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In the U.S. students in grades $\{9,10,11\}$ often learn long division of two polynomials, e.g.: $$ \frac{x^4 + 6x^2 + 2}{x^2 + 5} = x^2 + 1 - \frac{3}{x^2 + 5} \;. $$ I believe it is fair to say that almost never is any motivation provided: Why would anyone ever want to divide two polynomials?

Q. How can polynomial division be motivated at that grade level?

The dilemma seems to be that, where polynomial division is really needed—for example, in error-correcting codes (cyclic redundancy checking)—requires more advanced math to understand.

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    $\begingroup$ They help in factorizing polynomials and partial fractions. $\endgroup$
    – user5402
    Sep 1, 2018 at 13:56
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    $\begingroup$ I think this is one of many topics still taught in high school algebra 2 [= 2nd year algebra] classes whose inclusion has become more problematic over the last 50 some years as algebra 2 has transitioned (in the U.S., at least) from a course taken by relatively few students (maybe top 20%) to nearly all students. For what it's worth, I did make use of polynomial division in curve sketching in precalculus and first semester calculus classes to identify the linear/nonlinear asymptote of the graph of a rational function. $\endgroup$
    – Dave L Renfro
    Sep 1, 2018 at 15:25
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    $\begingroup$ In Canada, rational functions are taught in two forms, canonical ($\frac{a}{x-h}+k$) and general ($\frac{Ax+B}{Cx+D}$). Then, you can have a problem where two quantities vary linearly and you are interested in how their ratio varies. For example, you could have the mass and volume $m(t)$ and $v(t)$ vary linearly with time. Then the density is $d(t)=\frac{m(t)}{v(t)}$, where $d(t)$ will be rational in general form. Long division can then be used to get the canonical form which helps graphing the situation, but also get the $k$ parameter, which indicates what $d(t)$ tends to in the long term. $\endgroup$
    – orion2112
    Sep 1, 2018 at 17:40
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    $\begingroup$ I think a good answer is just to understand the asymptotic behavior of the graph. Likewise, if the formula has independent variable of time then the polynomial part is the non-transient feature of the model. So, it's part of a program of approximation and looking at the big-picture of models or graphs. But, perhaps you are looking for a quite different sort of answer than this. $\endgroup$
    – James S. Cook
    Sep 1, 2018 at 23:48
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    $\begingroup$ So that all four of the arithmetic operations on polynomials can be seen. (It does make it seem a bit more striking that $\mathbb{C}$ is closed under the operations...) As for hand-computing, it can also come up when integrating certain rational functions (especially when the divisor is linear or quadratic). But, I think it is essentially a vestige. Although, my experience suggests that [some] Algebra 2 students find it fun to an extent that has surprised me. +: I think that exploring why "synthetic division" works is a good opportunity to practice understanding and explaining algorithms. $\endgroup$
    – Benjamin Dickman
    Sep 2, 2018 at 3:53

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One reason is that it's essential to determining the slant asymptote of a rational expression.

Another reason is that it's a useful step in factoring large polynomials. For example, say you're trying to factor $x^3 + 4x^2 -4x-1$. You can determine, using the Rational Root Theorem, that x - 1 is a factor. Since x + 1 isn't, my next step would be to divide $x^3 + 4x^2 -4x-1$ by x - 1. That would tell me that the factorization is $(x - 1)(x^2 +5x + 1)$ and I could get the final result using the Quadratic Formula.

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  • $\begingroup$ i.e. if you know one factor $g$ of $f$ then you can compute its cofactor $f/g$ using the division algorithm. But that's a special case of division (with remainder) so it begs the principle regarding motivation.. $\endgroup$
    – Bill Dubuque
    Sep 2, 2018 at 23:14
  • $\begingroup$ How is this a "special case"? The question asked, "Why would you ever want to do this?" I believe I answered that. $\endgroup$
    – G. Allen
    Sep 3, 2018 at 0:39
  • $\begingroup$ It's the special case when the remainder $= 0,\,$ i.e. exact quotient. Motivating "quotient with remainder" via the special case of "exact quotient" seems circular to me. $\endgroup$
    – Bill Dubuque
    Sep 3, 2018 at 3:04
  • $\begingroup$ I don't see remainders mentioned anywhere in the question. It just asks about "long division of two polynomials". $\endgroup$
    – G. Allen
    Sep 3, 2018 at 4:19
  • $\begingroup$ They are equivalent, i.e. the OP simply states polynomial (long) division (with remainder) in an equivalent fractional form, i.e. $$\dfrac{f}g =\, q + \dfrac{r}g \iff f =\, q\,g + r,\quad \deg r < \deg g$$ What you wrote is true, i.e. factorizations can be computed by iterated (exact) divisions, assuming we are given an algorithm that can find a proper factor of a reducible polynomial. But I don't see how such iterated (exact) divisions provides much (external) motivation for division with remainder, i.e. long division. $\endgroup$
    – Bill Dubuque
    Sep 3, 2018 at 16:01
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There is no sane motivation for introducing this topic to kids in grade 9, 10, or 11. Not all these kids are even bound for college. Of those who are, only some will take calculus. This kind of manipulation of polynomials can be applied to certain things like curve sketching, but that's peripheral in a calculus course.

Topics like this just accumulate in math curricula like useless crap in someone's garage.

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    $\begingroup$ Agreed. A possible way to turn this into motivation is to be honest about it. Tell the students when they're learning to jump through hoops for no good reason rather than contriving some unconvincing motivation. Then, when you say that there /is/ convincing motivation for a topic, they'll be more likely to believe you and pay attention to it. $\endgroup$
    – Matt Ollis
    Sep 4, 2018 at 12:16
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There are many example of when this may be useful, I would point to integral calculus as a place where right hand side of your example is easier to evaluate with introductory tools than the left hand side.

Though I want to look past examples in application and speak to the idea of putting expressions into a particular form to "see" patterns.

If you look at the right hand side of the equation the form generated lets us talk about the end behavior of x visually.

The leading coefficient is positive. The highest degree term is even. So as x nears negative infinity this expression approaches positive infinity. As x nears positive infinity this expression approaches positive infinity. As x reaches zero this expression is at $\frac{2}{5}$.

This can be done visually and with minimal computation. This parallel structure in form helps us with most of the expressions we evaluate and to compare expressions.

So by doing this we can compare rational expressions to the polynomials and notice the similarities and the differences.

Also expressions can have asymptotic behavior to a line and polynomial division can show this.

Example: $f(x)=\frac{4x^3+x^2+3}{x^2-x+1}$

Long division gives us: $f(x)=4x+5+\frac{x-2}{x^2-x+1}$

The large x behavior of x has the fraction reaching 0 when x goes to either infinity.

So this function approaches $4x+5$ is has an asymptote to this line.

This means if we are talking generally about data if we are not in close to where this has curved behavior the linear model could work for us simplifying our calculations.

Graphing it we can see this:

enter image description here

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How about the mother of all elephants in the room... It forms the basis for our positional number system...in the sense that carrying out polynomial division and the substituting x=10 will give just the normal quotient and remainder....

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    $\begingroup$ Not quite, since - unlike decimals - there is no carrying in polynomial arithmetic, and coefficients can lie outside the interval $[0,9]$ for digits. $\endgroup$
    – Bill Dubuque
    Sep 2, 2018 at 20:27
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    $\begingroup$ @user683 It's not "minor". The claim is incorrect, e.g. in the OP substituting $x=10$ does not yield the quotient due to the negative coefficient in the remainder. You need to do further work to get the integer quotient and remainder. In more complex cases the quotient may be a higher degree polynomial with many negative coefficients, so there is much work left to do to massage that to the integer quotient. The division algorithms are related, but they are not the same. $\endgroup$
    – Bill Dubuque
    Sep 3, 2018 at 3:20
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Needed for partial fractions (calculus topic) and for LaPlace transforms (normal part of ODE course). In terms of applications, common need is in controls problems (very simple level systems engineering classes that most engineers take). EE will have more frequent needs.

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    $\begingroup$ The question is about motivating polynomial division to secondary students who are likely seeing it for the very first time. Many of these students probably won't ever take calculus, let alone DEs. For the ones who will eventually take more advanced classes, such classes are likely at least a year away (in the US secondary curriculum, probably more like two years away, given the usual pre-calc class which they'll have to take). Telling students that they have to learn something because they'll use it in a class they may-or-may-not take in two years doesn't seem like good motivation to me. $\endgroup$
    – Xander Henderson ♦
    Sep 2, 2018 at 16:01
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    $\begingroup$ I agree that it is not as strong as an immediate application. however it is something. And school is rather full of "you have to learn A to do B to eventually do X as a job". Note, I'm agreeing with you. But the situation is the situation. And it is still a very real motivation (and honest at least) to tell them that they need it for the next course. in fact, for school children it may even be more motivational than immediate links to the work world. But again...not ideal. $\endgroup$
    – guest
    Sep 2, 2018 at 16:22
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    $\begingroup$ Already minus 1ed. Should have stuck to a comment. Tommi, not falling for that again. $\endgroup$
    – guest
    Sep 2, 2018 at 16:23

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