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I saw one of my middle school students working on a problem that went something like this:

When $f(x)$ is divided by $x - \frac{1}{2}$, the quotient is $Q(x)$ and the remainder is $R$. What are the quotient and remainder, respectively, when $x f(x)$ is divided by $2x-1$?

Then you had to choose between several answers, each in terms of $Q(x)$ and $R$.

What do you call this kind of problem, and how can I learn how to solve things like this? I never saw anything like this when I did elementary math. For the integer case, it seems like you could use the quotient-remainder theorem, but I don't know how to generalize that to the function case.

Searching for "quotient-remainder theorem with functions" and such has gotten me nowhere.

Max
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    try searching with "polynomials" instead of "functions" – J. W. Tanner Nov 07 '19 at 02:28
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    See also Euclidean domains, and for higher-dimensional (multivariate) division algorithms see the literature on Grobner / standard basis algorithms – Bill Dubuque Nov 07 '19 at 02:32
  • Thanks @J.W.Tanner. The trick in this problem is to then use polynomial division. – Max Nov 07 '19 at 02:50
  • @Max Polynomial division is presumed known, cf. "when $f$ is divided by $,x - 1/2,$" so that's not the trick, Rather it's simply to scale the division by $x$ then $1/2,,$ - see my answer. – Bill Dubuque Nov 07 '19 at 03:48
  • @BillDubuque I used polynomial division to divide $x f(x) = x(Q(x) (x - 1/2) + R) = x^2 Q(x) - x (-1/2 + R)$ by $2x - 1$. This got me the correct answer without needing to know an external algorithm. – Max Nov 07 '19 at 03:59
  • @Max We don't need to do any further division - it suffices to scale the original division $\ f\div (x-1/2),,$ as I explained in my answer. That way clarifies the innate arithmetic, i.e. we are multiplying (scaling) it by $,x\equiv 1/2.\ $ It seems you are essentially doing the same thing (but I can't be sure since you didn't give details). – Bill Dubuque Nov 07 '19 at 04:17

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It's the Polynomial Division Algorithm. The solution involves only scaling that: $\ (a=\frac{1}2$ in OP)

$$\begin{align} \dfrac{f}{x-a} &\,=\, q + \dfrac{r}{x-a}\\[.4em] \Rightarrow\ \ \dfrac{xf}{x-a} &\,=\, xq + \dfrac{xr}{x-a}\\[.4em] &\,=\, xq + \dfrac{(x-a)r + ar}{x-a}\\[.4em] &\,=\, xq+r + \dfrac{\color{#c00}{ar}}{x-a}\\[.4em] \Rightarrow\ \ \dfrac{xf}{2(x-a)}&\, =\, \tfrac{1}2(xq+r) + \dfrac{\color{}{ar}}{2(x-a)} \end{align}\qquad$$

Simpler by congruences: $\bmod x\!-\!a\!:\ \ \begin{align}x\equiv a\\ f\equiv r\end{align}\ \Rightarrow\ xf\equiv \color{#c00}{ar}\ $ by the Congruence Product Rule

Bill Dubuque
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