problem you are working on. I don't suppose you've read this, have you?

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[Ognjen]

I really wish you would do as I asked and show the actual original; an image from book or screen would be good. I could help you far more easily if you showed exactly what you are asking about, rather than describe it in your own words.

In particular, it is nonsense to write "P(x) / M(x) = S'(x) + R'(x)" when you mean that S'(x) is the quotient and R'(x) is the remainder. You simply are not saying what you mean, which would be "P(x) = S'(x)*M(x) + R'(x)". Possibly if you wrote what you meant, you would see it. But presumably the author did, and you are just not understanding what you see! I want to point out what you should be seeing.

But I'll try to extract from this what is actually being said.

Can you see that when you write P(x) - R(x) = M(x) * N(x) * S(x), the right-hand side is a multiple of M(x), and also of N(x)? That is the key idea.

I would say it a little differently.

You are told that when you divide P(x) by [M(x)*N(x)], you get quotient S(x) and remainder R(x). This can be written as P(x) = [M(x)*N(x)] * S(x) + R(x). It also tells us that the degree of R(x) is less than that of the divisor, M(x)*N(x).

But this can also be written as P(x) = M(x)*[N(x)*S(x)] + R(x), which tells us that when you divide P(x) by M(x), you get quotient N(x)*S(x) and remainder R(x). There's a problem, however: In order to say that R(x) is still the remainder, we would need to know that its degree is less than that of M(x), our new divisor.

Now, in the image you gave, you had specific polynomials, and the remainder did in fact have a lower degree than either divisor, so the conclusion is valid in that case. Was the claim made in general, or just about this specific case? That's another reason we really need you to do as we ask, and show the exact problem you are working on. I don't suppose you've read this, have you?

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Welcome to FreeMathHelp.com! Please take the time to read the following before you make your first post. It will help you to get your math questions answered promptly and in the most helpful manner. A summary is available here, but please read the complete guidelines below at your earliest...

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Post the complete text of the exercise. This would include the full statement of the exercise including the instructions, so the tutors will know what you're working on. Typing exercises word-for-word also helps us identify possible mistakes in class materials. If there's a graphic or table or some other non-textual information necessary, please include a detailed description. You may upload images to our server.​

I am terribly sorry for angering you...
The reason why I didn't include text of the problem is because it is written in another language. I will attempt to translate it verbatim right away.

"Determine the remainder upon dividing polynomial P(x) by ( x^2 + x + 1 ) * ( x^2 + x + 2 ), if dividing P(x) by ( x^2 + x +1 ) gives the remainder of ( x + 1 ), and dividing P(x) by ( x^2 + x + 2 ) gives the remainder of ( x + 2 )."

I still don't quite understand your explanation... Why do the P(x) \ ( x^2 + x + 1 ) remainder and the R(x) \ ( x^2 + x + 1 ) remainder have to be the same number ? I understand why their remainder have to have the highest polynomial degree of one, but I don't understand why do the remainder have to be identical. For example, x + 7 is a polynomial with the degree of 1, but so is x + 3, and they aren't the same polynomials. How is it implied that R(x) / y remainder is the same as P(x) / y remainder ?

Thank you so much for the time invested and I once again apologize for my indecency. I will make sure to read the guidelines before submitting another thread in the future.

 

[Dr.Peterson]

I am terribly sorry for angering you...
The reason why I didn't include text of the problem is because it is written in another language. I will attempt to translate it verbatim right away.

I hope I don't appear too angry, just irritated.

But there's nothing wrong with showing us the original language (especially if it is in a form we can use Google to translate, or accompanied by a translation). Perhaps this should be mentioned in Read Before Posting.

The same is true of the solution you are asking about. We can talk better about something we can see than something we can't.

I still don't quite understand your explanation... Why do the P(x) \ ( x^2 + x + 1 ) remainder and the R(x) \ ( x^2 + x + 1 ) remainder have to be the same number ?

The remainder in these cases is not a number, of course, but a polynomial. What you ask here doesn't quite correspond to what you asked before; what are P(x) and R(x) here? I suppose you are referring to this part of what you wrote, which I haven't commented on yet because I didn't have enough context to know where it fits in:


This corresponds to a basic fact about division of integers: if the difference between two numbers is a multiple of the divisor, then their remainders must be the same. For example, if I divide 123 and 151 by 7, they will leave the same remainder, because 151 - 123 = 28 is a multiple of 7. Indeed, 123/7 = 17 R 4, and 151/7 = 21 R 4. Can you see why? The same reasoning applies to the polynomials.

As I said before, though, this would be easier to talk about if what you wrote made sense. What you really mean by the last line, for example, is [math]\frac{R(x)}{x^2+x+1} = S'(x) + \frac{x+1}{x^2+x+1}[/math] or equivalently [math]R(x) = S'(x)(x^2+x+1) + (x+1)[/math] Write both lines in the latter form (but with perhaps [imath]R_1(x)[/imath] and [imath]R_2(x)[/imath] in place of the remainders), then subtract one from the other and compare with the first line above.

 

[Ognjen]

I hope I don't appear too angry, just irritated.

But there's nothing wrong with showing us the original language (especially if it is in a form we can use Google to translate, or accompanied by a translation). Perhaps this should be mentioned in Read Before Posting.

The same is true of the solution you are asking about. We can talk better about something we can see than something we can't.


The remainder in these cases is not a number, of course, but a polynomial. What you ask here doesn't quite correspond to what you asked before; what are P(x) and R(x) here? I suppose you are referring to this part of what you wrote, which I haven't commented on yet because I didn't have enough context to know where it fits in:
View attachment 30472

This corresponds to a basic fact about division of integers: if the difference between two numbers is a multiple of the divisor, then their remainders must be the same. For example, if I divide 123 and 151 by 7, they will leave the same remainder, because 151 - 123 = 28 is a multiple of 7. Indeed, 123/7 = 17 R 4, and 151/7 = 21 R 4. Can you see why? The same reasoning applies to the polynomials.

As I said before, though, this would be easier to talk about if what you wrote made sense. What you really mean by the last line, for example, is [math]\frac{R(x)}{x^2+x+1} = S'(x) + \frac{x+1}{x^2+x+1}[/math] or equivalently [math]R(x) = S'(x)(x^2+x+1) + (x+1)[/math] Write both lines in the latter form (but with perhaps [imath]R_1(x)[/imath] and [imath]R_2(x)[/imath] in place of the remainders), then subtract one from the other and compare with the first line above.

But how do we know that in this concrete case it holds that the difference of the polynomials is a multiple of the divisor ? Just because we don't have the remainder as a separate term on the right hand side does not mean it's not incorporated in the ( x^2 + x + 2 ) * S(x) polynomial that is on that side, right ? Is it because we know that the ( x^2 + x + 2 ) * S(x) is of the power of 2 at least, and the remainder ( since R(x) is of the degree of 3 and its divisor of the degree of 2 ) would have to be of at least the power of 1 ? Still, it could be incorporated in the ''x + 2'' or whichever fitting part of the aforementioned expression either way, right ?

Thanks for the feedback though, I appreciate it immensely.

 

[Dr.Peterson]

But how do we know that in this concrete case it holds that the difference of the polynomials is a multiple of the divisor ? Just because we don't have the remainder as a separate term on the right hand side does not mean it's not incorporated in the ( x^2 + x + 2 ) * S(x) polynomial that is on that side, right ? Is it because we know that the ( x^2 + x + 2 ) * S(x) is of the power of 2 at least, and the remainder ( since R(x) is of the degree of 3 and its divisor of the degree of 2 ) would have to be of at least the power of 1 ? Still, it could be incorporated in the ''x + 2'' or whichever fitting part of the aforementioned expression either way, right ?

Thanks for the feedback though, I appreciate it immensely.

Forget degrees for the moment; that's a secondary issue. First, think about how polynomials and their factors work (and maybe go back to any theorems you've seen that they might be assuming you know). And think carefully about what you mean by "incorporated".

Did you do what I suggested, and "Write both lines in the latter form (but with perhaps [imath]R_1(x)[/imath] and [imath]R_2(x)[/imath] in place of the remainders), then subtract one from the other and compare with the first line above." (Secondarily, you'll want to observe that those two remainders must have degrees less than 2.)

Show me what you get. (I want you processing all this with your own mind, so you really learn it.)

 

[Ognjen]

Forget degrees for the moment; that's a secondary issue. First, think about how polynomials and their factors work (and maybe go back to any theorems you've seen that they might be assuming you know). And think carefully about what you mean by "incorporated".

Did you do what I suggested, and "Write both lines in the latter form (but with perhaps [imath]R_1(x)[/imath] and [imath]R_2(x)[/imath] in place of the remainders), then subtract one from the other and compare with the first line above." (Secondarily, you'll want to observe that those two remainders must have degrees less than 2.)

Show me what you get. (I want you processing all this with your own mind, so you really learn it.)

[math]\frac{P(x)}{ x^2 + x + 1 } - \frac{R(x)}{ x^2 + x + 1 } = S(x) + R_1(x) - S'(x) - R_2(x)[/math][math]\frac{P(x) - R(x)}{ x^2 + x + 1 } = S(x) - S'(x) + \frac{x + 1 - ( x + 1)}{ x^2 + x + 2 }[/math][math]\frac{P(x) - R(x)}{ x^2 + x + 1 } = S(x) - S'(x)[/math]
The problem is that I don't have the degree of P(x) ( or S(x) ), so this whole expression means very little to me, as far as I can see. The degree of S'(x) is definitely 1, but without knowing the degree of S(x) or P(x), I can't really spot anything new. What am I not seeing ? Or what am I supposed to see at the first place ?

It appears that (S(x) - S'(x)) equals ( x^2 + x + 2) * Q(x), which means that Q(x) is of the degree of zero, and thus not even a polynomial ( just a number factor ), right ? From that I can conclude that P(x) is... of the degree of 4 ? And thus S(x) is of the degree of 2, right ? But what is this all telling me ?

Also, I must have misunderstood you based on the nature of remainders given. (x + 1) / (x^2 + x + 1) is indivisible ( the degree of the dividend is lower than the degree of the divisor ).

Sorry for the late response, I was away for a day. And thank you for your incessant help through this.

 

[Dr.Peterson]

[math]\frac{P(x)}{ x^2 + x + 1 } - \frac{R(x)}{ x^2 + x + 1 } = S(x) + R_1(x) - S'(x) - R_2(x)[/math][math]\frac{P(x) - R(x)}{ x^2 + x + 1 } = S(x) - S'(x) + \frac{x + 1 - ( x + 1)}{ x^2 + x + 2 }[/math][math]\frac{P(x) - R(x)}{ x^2 + x + 1 } = S(x) - S'(x)[/math]
The problem is that I don't have the degree of P(x) ( or S(x) ), so this whole expression means very little to me, as far as I can see. The degree of S'(x) is definitely 1, but without knowing the degree of S(x) or P(x), I can't really spot anything new. What am I not seeing ? Or what am I supposed to see at the first place ?

It appears that (S(x) - S'(x)) equals ( x^2 + x + 2) * Q(x), which means that Q(x) is of the degree of zero, and thus not even a polynomial ( just a number factor ), right ? From that I can conclude that P(x) is... of the degree of 4 ? And thus S(x) is of the degree of 2, right ? But what is this all telling me ?

There are probably several things you're missing; I know I'm not explaining things very clearly, which is in part because I don't know your context, which would give me a better place to start.

But you still haven't seen one of the first things I tried to point out:

In particular, it is nonsense to write "P(x) / M(x) = S'(x) + R'(x)" when you mean that S'(x) is the quotient and R'(x) is the remainder. You simply are not saying what you mean, which would be "P(x) = S'(x)*M(x) + R'(x)".

When you now write [math]\frac{P(x)}{ x^2 + x + 1 } - \frac{R(x)}{ x^2 + x + 1 } = S(x) + R_1(x) - S'(x) - R_2(x)[/math] it is the same nonsense! The result of division is not the quotient plus the remainder; it's the quotient plus a fraction whose numerator is the remainder. As I said again later,

As I said before, though, this would be easier to talk about if what you wrote made sense. What you really mean by the last line, for example, is [math]\frac{R(x)}{x^2+x+1} = S'(x) + \frac{x+1}{x^2+x+1}[/math] or equivalently [math]R(x) = S'(x)(x^2+x+1) + (x+1)[/math] Write both lines in the latter form (but with perhaps [imath]R_1(x)[/imath] and [imath]R_2(x)[/imath] in place of the remainders), then subtract one from the other and compare with the first line above.

If you start with the correct form, things will be easier to see, I think.

Also, I must have misunderstood you based on the nature of remainders given. (x + 1) / (x^2 + x + 1) is indivisible ( the degree of the dividend is lower than the degree of the divisor ).

You're right that "(x + 1) / (x^2 + x + 1) is indivisible"; that is, it is a proper fraction, not a polynomial. That is the purpose of a remainder: its degree has to be lower than that of the divisor so that there is no more to divide.

Perhaps you are confusing the concepts of "remainder" and "fraction part", and you are taking [imath]R_1(x)[/imath] as the fraction rather than the polynomial ([imath]x+1[/imath] in the specific case we're talking about). This is parallel to the fact, using numbers, that if I divide 17 by 3 the quotient is 5 and the remainder is 2; I can write 17/3 = 5 2/3, or 17 = 3*5 + 2. With polynomials, when I divide [imath]x^2+3x-1[/imath] by [imath]x-2[/imath], the quotient is [imath]x+5[/imath] and the remainder is [imath]9[/imath]; I can write [imath]\frac{x^2+3x-1}{x-2}=x+5+\frac{9}{x-2}[/imath].

The point of what I asked you to do is obscured because you just replaced [imath]R_1(x)[/imath] and [imath]R_2(x)[/imath] with ([imath]x+1[/imath], when my intent was to show that these two remainders must be the same, from which you can conclude that they are both ([imath]x+1[/imath].

Would it be possible to show me an image of the original, just so I can be sure what the equations are, even if I can't translate the words? I really don't like working in the dark, without something to point to. (Tutoring face to face, I open a student's book to find the theorems he needs, and so on.) And a translation of the entire explanation you are trying to understand will help me tremendously.

 

[Ognjen]

I got the following:
[math]P(x) - R(x) = ( x^2 + x +1 ) * ( S(x) - S'(x) ) + R_2 - R_1[/math]Knowing that R_2 is is x + 1, I can put it into the equation, but I still don't get why exactly the remainders HAVE to cancel out, that was my sole question from the get-go, although I admittedly found a lot of trouble in expressing myself along the way. I probably misunderstood some part of what you just said again unfortunately, but I will take a picture of both the posed problem and the "solution" to it ( from the workbook ). The number of the problem is 13.

The reason why the solution isn't particularly helpful is because after the P(x) - R(x) expression, it says "and the degree of R(x) is 3." FOLLOWED BY "therefore we conclude that P(x) and R(x) divided by the same polynomial give the same remainder". It just seems like an odd logical leap to me as I read this part of the solution, like there's something elementary and thus not worthy of an explanation that I simply fail to notice, for some reason. Once again, everything else holds true and is clear, as long as I have a clear idea as to WHY the remainders have to be equal, which I don't...

Thank you so much for your help, I'm amazed by your level of dedication, especially considering your activity on this platform as a whole. I highly appreciate the existence of people like you, sincerely.

 

[Dr.Peterson]

I did some OCR and Google translation on the solution you provided; it's obviously not perfect, but here it is:

13. * Decision. The polynomial P(x) when divided by (x^2 + x + 1) (x^2 + x + 2) gives the quotient Q(x) and the remainder R(x) so that​

​

P(x) = (x^2 + x + 1)(x^2 + x + 2) Q(x) + R(x),​

​

that is, P(x) - R(x) = (x^2 + x + 1)(x^2 + x + 2) Q(x), where the degree of the polynomial R(x) is at most 3.​

Hence yes (x^2 + x + 1) | P(x) - R(x), i.e. the polynomials P(x) and R(x) when divided by x^2 + x + 1 give the same remainder.​

Since from the condition of the problem, the remainder when dividing the polynomial x^2 + x + 1 by P(x) is equal to x + 1, and it is the same for R(x), it is​

​

R(x) = (x^2 + x + 1) (ax + b) + x + 1 (1)​

​

where we have applied that the degree of the polynomial R(x) is at most 3. A similar conclusion can be reported when dividing P(x) - R(x) polynomial x^2 + x + 2, so we have​

​

R(x) = (x^2 + x + 2) (cx + d) + x + 2. (2)​

​

From (1) and (2) we find​

​

(x^2 + x + 1) (ax + b) + x + 1 = (x^2 + x + 2) (cx + d) + x + 2,​

​

and hence from the equality of two polynomials follows a = c = 0, b = d = -1. If a = 0 and b = -1 we replace in (1) we obtain the required remainder R(x) = -x^2.​

This is what I have been waiting for: the details of what it says, rather than little excerpts or paraphrases. I am no longer blind.

I put the key part in bold: the claim you are asking about, together with what it says implies that.

The statement that P(x) - R(x) = (x^2 + x + 1)(x^2 + x + 2) Q(x) means that the difference between the given polynomial P(x) and its remainder R(x) is a multiple of (x^2 + x + 1), and any two polynomials that differ by a multiple of the divisor must have the same remainder. Yes, this should be obvious; but I don't know what you have been taught from which it should be obvious!

If you had learned any number theory, which I assume you have not (though they use the symbol for divisibility, which is used there), this is a basic idea in modular congruence. This is what I was talking about in my first answer:

But if you are primarily asking how two polynomials can leave the same remainder, just think about the same question using numbers. If I divide 45 and 97 by 13, I get the same remainder: 45/13 = 3 R 6; 97/13 = 7 R 6. How did that happen (and how did I know they would have the same remainder without having done the divisions yet)?

In this example, 45 = 13*3 + 6, and 97 = 13*7 + 6; the difference between 45 and 97 is 97 - 45 = (13*7 + 6) - (13*3 + 6) = 13(7 - 4) = 13*3 = 39. If they had different remainders, then the difference could not be a multiple of 13.

The same is true for polynomials. Suppose that P(x) = (x^2 + x + 1)q(x) + r(x), so that its quotient is q with remainder r; and R(x) = (x^2 + x + 1)q'(x) + r'(x), with quotient q' and remainder r'. (Both remainders must have degree less than 2.) Then

P(x) - R(x) = [(x^2 + x + 1)q(x) + r(x)] - [(x^2 + x + 1)q'(x) + r'(x)] = (x^2 + x + 1)[q(x) - q'(x)] + [r(x) - r'(x)]​

But since we know P(x) - R(x) is a multiple of (x^2 + x + 1), we can conclude that r(x) - r'(x) = 0. Do you see that? So the two remainders must be the same.

The reason why the solution isn't particularly helpful is because after the P(x) - R(x) expression, it says "and the degree of R(x) is 3." FOLLOWED BY "therefore we conclude that P(x) and R(x) divided by the same polynomial give the same remainder". It just seems like an odd logical leap to me as I read this part of the solution, like there's something elementary and thus not worthy of an explanation that I simply fail to notice, for some reason. Once again, everything else holds true and is clear, as long as I have a clear idea as to WHY the remainders have to be equal, which I don't...

As I've said, the degree is not the central issue; I think you have been distracted by that. The fact that the degree of R(x) is at most 3 is not a condition of the fact you are asking about! It is used later. The equality of the two remainders follows only from the divisibility of the difference, which is the last thing that had been mentioned.

Can you see why I have been begging to see the entire solution you were asking about? It is quite common that students don't give us everything we need, because they don't understand how important the entire context can be to communication. The added difficulty of language differences makes this harder, but it doesn't make it any less necessary.

There's a good chance that you still have questions, so ask them. I'll keep working at it, because I'm stubborn.

 

[blamocur]

This problem has special properties which allow for an easier but ad hoc solution:

[math]P = S (x^2+x+1) + (x+1)[/math][math]P = F (x^2+x+2) + (x+2)[/math]
Since [imath](x^2+x+2) - (x^2+x+1) = 1[/imath] we can do this:

[math]P = P \left((x^2+x+2) - (x^2+x+1)\right)[/math][math]= (S-F)\left[(x^2+x+1)(x^2+x+2)\right] + \left[(x+1)(x^2+x+2) - (x+2)(x^2+x+1)\right][/math][math]= (S-F)\left[(x^2+x+1)(x^2+x+2)\right] + R(x)[/math] where
[math]R(x) = (x+1)(x^2+x+2) - (x+2)(x^2+x+1) = -x^2[/math]

 

[Ognjen]

I did some OCR and Google translation on the solution you provided; it's obviously not perfect, but here it is:

13. * Decision. The polynomial P(x) when divided by (x^2 + x + 1) (x^2 + x + 2) gives the quotient Q(x) and the remainder R(x) so that​

​

P(x) = (x^2 + x + 1)(x^2 + x + 2) Q(x) + R(x),​

​

that is, P(x) - R(x) = (x^2 + x + 1)(x^2 + x + 2) Q(x), where the degree of the polynomial R(x) is at most 3.​

Hence yes (x^2 + x + 1) | P(x) - R(x), i.e. the polynomials P(x) and R(x) when divided by x^2 + x + 1 give the same remainder.​

Since from the condition of the problem, the remainder when dividing the polynomial x^2 + x + 1 by P(x) is equal to x + 1, and it is the same for R(x), it is​

​

R(x) = (x^2 + x + 1) (ax + b) + x + 1 (1)​

​

where we have applied that the degree of the polynomial R(x) is at most 3. A similar conclusion can be reported when dividing P(x) - R(x) polynomial x^2 + x + 2, so we have​

​

R(x) = (x^2 + x + 2) (cx + d) + x + 2. (2)​

​

From (1) and (2) we find​

​

(x^2 + x + 1) (ax + b) + x + 1 = (x^2 + x + 2) (cx + d) + x + 2,​

​

and hence from the equality of two polynomials follows a = c = 0, b = d = -1. If a = 0 and b = -1 we replace in (1) we obtain the required remainder R(x) = -x^2.​

This is what I have been waiting for: the details of what it says, rather than little excerpts or paraphrases. I am no longer blind.

I put the key part in bold: the claim you are asking about, together with what it says implies that.

The statement that P(x) - R(x) = (x^2 + x + 1)(x^2 + x + 2) Q(x) means that the difference between the given polynomial P(x) and its remainder R(x) is a multiple of (x^2 + x + 1), and any two polynomials that differ by a multiple of the divisor must have the same remainder. Yes, this should be obvious; but I don't know what you have been taught from which it should be obvious!

If you had learned any number theory, which I assume you have not (though they use the symbol for divisibility, which is used there), this is a basic idea in modular congruence. This is what I was talking about in my first answer:

In this example, 45 = 13*3 + 6, and 97 = 13*7 + 6; the difference between 45 and 97 is 97 - 45 = (13*7 + 6) - (13*3 + 6) = 13(7 - 4) = 13*3 = 39. If they had different remainders, then the difference could not be a multiple of 13.

The same is true for polynomials. Suppose that P(x) = (x^2 + x + 1)q(x) + r(x), so that its quotient is q with remainder r; and R(x) = (x^2 + x + 1)q'(x) + r'(x), with quotient q' and remainder r'. (Both remainders must have degree less than 2.) Then

P(x) - R(x) = [(x^2 + x + 1)q(x) + r(x)] - [(x^2 + x + 1)q'(x) + r'(x)] = (x^2 + x + 1)[q(x) - q'(x)] + [r(x) - r'(x)]​

But since we know P(x) - R(x) is a multiple of (x^2 + x + 1), we can conclude that r(x) - r'(x) = 0. Do you see that? So the two remainders must be the same.


As I've said, the degree is not the central issue; I think you have been distracted by that. The fact that the degree of R(x) is at most 3 is not a condition of the fact you are asking about! It is used later. The equality of the two remainders follows only from the divisibility of the difference, which is the last thing that had been mentioned.

Can you see why I have been begging to see the entire solution you were asking about? It is quite common that students don't give us everything we need, because they don't understand how important the entire context can be to communication. The added difficulty of language differences makes this harder, but it doesn't make it any less necessary.

There's a good chance that you still have questions, so ask them. I'll keep working at it, because I'm stubborn.

Oh I get it now !

Thank you so much for your devotion and I once again apologize for the inconvenience caused by my ignorance.

 

[Ognjen]

This problem has special properties which allow for an easier but ad hoc solution:

[math]P = S (x^2+x+1) + (x+1)[/math][math]P = F (x^2+x+2) + (x+2)[/math]
Since [imath](x^2+x+2) - (x^2+x+1) = 1[/imath] we can do this:

[math]P = P \left((x^2+x+2) - (x^2+x+1)\right)[/math][math]= (S-F)\left[(x^2+x+1)(x^2+x+2)\right] + \left[(x+1)(x^2+x+2) - (x+2)(x^2+x+1)\right][/math][math]= (S-F)\left[(x^2+x+1)(x^2+x+2)\right] + R(x)[/math] where
[math]R(x) = (x+1)(x^2+x+2) - (x+2)(x^2+x+1) = -x^2[/math]

Thank you ! This is a very neat solution.