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This question arose while I was tutoring a grade 11 student on the subjects of polynomial division and the Remainder Theorem, and I could not provide a satisfactory answer to it. Presumably, $f(x)$ is a polynomial. I am aware we can write $f(x)$ in the form $$ f(x) = d(x)q(x) + r(x) $$ representing its division by $d(x)$, where $q(x)$ is the quotient of the division and $r(x)$ is the remainder, with deg $r(x) < $ deg $d(x)$. But for $$ f(x) = (x+5)q(x) + (x+3), $$ assuming $r(x) = (x+3)$, the strict inequality does not hold. I tried analogously shifting the question to integers, where, for example, we can write 27 upon the division of 6 as $$ 27 = (6)(4) + 3 $$ and we would then have the smallest multiple of 6 greater than 27 is one more than the quotient, so 5. But I am not sure how to extend this to the question posed.

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    $\begingroup$ What is the question? There are many ways to find $q,r$, but only one where the $r$ has degree smaller than $d$. $\endgroup$ – vadim123 Nov 19 '16 at 19:07
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    $\begingroup$ What is the order relation on polynomials? $\endgroup$ – Bernard Nov 19 '16 at 19:10
  • $\begingroup$ What do you mean by "multiple"? What do you mean by "first" and what do you mean by $> f(x)$? None of those have clear and unambiguous meanings. It is obvious that $(x+5)(q(x) + 1) = f(x)+2 > f(x)$, but it isn't clear that this is a multiple (it's not a scalar multiple), that it is the "first" and it's worth noting that $(x+5)(q(x) + 2) =f(x) + x+7 \not > f(x)$ . Your question may or may not be meaningful. $\endgroup$ – fleablood Nov 20 '16 at 1:09
  • $\begingroup$ It might help to know how this question arose--was it part of a homework exercise, something the student thought of, or something else? $\endgroup$ – David K Nov 20 '16 at 2:53
  • $\begingroup$ @fleablood I am not sure what I mean. The question was exactly as posed. Glad to see I am not the only one who found it ambiguous. $\endgroup$ – Shaun Nov 20 '16 at 21:08
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Observe that \begin{align} (x+5)(q(x) + 1) &= (x+5)q(x) + (x+5) \\ &= (x+5)q(x) + (x+3) + 2 \\ &= f(x) + 2 \\ &> f(x). \end{align} So $(x+5)(q(x) + 1)$ is a multiple of $x+5$ that is greater than $f(x)$.

Whether this is the "first" such multiple of $x+5$ is a matter of interpretation, because for $x=-4$, we have $f(x) < (x+5)q(x) < (x+5)(q(x) + 1)$. I suppose the idea of the exercise was that $q(x)$ is not the quotient in the polynomial division of $f(x)$ by $x+5$; $$ f(x) = (x+5)(q(x) + 1) - 2, $$ and therefore the remainder is $-2$, and the quotient is $q(x) + 1$ which is the other factor in the "multiple" of $q+5$ that you were asked to find.

Coincidence? I think not. Admittedly, it does seem like a rather awkward way to think about polynomial division.

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  • $\begingroup$ Yeah I'm going to say the question itself is pretty ambiguous. But thank you for your insight! I did end up figuring out that the intended approach was probably manipulating the expression to obtain the correct quotient for polynomial division of $f(x)$ by $(x+5)$. $\endgroup$ – Shaun Nov 20 '16 at 21:11
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I am not clear with your question, but answering on what i have understood from your argument.

I guess your question is not correct. For polynomial division, remainder theorem implies that for f(x) and d(x) there exists some q(x) and r(x) such that f(x) = d(x)*q(x) + r(x) where deg(r(x)) < deg(d(x)) is satisfied.

For the question you are choosing r(x) and d(x) on your own and trying to violate the theorem which is not correct.

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  • $\begingroup$ The question given was exactly as I stated. So are you saying the question is incorrect? $\endgroup$ – Shaun Nov 19 '16 at 19:39
  • $\begingroup$ it is incorrect way of interpreting the theorem i am saying, theorem says with f(x) and (x+5) there is some q(x) and r(x) such that f(x) = (x+5)q(x) + r(x) where deg(r(x)) < deg(x+5) like f(x)=x^2-25 we can have q(x) = x-5 and r(x) =0 satisfies that theorem, you are chosing d(x) and r(x) and finding the f(x) and q(x) which satisfies the theorem which is wrong, you can only chose f(x) and d(x) and find the q(x) and r(x) which satisfies it. $\endgroup$ – Ankit Vallecha Nov 19 '16 at 19:51
  • $\begingroup$ So in this case, we can rearrange the given expression to be (x+5)(q(x)+1) - 2. And extending from my analogy to integer division, the next multiple of (x+5) greater than f(x) should be (q(x)+1)+1, or q(x)+2? $\endgroup$ – Shaun Nov 19 '16 at 20:00
  • $\begingroup$ first of all there is nothing greater or smaller in polynomials, eg we cant say x^2 > x means x^2 is greater than x, these are inequalities that are true for certain values of x, like this one is true for x>1 or x<-1 and false for 0<=x<=1. so for x>1 or x<-1 x^2 is greater than x, but for 0<=x<=1 x^2 is smaller than x $\endgroup$ – Ankit Vallecha Nov 19 '16 at 20:15
  • $\begingroup$ got your doubt cleared? $\endgroup$ – Ankit Vallecha Nov 19 '16 at 21:09

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