Why do we automatically assume that when we divide a polynomial by a second degree polynomial the remainder...












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Given question:




If a polynomial leaves a remainder of $5$ when divided by $x − 3$ and a remainder of $−7$ when divided by $x + 1$,
what is the remainder when the polynomial is divided by $x^2 − 2x − 3$?




Solution:




We observe that when we divide by a second degree polynomial the remainder will generally be linear. Thus
the division statement becomes
$p(x) = (x^2 − 2x − 3)q(x) + ax + b $




Can someone please explain at a PRE-CALCULUS level? Thanks






algebra-precalculus polynomials







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  • 21




    $begingroup$
    Compare to long division with whole numbers: if you divide by $10$, say, then the remainder will be less than $10$. Why? Because if not, then you stopped dividing too early.
    $endgroup$
    – Théophile
    Dec 12 '18 at 18:50
















15












$begingroup$


Given question:




If a polynomial leaves a remainder of $5$ when divided by $x − 3$ and a remainder of $−7$ when divided by $x + 1$,
what is the remainder when the polynomial is divided by $x^2 − 2x − 3$?




Solution:




We observe that when we divide by a second degree polynomial the remainder will generally be linear. Thus
the division statement becomes
$p(x) = (x^2 − 2x − 3)q(x) + ax + b $




Can someone please explain at a PRE-CALCULUS level? Thanks






algebra-precalculus polynomials







share|cite|improve this question











$endgroup$








  • 21




    $begingroup$
    Compare to long division with whole numbers: if you divide by $10$, say, then the remainder will be less than $10$. Why? Because if not, then you stopped dividing too early.
    $endgroup$
    – Théophile
    Dec 12 '18 at 18:50














15












15








15


2



$begingroup$


Given question:




If a polynomial leaves a remainder of $5$ when divided by $x − 3$ and a remainder of $−7$ when divided by $x + 1$,
what is the remainder when the polynomial is divided by $x^2 − 2x − 3$?




Solution:




We observe that when we divide by a second degree polynomial the remainder will generally be linear. Thus
the division statement becomes
$p(x) = (x^2 − 2x − 3)q(x) + ax + b $




Can someone please explain at a PRE-CALCULUS level? Thanks






algebra-precalculus polynomials







share|cite|improve this question











$endgroup$




Given question:




If a polynomial leaves a remainder of $5$ when divided by $x − 3$ and a remainder of $−7$ when divided by $x + 1$,
what is the remainder when the polynomial is divided by $x^2 − 2x − 3$?




Solution:




We observe that when we divide by a second degree polynomial the remainder will generally be linear. Thus
the division statement becomes
$p(x) = (x^2 − 2x − 3)q(x) + ax + b $




Can someone please explain at a PRE-CALCULUS level? Thanks






algebra-precalculus polynomials




algebra-precalculus polynomials






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edited Dec 12 '18 at 23:24









Eric Wofsey

192k14217351




192k14217351










asked Dec 12 '18 at 15:50









Ke KeKe Ke

1028




1028








  • 21




    $begingroup$
    Compare to long division with whole numbers: if you divide by $10$, say, then the remainder will be less than $10$. Why? Because if not, then you stopped dividing too early.
    $endgroup$
    – Théophile
    Dec 12 '18 at 18:50














  • 21




    $begingroup$
    Compare to long division with whole numbers: if you divide by $10$, say, then the remainder will be less than $10$. Why? Because if not, then you stopped dividing too early.
    $endgroup$
    – Théophile
    Dec 12 '18 at 18:50








21




21




$begingroup$
Compare to long division with whole numbers: if you divide by $10$, say, then the remainder will be less than $10$. Why? Because if not, then you stopped dividing too early.
$endgroup$
– Théophile
Dec 12 '18 at 18:50




$begingroup$
Compare to long division with whole numbers: if you divide by $10$, say, then the remainder will be less than $10$. Why? Because if not, then you stopped dividing too early.
$endgroup$
– Théophile
Dec 12 '18 at 18:50










6 Answers
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active

oldest

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Because by definition the quotient and the remainder of the division of a polynomial $p_1(x)$ by a polynomial $p_2(x)$ are polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that





  1. $p_1(x)=p_2(x)q(x)+r(x)$;


  2. $r(x)=0$ or its degree is smaller than the degree of $p_2(x)$.


In particular, if $p_2(x)$ is quadratic, then the degree of $r(x)$ is at most $1$.






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  • 6




    $begingroup$
    This answer assumes that $p_2$ and $r$ exist, which is not trivial.
    $endgroup$
    – Solomonoff's Secret
    Dec 12 '18 at 20:31








  • 2




    $begingroup$
    @Solomonoff'sSecret: How does this answer assume that $q$ (I don't think you meant $p_2$) and $r$ exist?
    $endgroup$
    – Pakk
    Dec 12 '18 at 21:14






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    @Solomonoff'sSecret I would say it's pretty trivial that $p_2$ exists...
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    – HelloGoodbye
    Dec 12 '18 at 22:48






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    $begingroup$
    @Solomonoff'sSecret Actually, the answer doesn't assume that it exists. The answer just states that: if it exists, then it's true by definition.
    $endgroup$
    – user202729
    Dec 13 '18 at 11:22








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    $begingroup$
    @user202729 It uses the word "the" which suggests both existence and uniqueness.
    $endgroup$
    – Oscar Cunningham
    Dec 13 '18 at 15:41



















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While José Carlos Santos's answer is correct, I think it might be useful to talk about why $r(x)$ is defined to have degree less than $p_2(x)$. (J.G. explains this, but at a higher level than I suspect someone wanting "pre-calculus" would follow.)



If the degree of $r(x)$ is the same size or greater than $p_2(x)$, then you can subtract off a multiple of $p_2(x)$ from $r(x)$ to get a new remainder of lower degree. I.e., we can continue the division process until $r(x)$ is of lower degree than $p_2(x)$. It is only then that we have to stop and call anything left over the remainder.



For example, dividing $3x^3 - 2x^2 + x - 5$ by $x^2 + 1$ follows these steps:




  1. Note that the ratio of the leading terms is $frac {3x^3}{x^2} = 3x$

  2. Multiply $x^2 + 1$ by $3x$ and subtract the result ($3x^3 + 3x$) from $3x^3 - 2x^2 + x - 5$, leaving $-2x^2 - 2x - 5$. I.e.,$$3x^3 - 2x^2 + x - 5 = (3x)(x^2 + 1) + (-2x^2 - 2x - 5)$$

  3. Note that the ratio of the leading terms of the remainder and $x^2 + 1$ is $frac {-2x^2}{x^2} = -2$.

  4. Multiply $x^2 + 1$ by $-2$ and subtract the result ($-2x^2 - 2$) from $-2x^2 - 2x - 5$, leaving $-2x - 3$. I.e., $$-2x^2 - 2x - 5 = (-2)(x^2 + 1) + (-2x-3)$$

  5. Note that the ratio of the leading terms is now $frac {-2x}{x^2} = frac {-2}x$, which is not a polynomial, so we cannot continue.


Combining the results from steps 2 and 4:
$$begin{align}3x^3 - 2x^2 + x - 5 &= (3x)(x^2 + 1) + (-2)(x^2 + 1) + (-2x-3)\&=(3x - 2)(x^2 + 1) + (-2x-3)end{align}$$



If you stop at step 2, then $r(x)$ is indeed not linear. But at that stage, we could continue the division to get something smaller. It was only when the remainder was of lower degree that we had to stop.






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  • $begingroup$
    I'm not American, but from what I've read pre-calc includes induction. Was my use of modulo arithmetic the comprehensibility issue?
    $endgroup$
    – J.G.
    Dec 12 '18 at 23:35






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    @J.G. - It is true that both induction and modular arithmetic are both taught pre-calculus, But not to everyone, and many of those who have learned them have never truly mastered them. I took the "pre-calculus" request not as a specific level to be met, but rather to mean that the OP wanted as simple an explanation as possible. Since it is an algebra question, and the explanation only requires simple algebra, that is what I gave. "Master the minimum" is a teaching maxim that applies here, in my opinion.
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    – Paul Sinclair
    Dec 12 '18 at 23:52










  • $begingroup$
    Thanks for that clarification.
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    – J.G.
    Dec 13 '18 at 7:14










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    @PaulSinclair That's also why I posted my answer. I wanted to get the principle across at the most beaux level I could.
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    – timtfj
    Dec 13 '18 at 23:24










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    @timtfj - but what I don't understand is what your post added that mine did not contain. I mentioned J.G.'s post because I wanted to acknowledge that he had already addressed the issue, and my post was just to explain it in a more simple manner. But I don't see anything in your post 2.5 hours later that I didn't cover as or more simply.
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    – Paul Sinclair
    Dec 14 '18 at 0:30



















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The point is we can prove as much. Let the quadratic be $q:=ax^2+bx+c$. Working modulo $q$, $x^2=-(bx+c)/a$. Each $x^n$ is then of the form $cx+d$ modulo $q$; you can prove this by induction (you can even get recursion relations on the coefficients). For example, $x^3=-(bx^2+cx)/a$, but we can turn replace $x^2$ as above.






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  • $begingroup$
    To PROVE you need to have a degree-n polynomial, not a quadratic. But it works exactly like that. so it's a_n x^n + a_(n-1) x^(n-1) + .. + a_0, etc.
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    – user3445853
    Dec 13 '18 at 11:12










  • $begingroup$
    @user3445853 I don't understand what point you're trying to make. Are you emphasising that the remainder on division by a degree $n$ polynomial is, provably by this strategy, of degree $<n$?
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    – J.G.
    Dec 13 '18 at 11:23



















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An explanation rather than a full proof. Trying to describe it more simply:



Suppose your first polynomial is
$$P=ax^4+bx^3+cx^2+dx+e$$
and you want to divide it by
$$Q=fx^2+gx+h$$



The first step is to multiply $Q$ by something which will turn its first term into $ax^4$. You've now got a polynomial which, when you subtract it from $P$, will make $ax^4$ disappear.



Then you do the same again, except you arrange for the $x^3$ term to disappear.



Now you have two quadratics: what's left of $P$, and $Q$.



So, multiply $Q$ by a suitable constant to make the $x^2$ terms match, and subtract it. Now all that's left is a multiple of $x$ and a constant, since you've got rid of all the higher order terms—and that's your linear remainder.






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    Simply explaining: the key IDEA is that you are not done dividing until your remainder is not further divisible by your divisor. The key WORD in your post is "GENERALLY" because the conclusion is not hard and fast, it is simply "usually" so.



    IDEA first. Consider simple numbers, perhaps dividing 1,034 by 10. You begin at the left and work right (this is taught in first grade of course so I won't belabor the process) until you cannot usefully continue. So you get 100 R 34, and 3 R 4, and stop when left with a remainder (4) less than your divisor, 10. Easy enough to figure when it's "less than" with numbers, but slightly less so with polynomials. But we see it slightly differently with polynomials: we stop when the order of the remainder is lower than the order of the divisor. So dividing by 7x^2, and having a remainder with just x^2, we usually grit our teeth (we DO love our integers...) and use "1/7" eliminating the remainder's 2nd order term (x^2). At that point, coninuing becomes difficult even with gritted teeth.



    And so we stop. Notice that it is not that the remainder is now linear with that being special in some way: it is simply of lower order than the divisor. In the case of dividing by a 2nd order polynomial, that means it is 1st order, or simply a constant (number).



    Which brings in the "generally" part: in practice, it COULD be just the constant, and not a linear equation. "x^2 + 8" divided by x^2, for example. But not usually. Hence "generally" in your post's statement. And moving up in order, say a fifth order divisor, while you'd usually expect, just out of probability, a fourth order remainder, it could also be a third, second, or first order remainder, or just a constant.



    So it's not a great theorem with a stunning proof, just a practical observation based upon the mechanics we use for the division.



    (And it adds an "idiot check" in case one stops early, thinking he's finished. "Wait, I can't be, it ought to be linear and it isn't so maybe I need to keep working...")






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      0












      $begingroup$

      The accepted answer is perfect, of course, and I would not have posted this as an answer if I had enough reputation to comment... but I haven't, so this is my answer (or really a follow-up-question).



      I wonder if the solution provided in the question $p(x)=(x^2−2x−3)q(x)+ax+b$ is the "full" solution. This solution describes only the form of the remainder: a linear term.



      With the actual terms given in the question, it is possible to determine actual values for $a$ and $b$, being $3$ and $-4$. The remainder therefore is the term $3x-4$.



      I found this solution by setting up an Excel sheet and trying out a couple (hundred) of combinations, but I now wonder if there is a way of finding this term without using "brute force".



      Note: if you multiply the two terms given in the question $(x-3)$ and $(x+1)$ you get the third term $(x^2-2x-3)$, so this question boils down to something like "remainder when dividing by 3, remainder when dividing by 5, remainder when dividing by 15", which might have an impact on solving it.






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      • $begingroup$
        Ke Ke only quoted the part of the solution relevant to the question they asked. If you want to know how to finish solving the problem, your question, even through it relates to this one, should be asked in a new thread. You can link back to this thread in your post, as relevant information, but it is preferable that separate questions have their own threads. This allows the added question to be searchable, and keeps it clear which answers apply to which questions.
        $endgroup$
        – Paul Sinclair
        Dec 14 '18 at 14:00














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      6 Answers
      6






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      6 Answers
      6






      active

      oldest

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      active

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      active

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      21












      $begingroup$

      Because by definition the quotient and the remainder of the division of a polynomial $p_1(x)$ by a polynomial $p_2(x)$ are polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that





      1. $p_1(x)=p_2(x)q(x)+r(x)$;


      2. $r(x)=0$ or its degree is smaller than the degree of $p_2(x)$.


      In particular, if $p_2(x)$ is quadratic, then the degree of $r(x)$ is at most $1$.






      share|cite|improve this answer









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      • 6




        $begingroup$
        This answer assumes that $p_2$ and $r$ exist, which is not trivial.
        $endgroup$
        – Solomonoff's Secret
        Dec 12 '18 at 20:31








      • 2




        $begingroup$
        @Solomonoff'sSecret: How does this answer assume that $q$ (I don't think you meant $p_2$) and $r$ exist?
        $endgroup$
        – Pakk
        Dec 12 '18 at 21:14






      • 2




        $begingroup$
        @Solomonoff'sSecret I would say it's pretty trivial that $p_2$ exists...
        $endgroup$
        – HelloGoodbye
        Dec 12 '18 at 22:48






      • 1




        $begingroup$
        @Solomonoff'sSecret Actually, the answer doesn't assume that it exists. The answer just states that: if it exists, then it's true by definition.
        $endgroup$
        – user202729
        Dec 13 '18 at 11:22








      • 1




        $begingroup$
        @user202729 It uses the word "the" which suggests both existence and uniqueness.
        $endgroup$
        – Oscar Cunningham
        Dec 13 '18 at 15:41
















      21












      $begingroup$

      Because by definition the quotient and the remainder of the division of a polynomial $p_1(x)$ by a polynomial $p_2(x)$ are polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that





      1. $p_1(x)=p_2(x)q(x)+r(x)$;


      2. $r(x)=0$ or its degree is smaller than the degree of $p_2(x)$.


      In particular, if $p_2(x)$ is quadratic, then the degree of $r(x)$ is at most $1$.






      share|cite|improve this answer









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      • 6




        $begingroup$
        This answer assumes that $p_2$ and $r$ exist, which is not trivial.
        $endgroup$
        – Solomonoff's Secret
        Dec 12 '18 at 20:31








      • 2




        $begingroup$
        @Solomonoff'sSecret: How does this answer assume that $q$ (I don't think you meant $p_2$) and $r$ exist?
        $endgroup$
        – Pakk
        Dec 12 '18 at 21:14






      • 2




        $begingroup$
        @Solomonoff'sSecret I would say it's pretty trivial that $p_2$ exists...
        $endgroup$
        – HelloGoodbye
        Dec 12 '18 at 22:48






      • 1




        $begingroup$
        @Solomonoff'sSecret Actually, the answer doesn't assume that it exists. The answer just states that: if it exists, then it's true by definition.
        $endgroup$
        – user202729
        Dec 13 '18 at 11:22








      • 1




        $begingroup$
        @user202729 It uses the word "the" which suggests both existence and uniqueness.
        $endgroup$
        – Oscar Cunningham
        Dec 13 '18 at 15:41














      21












      21








      21





      $begingroup$

      Because by definition the quotient and the remainder of the division of a polynomial $p_1(x)$ by a polynomial $p_2(x)$ are polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that





      1. $p_1(x)=p_2(x)q(x)+r(x)$;


      2. $r(x)=0$ or its degree is smaller than the degree of $p_2(x)$.


      In particular, if $p_2(x)$ is quadratic, then the degree of $r(x)$ is at most $1$.






      share|cite|improve this answer









      $endgroup$



      Because by definition the quotient and the remainder of the division of a polynomial $p_1(x)$ by a polynomial $p_2(x)$ are polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that





      1. $p_1(x)=p_2(x)q(x)+r(x)$;


      2. $r(x)=0$ or its degree is smaller than the degree of $p_2(x)$.


      In particular, if $p_2(x)$ is quadratic, then the degree of $r(x)$ is at most $1$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 12 '18 at 15:53









      José Carlos SantosJosé Carlos Santos

      172k22132239




      172k22132239








      • 6




        $begingroup$
        This answer assumes that $p_2$ and $r$ exist, which is not trivial.
        $endgroup$
        – Solomonoff's Secret
        Dec 12 '18 at 20:31








      • 2




        $begingroup$
        @Solomonoff'sSecret: How does this answer assume that $q$ (I don't think you meant $p_2$) and $r$ exist?
        $endgroup$
        – Pakk
        Dec 12 '18 at 21:14






      • 2




        $begingroup$
        @Solomonoff'sSecret I would say it's pretty trivial that $p_2$ exists...
        $endgroup$
        – HelloGoodbye
        Dec 12 '18 at 22:48






      • 1




        $begingroup$
        @Solomonoff'sSecret Actually, the answer doesn't assume that it exists. The answer just states that: if it exists, then it's true by definition.
        $endgroup$
        – user202729
        Dec 13 '18 at 11:22








      • 1




        $begingroup$
        @user202729 It uses the word "the" which suggests both existence and uniqueness.
        $endgroup$
        – Oscar Cunningham
        Dec 13 '18 at 15:41














      • 6




        $begingroup$
        This answer assumes that $p_2$ and $r$ exist, which is not trivial.
        $endgroup$
        – Solomonoff's Secret
        Dec 12 '18 at 20:31








      • 2




        $begingroup$
        @Solomonoff'sSecret: How does this answer assume that $q$ (I don't think you meant $p_2$) and $r$ exist?
        $endgroup$
        – Pakk
        Dec 12 '18 at 21:14






      • 2




        $begingroup$
        @Solomonoff'sSecret I would say it's pretty trivial that $p_2$ exists...
        $endgroup$
        – HelloGoodbye
        Dec 12 '18 at 22:48






      • 1




        $begingroup$
        @Solomonoff'sSecret Actually, the answer doesn't assume that it exists. The answer just states that: if it exists, then it's true by definition.
        $endgroup$
        – user202729
        Dec 13 '18 at 11:22








      • 1




        $begingroup$
        @user202729 It uses the word "the" which suggests both existence and uniqueness.
        $endgroup$
        – Oscar Cunningham
        Dec 13 '18 at 15:41








      6




      6




      $begingroup$
      This answer assumes that $p_2$ and $r$ exist, which is not trivial.
      $endgroup$
      – Solomonoff's Secret
      Dec 12 '18 at 20:31






      $begingroup$
      This answer assumes that $p_2$ and $r$ exist, which is not trivial.
      $endgroup$
      – Solomonoff's Secret
      Dec 12 '18 at 20:31






      2




      2




      $begingroup$
      @Solomonoff'sSecret: How does this answer assume that $q$ (I don't think you meant $p_2$) and $r$ exist?
      $endgroup$
      – Pakk
      Dec 12 '18 at 21:14




      $begingroup$
      @Solomonoff'sSecret: How does this answer assume that $q$ (I don't think you meant $p_2$) and $r$ exist?
      $endgroup$
      – Pakk
      Dec 12 '18 at 21:14




      2




      2




      $begingroup$
      @Solomonoff'sSecret I would say it's pretty trivial that $p_2$ exists...
      $endgroup$
      – HelloGoodbye
      Dec 12 '18 at 22:48




      $begingroup$
      @Solomonoff'sSecret I would say it's pretty trivial that $p_2$ exists...
      $endgroup$
      – HelloGoodbye
      Dec 12 '18 at 22:48




      1




      1




      $begingroup$
      @Solomonoff'sSecret Actually, the answer doesn't assume that it exists. The answer just states that: if it exists, then it's true by definition.
      $endgroup$
      – user202729
      Dec 13 '18 at 11:22






      $begingroup$
      @Solomonoff'sSecret Actually, the answer doesn't assume that it exists. The answer just states that: if it exists, then it's true by definition.
      $endgroup$
      – user202729
      Dec 13 '18 at 11:22






      1




      1




      $begingroup$
      @user202729 It uses the word "the" which suggests both existence and uniqueness.
      $endgroup$
      – Oscar Cunningham
      Dec 13 '18 at 15:41




      $begingroup$
      @user202729 It uses the word "the" which suggests both existence and uniqueness.
      $endgroup$
      – Oscar Cunningham
      Dec 13 '18 at 15:41











      12












      $begingroup$

      While José Carlos Santos's answer is correct, I think it might be useful to talk about why $r(x)$ is defined to have degree less than $p_2(x)$. (J.G. explains this, but at a higher level than I suspect someone wanting "pre-calculus" would follow.)



      If the degree of $r(x)$ is the same size or greater than $p_2(x)$, then you can subtract off a multiple of $p_2(x)$ from $r(x)$ to get a new remainder of lower degree. I.e., we can continue the division process until $r(x)$ is of lower degree than $p_2(x)$. It is only then that we have to stop and call anything left over the remainder.



      For example, dividing $3x^3 - 2x^2 + x - 5$ by $x^2 + 1$ follows these steps:




      1. Note that the ratio of the leading terms is $frac {3x^3}{x^2} = 3x$

      2. Multiply $x^2 + 1$ by $3x$ and subtract the result ($3x^3 + 3x$) from $3x^3 - 2x^2 + x - 5$, leaving $-2x^2 - 2x - 5$. I.e.,$$3x^3 - 2x^2 + x - 5 = (3x)(x^2 + 1) + (-2x^2 - 2x - 5)$$

      3. Note that the ratio of the leading terms of the remainder and $x^2 + 1$ is $frac {-2x^2}{x^2} = -2$.

      4. Multiply $x^2 + 1$ by $-2$ and subtract the result ($-2x^2 - 2$) from $-2x^2 - 2x - 5$, leaving $-2x - 3$. I.e., $$-2x^2 - 2x - 5 = (-2)(x^2 + 1) + (-2x-3)$$

      5. Note that the ratio of the leading terms is now $frac {-2x}{x^2} = frac {-2}x$, which is not a polynomial, so we cannot continue.


      Combining the results from steps 2 and 4:
      $$begin{align}3x^3 - 2x^2 + x - 5 &= (3x)(x^2 + 1) + (-2)(x^2 + 1) + (-2x-3)\&=(3x - 2)(x^2 + 1) + (-2x-3)end{align}$$



      If you stop at step 2, then $r(x)$ is indeed not linear. But at that stage, we could continue the division to get something smaller. It was only when the remainder was of lower degree that we had to stop.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I'm not American, but from what I've read pre-calc includes induction. Was my use of modulo arithmetic the comprehensibility issue?
        $endgroup$
        – J.G.
        Dec 12 '18 at 23:35






      • 1




        $begingroup$
        @J.G. - It is true that both induction and modular arithmetic are both taught pre-calculus, But not to everyone, and many of those who have learned them have never truly mastered them. I took the "pre-calculus" request not as a specific level to be met, but rather to mean that the OP wanted as simple an explanation as possible. Since it is an algebra question, and the explanation only requires simple algebra, that is what I gave. "Master the minimum" is a teaching maxim that applies here, in my opinion.
        $endgroup$
        – Paul Sinclair
        Dec 12 '18 at 23:52










      • $begingroup$
        Thanks for that clarification.
        $endgroup$
        – J.G.
        Dec 13 '18 at 7:14










      • $begingroup$
        @PaulSinclair That's also why I posted my answer. I wanted to get the principle across at the most beaux level I could.
        $endgroup$
        – timtfj
        Dec 13 '18 at 23:24










      • $begingroup$
        @timtfj - but what I don't understand is what your post added that mine did not contain. I mentioned J.G.'s post because I wanted to acknowledge that he had already addressed the issue, and my post was just to explain it in a more simple manner. But I don't see anything in your post 2.5 hours later that I didn't cover as or more simply.
        $endgroup$
        – Paul Sinclair
        Dec 14 '18 at 0:30
















      12












      $begingroup$

      While José Carlos Santos's answer is correct, I think it might be useful to talk about why $r(x)$ is defined to have degree less than $p_2(x)$. (J.G. explains this, but at a higher level than I suspect someone wanting "pre-calculus" would follow.)



      If the degree of $r(x)$ is the same size or greater than $p_2(x)$, then you can subtract off a multiple of $p_2(x)$ from $r(x)$ to get a new remainder of lower degree. I.e., we can continue the division process until $r(x)$ is of lower degree than $p_2(x)$. It is only then that we have to stop and call anything left over the remainder.



      For example, dividing $3x^3 - 2x^2 + x - 5$ by $x^2 + 1$ follows these steps:




      1. Note that the ratio of the leading terms is $frac {3x^3}{x^2} = 3x$

      2. Multiply $x^2 + 1$ by $3x$ and subtract the result ($3x^3 + 3x$) from $3x^3 - 2x^2 + x - 5$, leaving $-2x^2 - 2x - 5$. I.e.,$$3x^3 - 2x^2 + x - 5 = (3x)(x^2 + 1) + (-2x^2 - 2x - 5)$$

      3. Note that the ratio of the leading terms of the remainder and $x^2 + 1$ is $frac {-2x^2}{x^2} = -2$.

      4. Multiply $x^2 + 1$ by $-2$ and subtract the result ($-2x^2 - 2$) from $-2x^2 - 2x - 5$, leaving $-2x - 3$. I.e., $$-2x^2 - 2x - 5 = (-2)(x^2 + 1) + (-2x-3)$$

      5. Note that the ratio of the leading terms is now $frac {-2x}{x^2} = frac {-2}x$, which is not a polynomial, so we cannot continue.


      Combining the results from steps 2 and 4:
      $$begin{align}3x^3 - 2x^2 + x - 5 &= (3x)(x^2 + 1) + (-2)(x^2 + 1) + (-2x-3)\&=(3x - 2)(x^2 + 1) + (-2x-3)end{align}$$



      If you stop at step 2, then $r(x)$ is indeed not linear. But at that stage, we could continue the division to get something smaller. It was only when the remainder was of lower degree that we had to stop.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I'm not American, but from what I've read pre-calc includes induction. Was my use of modulo arithmetic the comprehensibility issue?
        $endgroup$
        – J.G.
        Dec 12 '18 at 23:35






      • 1




        $begingroup$
        @J.G. - It is true that both induction and modular arithmetic are both taught pre-calculus, But not to everyone, and many of those who have learned them have never truly mastered them. I took the "pre-calculus" request not as a specific level to be met, but rather to mean that the OP wanted as simple an explanation as possible. Since it is an algebra question, and the explanation only requires simple algebra, that is what I gave. "Master the minimum" is a teaching maxim that applies here, in my opinion.
        $endgroup$
        – Paul Sinclair
        Dec 12 '18 at 23:52










      • $begingroup$
        Thanks for that clarification.
        $endgroup$
        – J.G.
        Dec 13 '18 at 7:14










      • $begingroup$
        @PaulSinclair That's also why I posted my answer. I wanted to get the principle across at the most beaux level I could.
        $endgroup$
        – timtfj
        Dec 13 '18 at 23:24










      • $begingroup$
        @timtfj - but what I don't understand is what your post added that mine did not contain. I mentioned J.G.'s post because I wanted to acknowledge that he had already addressed the issue, and my post was just to explain it in a more simple manner. But I don't see anything in your post 2.5 hours later that I didn't cover as or more simply.
        $endgroup$
        – Paul Sinclair
        Dec 14 '18 at 0:30














      12












      12








      12





      $begingroup$

      While José Carlos Santos's answer is correct, I think it might be useful to talk about why $r(x)$ is defined to have degree less than $p_2(x)$. (J.G. explains this, but at a higher level than I suspect someone wanting "pre-calculus" would follow.)



      If the degree of $r(x)$ is the same size or greater than $p_2(x)$, then you can subtract off a multiple of $p_2(x)$ from $r(x)$ to get a new remainder of lower degree. I.e., we can continue the division process until $r(x)$ is of lower degree than $p_2(x)$. It is only then that we have to stop and call anything left over the remainder.



      For example, dividing $3x^3 - 2x^2 + x - 5$ by $x^2 + 1$ follows these steps:




      1. Note that the ratio of the leading terms is $frac {3x^3}{x^2} = 3x$

      2. Multiply $x^2 + 1$ by $3x$ and subtract the result ($3x^3 + 3x$) from $3x^3 - 2x^2 + x - 5$, leaving $-2x^2 - 2x - 5$. I.e.,$$3x^3 - 2x^2 + x - 5 = (3x)(x^2 + 1) + (-2x^2 - 2x - 5)$$

      3. Note that the ratio of the leading terms of the remainder and $x^2 + 1$ is $frac {-2x^2}{x^2} = -2$.

      4. Multiply $x^2 + 1$ by $-2$ and subtract the result ($-2x^2 - 2$) from $-2x^2 - 2x - 5$, leaving $-2x - 3$. I.e., $$-2x^2 - 2x - 5 = (-2)(x^2 + 1) + (-2x-3)$$

      5. Note that the ratio of the leading terms is now $frac {-2x}{x^2} = frac {-2}x$, which is not a polynomial, so we cannot continue.


      Combining the results from steps 2 and 4:
      $$begin{align}3x^3 - 2x^2 + x - 5 &= (3x)(x^2 + 1) + (-2)(x^2 + 1) + (-2x-3)\&=(3x - 2)(x^2 + 1) + (-2x-3)end{align}$$



      If you stop at step 2, then $r(x)$ is indeed not linear. But at that stage, we could continue the division to get something smaller. It was only when the remainder was of lower degree that we had to stop.






      share|cite|improve this answer









      $endgroup$



      While José Carlos Santos's answer is correct, I think it might be useful to talk about why $r(x)$ is defined to have degree less than $p_2(x)$. (J.G. explains this, but at a higher level than I suspect someone wanting "pre-calculus" would follow.)



      If the degree of $r(x)$ is the same size or greater than $p_2(x)$, then you can subtract off a multiple of $p_2(x)$ from $r(x)$ to get a new remainder of lower degree. I.e., we can continue the division process until $r(x)$ is of lower degree than $p_2(x)$. It is only then that we have to stop and call anything left over the remainder.



      For example, dividing $3x^3 - 2x^2 + x - 5$ by $x^2 + 1$ follows these steps:




      1. Note that the ratio of the leading terms is $frac {3x^3}{x^2} = 3x$

      2. Multiply $x^2 + 1$ by $3x$ and subtract the result ($3x^3 + 3x$) from $3x^3 - 2x^2 + x - 5$, leaving $-2x^2 - 2x - 5$. I.e.,$$3x^3 - 2x^2 + x - 5 = (3x)(x^2 + 1) + (-2x^2 - 2x - 5)$$

      3. Note that the ratio of the leading terms of the remainder and $x^2 + 1$ is $frac {-2x^2}{x^2} = -2$.

      4. Multiply $x^2 + 1$ by $-2$ and subtract the result ($-2x^2 - 2$) from $-2x^2 - 2x - 5$, leaving $-2x - 3$. I.e., $$-2x^2 - 2x - 5 = (-2)(x^2 + 1) + (-2x-3)$$

      5. Note that the ratio of the leading terms is now $frac {-2x}{x^2} = frac {-2}x$, which is not a polynomial, so we cannot continue.


      Combining the results from steps 2 and 4:
      $$begin{align}3x^3 - 2x^2 + x - 5 &= (3x)(x^2 + 1) + (-2)(x^2 + 1) + (-2x-3)\&=(3x - 2)(x^2 + 1) + (-2x-3)end{align}$$



      If you stop at step 2, then $r(x)$ is indeed not linear. But at that stage, we could continue the division to get something smaller. It was only when the remainder was of lower degree that we had to stop.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 12 '18 at 17:52









      Paul SinclairPaul Sinclair

      20.7k21543




      20.7k21543












      • $begingroup$
        I'm not American, but from what I've read pre-calc includes induction. Was my use of modulo arithmetic the comprehensibility issue?
        $endgroup$
        – J.G.
        Dec 12 '18 at 23:35






      • 1




        $begingroup$
        @J.G. - It is true that both induction and modular arithmetic are both taught pre-calculus, But not to everyone, and many of those who have learned them have never truly mastered them. I took the "pre-calculus" request not as a specific level to be met, but rather to mean that the OP wanted as simple an explanation as possible. Since it is an algebra question, and the explanation only requires simple algebra, that is what I gave. "Master the minimum" is a teaching maxim that applies here, in my opinion.
        $endgroup$
        – Paul Sinclair
        Dec 12 '18 at 23:52










      • $begingroup$
        Thanks for that clarification.
        $endgroup$
        – J.G.
        Dec 13 '18 at 7:14










      • $begingroup$
        @PaulSinclair That's also why I posted my answer. I wanted to get the principle across at the most beaux level I could.
        $endgroup$
        – timtfj
        Dec 13 '18 at 23:24










      • $begingroup$
        @timtfj - but what I don't understand is what your post added that mine did not contain. I mentioned J.G.'s post because I wanted to acknowledge that he had already addressed the issue, and my post was just to explain it in a more simple manner. But I don't see anything in your post 2.5 hours later that I didn't cover as or more simply.
        $endgroup$
        – Paul Sinclair
        Dec 14 '18 at 0:30


















      • $begingroup$
        I'm not American, but from what I've read pre-calc includes induction. Was my use of modulo arithmetic the comprehensibility issue?
        $endgroup$
        – J.G.
        Dec 12 '18 at 23:35






      • 1




        $begingroup$
        @J.G. - It is true that both induction and modular arithmetic are both taught pre-calculus, But not to everyone, and many of those who have learned them have never truly mastered them. I took the "pre-calculus" request not as a specific level to be met, but rather to mean that the OP wanted as simple an explanation as possible. Since it is an algebra question, and the explanation only requires simple algebra, that is what I gave. "Master the minimum" is a teaching maxim that applies here, in my opinion.
        $endgroup$
        – Paul Sinclair
        Dec 12 '18 at 23:52










      • $begingroup$
        Thanks for that clarification.
        $endgroup$
        – J.G.
        Dec 13 '18 at 7:14










      • $begingroup$
        @PaulSinclair That's also why I posted my answer. I wanted to get the principle across at the most beaux level I could.
        $endgroup$
        – timtfj
        Dec 13 '18 at 23:24










      • $begingroup$
        @timtfj - but what I don't understand is what your post added that mine did not contain. I mentioned J.G.'s post because I wanted to acknowledge that he had already addressed the issue, and my post was just to explain it in a more simple manner. But I don't see anything in your post 2.5 hours later that I didn't cover as or more simply.
        $endgroup$
        – Paul Sinclair
        Dec 14 '18 at 0:30
















      $begingroup$
      I'm not American, but from what I've read pre-calc includes induction. Was my use of modulo arithmetic the comprehensibility issue?
      $endgroup$
      – J.G.
      Dec 12 '18 at 23:35




      $begingroup$
      I'm not American, but from what I've read pre-calc includes induction. Was my use of modulo arithmetic the comprehensibility issue?
      $endgroup$
      – J.G.
      Dec 12 '18 at 23:35




      1




      1




      $begingroup$
      @J.G. - It is true that both induction and modular arithmetic are both taught pre-calculus, But not to everyone, and many of those who have learned them have never truly mastered them. I took the "pre-calculus" request not as a specific level to be met, but rather to mean that the OP wanted as simple an explanation as possible. Since it is an algebra question, and the explanation only requires simple algebra, that is what I gave. "Master the minimum" is a teaching maxim that applies here, in my opinion.
      $endgroup$
      – Paul Sinclair
      Dec 12 '18 at 23:52




      $begingroup$
      @J.G. - It is true that both induction and modular arithmetic are both taught pre-calculus, But not to everyone, and many of those who have learned them have never truly mastered them. I took the "pre-calculus" request not as a specific level to be met, but rather to mean that the OP wanted as simple an explanation as possible. Since it is an algebra question, and the explanation only requires simple algebra, that is what I gave. "Master the minimum" is a teaching maxim that applies here, in my opinion.
      $endgroup$
      – Paul Sinclair
      Dec 12 '18 at 23:52












      $begingroup$
      Thanks for that clarification.
      $endgroup$
      – J.G.
      Dec 13 '18 at 7:14




      $begingroup$
      Thanks for that clarification.
      $endgroup$
      – J.G.
      Dec 13 '18 at 7:14












      $begingroup$
      @PaulSinclair That's also why I posted my answer. I wanted to get the principle across at the most beaux level I could.
      $endgroup$
      – timtfj
      Dec 13 '18 at 23:24




      $begingroup$
      @PaulSinclair That's also why I posted my answer. I wanted to get the principle across at the most beaux level I could.
      $endgroup$
      – timtfj
      Dec 13 '18 at 23:24












      $begingroup$
      @timtfj - but what I don't understand is what your post added that mine did not contain. I mentioned J.G.'s post because I wanted to acknowledge that he had already addressed the issue, and my post was just to explain it in a more simple manner. But I don't see anything in your post 2.5 hours later that I didn't cover as or more simply.
      $endgroup$
      – Paul Sinclair
      Dec 14 '18 at 0:30




      $begingroup$
      @timtfj - but what I don't understand is what your post added that mine did not contain. I mentioned J.G.'s post because I wanted to acknowledge that he had already addressed the issue, and my post was just to explain it in a more simple manner. But I don't see anything in your post 2.5 hours later that I didn't cover as or more simply.
      $endgroup$
      – Paul Sinclair
      Dec 14 '18 at 0:30











      3












      $begingroup$

      The point is we can prove as much. Let the quadratic be $q:=ax^2+bx+c$. Working modulo $q$, $x^2=-(bx+c)/a$. Each $x^n$ is then of the form $cx+d$ modulo $q$; you can prove this by induction (you can even get recursion relations on the coefficients). For example, $x^3=-(bx^2+cx)/a$, but we can turn replace $x^2$ as above.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        To PROVE you need to have a degree-n polynomial, not a quadratic. But it works exactly like that. so it's a_n x^n + a_(n-1) x^(n-1) + .. + a_0, etc.
        $endgroup$
        – user3445853
        Dec 13 '18 at 11:12










      • $begingroup$
        @user3445853 I don't understand what point you're trying to make. Are you emphasising that the remainder on division by a degree $n$ polynomial is, provably by this strategy, of degree $<n$?
        $endgroup$
        – J.G.
        Dec 13 '18 at 11:23
















      3












      $begingroup$

      The point is we can prove as much. Let the quadratic be $q:=ax^2+bx+c$. Working modulo $q$, $x^2=-(bx+c)/a$. Each $x^n$ is then of the form $cx+d$ modulo $q$; you can prove this by induction (you can even get recursion relations on the coefficients). For example, $x^3=-(bx^2+cx)/a$, but we can turn replace $x^2$ as above.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        To PROVE you need to have a degree-n polynomial, not a quadratic. But it works exactly like that. so it's a_n x^n + a_(n-1) x^(n-1) + .. + a_0, etc.
        $endgroup$
        – user3445853
        Dec 13 '18 at 11:12










      • $begingroup$
        @user3445853 I don't understand what point you're trying to make. Are you emphasising that the remainder on division by a degree $n$ polynomial is, provably by this strategy, of degree $<n$?
        $endgroup$
        – J.G.
        Dec 13 '18 at 11:23














      3












      3








      3





      $begingroup$

      The point is we can prove as much. Let the quadratic be $q:=ax^2+bx+c$. Working modulo $q$, $x^2=-(bx+c)/a$. Each $x^n$ is then of the form $cx+d$ modulo $q$; you can prove this by induction (you can even get recursion relations on the coefficients). For example, $x^3=-(bx^2+cx)/a$, but we can turn replace $x^2$ as above.






      share|cite|improve this answer











      $endgroup$



      The point is we can prove as much. Let the quadratic be $q:=ax^2+bx+c$. Working modulo $q$, $x^2=-(bx+c)/a$. Each $x^n$ is then of the form $cx+d$ modulo $q$; you can prove this by induction (you can even get recursion relations on the coefficients). For example, $x^3=-(bx^2+cx)/a$, but we can turn replace $x^2$ as above.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 12 '18 at 23:33

























      answered Dec 12 '18 at 15:54









      J.G.J.G.

      32.6k23250




      32.6k23250












      • $begingroup$
        To PROVE you need to have a degree-n polynomial, not a quadratic. But it works exactly like that. so it's a_n x^n + a_(n-1) x^(n-1) + .. + a_0, etc.
        $endgroup$
        – user3445853
        Dec 13 '18 at 11:12










      • $begingroup$
        @user3445853 I don't understand what point you're trying to make. Are you emphasising that the remainder on division by a degree $n$ polynomial is, provably by this strategy, of degree $<n$?
        $endgroup$
        – J.G.
        Dec 13 '18 at 11:23


















      • $begingroup$
        To PROVE you need to have a degree-n polynomial, not a quadratic. But it works exactly like that. so it's a_n x^n + a_(n-1) x^(n-1) + .. + a_0, etc.
        $endgroup$
        – user3445853
        Dec 13 '18 at 11:12










      • $begingroup$
        @user3445853 I don't understand what point you're trying to make. Are you emphasising that the remainder on division by a degree $n$ polynomial is, provably by this strategy, of degree $<n$?
        $endgroup$
        – J.G.
        Dec 13 '18 at 11:23
















      $begingroup$
      To PROVE you need to have a degree-n polynomial, not a quadratic. But it works exactly like that. so it's a_n x^n + a_(n-1) x^(n-1) + .. + a_0, etc.
      $endgroup$
      – user3445853
      Dec 13 '18 at 11:12




      $begingroup$
      To PROVE you need to have a degree-n polynomial, not a quadratic. But it works exactly like that. so it's a_n x^n + a_(n-1) x^(n-1) + .. + a_0, etc.
      $endgroup$
      – user3445853
      Dec 13 '18 at 11:12












      $begingroup$
      @user3445853 I don't understand what point you're trying to make. Are you emphasising that the remainder on division by a degree $n$ polynomial is, provably by this strategy, of degree $<n$?
      $endgroup$
      – J.G.
      Dec 13 '18 at 11:23




      $begingroup$
      @user3445853 I don't understand what point you're trying to make. Are you emphasising that the remainder on division by a degree $n$ polynomial is, provably by this strategy, of degree $<n$?
      $endgroup$
      – J.G.
      Dec 13 '18 at 11:23











      2












      $begingroup$

      An explanation rather than a full proof. Trying to describe it more simply:



      Suppose your first polynomial is
      $$P=ax^4+bx^3+cx^2+dx+e$$
      and you want to divide it by
      $$Q=fx^2+gx+h$$



      The first step is to multiply $Q$ by something which will turn its first term into $ax^4$. You've now got a polynomial which, when you subtract it from $P$, will make $ax^4$ disappear.



      Then you do the same again, except you arrange for the $x^3$ term to disappear.



      Now you have two quadratics: what's left of $P$, and $Q$.



      So, multiply $Q$ by a suitable constant to make the $x^2$ terms match, and subtract it. Now all that's left is a multiple of $x$ and a constant, since you've got rid of all the higher order terms—and that's your linear remainder.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        An explanation rather than a full proof. Trying to describe it more simply:



        Suppose your first polynomial is
        $$P=ax^4+bx^3+cx^2+dx+e$$
        and you want to divide it by
        $$Q=fx^2+gx+h$$



        The first step is to multiply $Q$ by something which will turn its first term into $ax^4$. You've now got a polynomial which, when you subtract it from $P$, will make $ax^4$ disappear.



        Then you do the same again, except you arrange for the $x^3$ term to disappear.



        Now you have two quadratics: what's left of $P$, and $Q$.



        So, multiply $Q$ by a suitable constant to make the $x^2$ terms match, and subtract it. Now all that's left is a multiple of $x$ and a constant, since you've got rid of all the higher order terms—and that's your linear remainder.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          An explanation rather than a full proof. Trying to describe it more simply:



          Suppose your first polynomial is
          $$P=ax^4+bx^3+cx^2+dx+e$$
          and you want to divide it by
          $$Q=fx^2+gx+h$$



          The first step is to multiply $Q$ by something which will turn its first term into $ax^4$. You've now got a polynomial which, when you subtract it from $P$, will make $ax^4$ disappear.



          Then you do the same again, except you arrange for the $x^3$ term to disappear.



          Now you have two quadratics: what's left of $P$, and $Q$.



          So, multiply $Q$ by a suitable constant to make the $x^2$ terms match, and subtract it. Now all that's left is a multiple of $x$ and a constant, since you've got rid of all the higher order terms—and that's your linear remainder.






          share|cite|improve this answer











          $endgroup$



          An explanation rather than a full proof. Trying to describe it more simply:



          Suppose your first polynomial is
          $$P=ax^4+bx^3+cx^2+dx+e$$
          and you want to divide it by
          $$Q=fx^2+gx+h$$



          The first step is to multiply $Q$ by something which will turn its first term into $ax^4$. You've now got a polynomial which, when you subtract it from $P$, will make $ax^4$ disappear.



          Then you do the same again, except you arrange for the $x^3$ term to disappear.



          Now you have two quadratics: what's left of $P$, and $Q$.



          So, multiply $Q$ by a suitable constant to make the $x^2$ terms match, and subtract it. Now all that's left is a multiple of $x$ and a constant, since you've got rid of all the higher order terms—and that's your linear remainder.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 23:38

























          answered Dec 12 '18 at 20:13









          timtfjtimtfj

          2,503420




          2,503420























              0












              $begingroup$

              Simply explaining: the key IDEA is that you are not done dividing until your remainder is not further divisible by your divisor. The key WORD in your post is "GENERALLY" because the conclusion is not hard and fast, it is simply "usually" so.



              IDEA first. Consider simple numbers, perhaps dividing 1,034 by 10. You begin at the left and work right (this is taught in first grade of course so I won't belabor the process) until you cannot usefully continue. So you get 100 R 34, and 3 R 4, and stop when left with a remainder (4) less than your divisor, 10. Easy enough to figure when it's "less than" with numbers, but slightly less so with polynomials. But we see it slightly differently with polynomials: we stop when the order of the remainder is lower than the order of the divisor. So dividing by 7x^2, and having a remainder with just x^2, we usually grit our teeth (we DO love our integers...) and use "1/7" eliminating the remainder's 2nd order term (x^2). At that point, coninuing becomes difficult even with gritted teeth.



              And so we stop. Notice that it is not that the remainder is now linear with that being special in some way: it is simply of lower order than the divisor. In the case of dividing by a 2nd order polynomial, that means it is 1st order, or simply a constant (number).



              Which brings in the "generally" part: in practice, it COULD be just the constant, and not a linear equation. "x^2 + 8" divided by x^2, for example. But not usually. Hence "generally" in your post's statement. And moving up in order, say a fifth order divisor, while you'd usually expect, just out of probability, a fourth order remainder, it could also be a third, second, or first order remainder, or just a constant.



              So it's not a great theorem with a stunning proof, just a practical observation based upon the mechanics we use for the division.



              (And it adds an "idiot check" in case one stops early, thinking he's finished. "Wait, I can't be, it ought to be linear and it isn't so maybe I need to keep working...")






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Simply explaining: the key IDEA is that you are not done dividing until your remainder is not further divisible by your divisor. The key WORD in your post is "GENERALLY" because the conclusion is not hard and fast, it is simply "usually" so.



                IDEA first. Consider simple numbers, perhaps dividing 1,034 by 10. You begin at the left and work right (this is taught in first grade of course so I won't belabor the process) until you cannot usefully continue. So you get 100 R 34, and 3 R 4, and stop when left with a remainder (4) less than your divisor, 10. Easy enough to figure when it's "less than" with numbers, but slightly less so with polynomials. But we see it slightly differently with polynomials: we stop when the order of the remainder is lower than the order of the divisor. So dividing by 7x^2, and having a remainder with just x^2, we usually grit our teeth (we DO love our integers...) and use "1/7" eliminating the remainder's 2nd order term (x^2). At that point, coninuing becomes difficult even with gritted teeth.



                And so we stop. Notice that it is not that the remainder is now linear with that being special in some way: it is simply of lower order than the divisor. In the case of dividing by a 2nd order polynomial, that means it is 1st order, or simply a constant (number).



                Which brings in the "generally" part: in practice, it COULD be just the constant, and not a linear equation. "x^2 + 8" divided by x^2, for example. But not usually. Hence "generally" in your post's statement. And moving up in order, say a fifth order divisor, while you'd usually expect, just out of probability, a fourth order remainder, it could also be a third, second, or first order remainder, or just a constant.



                So it's not a great theorem with a stunning proof, just a practical observation based upon the mechanics we use for the division.



                (And it adds an "idiot check" in case one stops early, thinking he's finished. "Wait, I can't be, it ought to be linear and it isn't so maybe I need to keep working...")






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Simply explaining: the key IDEA is that you are not done dividing until your remainder is not further divisible by your divisor. The key WORD in your post is "GENERALLY" because the conclusion is not hard and fast, it is simply "usually" so.



                  IDEA first. Consider simple numbers, perhaps dividing 1,034 by 10. You begin at the left and work right (this is taught in first grade of course so I won't belabor the process) until you cannot usefully continue. So you get 100 R 34, and 3 R 4, and stop when left with a remainder (4) less than your divisor, 10. Easy enough to figure when it's "less than" with numbers, but slightly less so with polynomials. But we see it slightly differently with polynomials: we stop when the order of the remainder is lower than the order of the divisor. So dividing by 7x^2, and having a remainder with just x^2, we usually grit our teeth (we DO love our integers...) and use "1/7" eliminating the remainder's 2nd order term (x^2). At that point, coninuing becomes difficult even with gritted teeth.



                  And so we stop. Notice that it is not that the remainder is now linear with that being special in some way: it is simply of lower order than the divisor. In the case of dividing by a 2nd order polynomial, that means it is 1st order, or simply a constant (number).



                  Which brings in the "generally" part: in practice, it COULD be just the constant, and not a linear equation. "x^2 + 8" divided by x^2, for example. But not usually. Hence "generally" in your post's statement. And moving up in order, say a fifth order divisor, while you'd usually expect, just out of probability, a fourth order remainder, it could also be a third, second, or first order remainder, or just a constant.



                  So it's not a great theorem with a stunning proof, just a practical observation based upon the mechanics we use for the division.



                  (And it adds an "idiot check" in case one stops early, thinking he's finished. "Wait, I can't be, it ought to be linear and it isn't so maybe I need to keep working...")






                  share|cite|improve this answer









                  $endgroup$



                  Simply explaining: the key IDEA is that you are not done dividing until your remainder is not further divisible by your divisor. The key WORD in your post is "GENERALLY" because the conclusion is not hard and fast, it is simply "usually" so.



                  IDEA first. Consider simple numbers, perhaps dividing 1,034 by 10. You begin at the left and work right (this is taught in first grade of course so I won't belabor the process) until you cannot usefully continue. So you get 100 R 34, and 3 R 4, and stop when left with a remainder (4) less than your divisor, 10. Easy enough to figure when it's "less than" with numbers, but slightly less so with polynomials. But we see it slightly differently with polynomials: we stop when the order of the remainder is lower than the order of the divisor. So dividing by 7x^2, and having a remainder with just x^2, we usually grit our teeth (we DO love our integers...) and use "1/7" eliminating the remainder's 2nd order term (x^2). At that point, coninuing becomes difficult even with gritted teeth.



                  And so we stop. Notice that it is not that the remainder is now linear with that being special in some way: it is simply of lower order than the divisor. In the case of dividing by a 2nd order polynomial, that means it is 1st order, or simply a constant (number).



                  Which brings in the "generally" part: in practice, it COULD be just the constant, and not a linear equation. "x^2 + 8" divided by x^2, for example. But not usually. Hence "generally" in your post's statement. And moving up in order, say a fifth order divisor, while you'd usually expect, just out of probability, a fourth order remainder, it could also be a third, second, or first order remainder, or just a constant.



                  So it's not a great theorem with a stunning proof, just a practical observation based upon the mechanics we use for the division.



                  (And it adds an "idiot check" in case one stops early, thinking he's finished. "Wait, I can't be, it ought to be linear and it isn't so maybe I need to keep working...")







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 16:13









                  RoyRoy

                  1




                  1























                      0












                      $begingroup$

                      The accepted answer is perfect, of course, and I would not have posted this as an answer if I had enough reputation to comment... but I haven't, so this is my answer (or really a follow-up-question).



                      I wonder if the solution provided in the question $p(x)=(x^2−2x−3)q(x)+ax+b$ is the "full" solution. This solution describes only the form of the remainder: a linear term.



                      With the actual terms given in the question, it is possible to determine actual values for $a$ and $b$, being $3$ and $-4$. The remainder therefore is the term $3x-4$.



                      I found this solution by setting up an Excel sheet and trying out a couple (hundred) of combinations, but I now wonder if there is a way of finding this term without using "brute force".



                      Note: if you multiply the two terms given in the question $(x-3)$ and $(x+1)$ you get the third term $(x^2-2x-3)$, so this question boils down to something like "remainder when dividing by 3, remainder when dividing by 5, remainder when dividing by 15", which might have an impact on solving it.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Ke Ke only quoted the part of the solution relevant to the question they asked. If you want to know how to finish solving the problem, your question, even through it relates to this one, should be asked in a new thread. You can link back to this thread in your post, as relevant information, but it is preferable that separate questions have their own threads. This allows the added question to be searchable, and keeps it clear which answers apply to which questions.
                        $endgroup$
                        – Paul Sinclair
                        Dec 14 '18 at 14:00


















                      0












                      $begingroup$

                      The accepted answer is perfect, of course, and I would not have posted this as an answer if I had enough reputation to comment... but I haven't, so this is my answer (or really a follow-up-question).



                      I wonder if the solution provided in the question $p(x)=(x^2−2x−3)q(x)+ax+b$ is the "full" solution. This solution describes only the form of the remainder: a linear term.



                      With the actual terms given in the question, it is possible to determine actual values for $a$ and $b$, being $3$ and $-4$. The remainder therefore is the term $3x-4$.



                      I found this solution by setting up an Excel sheet and trying out a couple (hundred) of combinations, but I now wonder if there is a way of finding this term without using "brute force".



                      Note: if you multiply the two terms given in the question $(x-3)$ and $(x+1)$ you get the third term $(x^2-2x-3)$, so this question boils down to something like "remainder when dividing by 3, remainder when dividing by 5, remainder when dividing by 15", which might have an impact on solving it.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Ke Ke only quoted the part of the solution relevant to the question they asked. If you want to know how to finish solving the problem, your question, even through it relates to this one, should be asked in a new thread. You can link back to this thread in your post, as relevant information, but it is preferable that separate questions have their own threads. This allows the added question to be searchable, and keeps it clear which answers apply to which questions.
                        $endgroup$
                        – Paul Sinclair
                        Dec 14 '18 at 14:00
















                      0












                      0








                      0





                      $begingroup$

                      The accepted answer is perfect, of course, and I would not have posted this as an answer if I had enough reputation to comment... but I haven't, so this is my answer (or really a follow-up-question).



                      I wonder if the solution provided in the question $p(x)=(x^2−2x−3)q(x)+ax+b$ is the "full" solution. This solution describes only the form of the remainder: a linear term.



                      With the actual terms given in the question, it is possible to determine actual values for $a$ and $b$, being $3$ and $-4$. The remainder therefore is the term $3x-4$.



                      I found this solution by setting up an Excel sheet and trying out a couple (hundred) of combinations, but I now wonder if there is a way of finding this term without using "brute force".



                      Note: if you multiply the two terms given in the question $(x-3)$ and $(x+1)$ you get the third term $(x^2-2x-3)$, so this question boils down to something like "remainder when dividing by 3, remainder when dividing by 5, remainder when dividing by 15", which might have an impact on solving it.






                      share|cite|improve this answer









                      $endgroup$



                      The accepted answer is perfect, of course, and I would not have posted this as an answer if I had enough reputation to comment... but I haven't, so this is my answer (or really a follow-up-question).



                      I wonder if the solution provided in the question $p(x)=(x^2−2x−3)q(x)+ax+b$ is the "full" solution. This solution describes only the form of the remainder: a linear term.



                      With the actual terms given in the question, it is possible to determine actual values for $a$ and $b$, being $3$ and $-4$. The remainder therefore is the term $3x-4$.



                      I found this solution by setting up an Excel sheet and trying out a couple (hundred) of combinations, but I now wonder if there is a way of finding this term without using "brute force".



                      Note: if you multiply the two terms given in the question $(x-3)$ and $(x+1)$ you get the third term $(x^2-2x-3)$, so this question boils down to something like "remainder when dividing by 3, remainder when dividing by 5, remainder when dividing by 15", which might have an impact on solving it.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 13 '18 at 17:24









                      RalfRalf

                      172




                      172












                      • $begingroup$
                        Ke Ke only quoted the part of the solution relevant to the question they asked. If you want to know how to finish solving the problem, your question, even through it relates to this one, should be asked in a new thread. You can link back to this thread in your post, as relevant information, but it is preferable that separate questions have their own threads. This allows the added question to be searchable, and keeps it clear which answers apply to which questions.
                        $endgroup$
                        – Paul Sinclair
                        Dec 14 '18 at 14:00




















                      • $begingroup$
                        Ke Ke only quoted the part of the solution relevant to the question they asked. If you want to know how to finish solving the problem, your question, even through it relates to this one, should be asked in a new thread. You can link back to this thread in your post, as relevant information, but it is preferable that separate questions have their own threads. This allows the added question to be searchable, and keeps it clear which answers apply to which questions.
                        $endgroup$
                        – Paul Sinclair
                        Dec 14 '18 at 14:00


















                      $begingroup$
                      Ke Ke only quoted the part of the solution relevant to the question they asked. If you want to know how to finish solving the problem, your question, even through it relates to this one, should be asked in a new thread. You can link back to this thread in your post, as relevant information, but it is preferable that separate questions have their own threads. This allows the added question to be searchable, and keeps it clear which answers apply to which questions.
                      $endgroup$
                      – Paul Sinclair
                      Dec 14 '18 at 14:00






                      $begingroup$
                      Ke Ke only quoted the part of the solution relevant to the question they asked. If you want to know how to finish solving the problem, your question, even through it relates to this one, should be asked in a new thread. You can link back to this thread in your post, as relevant information, but it is preferable that separate questions have their own threads. This allows the added question to be searchable, and keeps it clear which answers apply to which questions.
                      $endgroup$
                      – Paul Sinclair
                      Dec 14 '18 at 14:00




















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