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I am an economist, so I use various kinds of mathematics. My daughter is learning synthetic division in high school algebra. I have never come across it in my 30 years in economics. Does it have a use in higher mathematics? Is it worth teaching in high school?

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    $\begingroup$ This belongs more on matheducators than here. That being said, to echo a commonly expressed sentiment here, the purpose of teaching algorithms like synthetic division is that they teach the process of learning a procedural methodology and applying it in a new setting, a skillset which is immensely valuable. And I do use synthetic division when I'm in a crunch, although calculators have certainly obviated the necessity of the procedure. $\endgroup$ Nov 16, 2021 at 1:42
  • $\begingroup$ I just want to add that while synthetic division (the particular algorithm) might not have much application in higher mathematics, the fact that one can divide polynomials with remainder is very useful in many fields. $\endgroup$
    – podiki
    Nov 16, 2021 at 2:08
  • $\begingroup$ For a practical example, polynomial division is the first step in partial fractions decomposition, where the link lists several real life applications. $\endgroup$
    – dxiv
    Nov 16, 2021 at 3:11
  • $\begingroup$ If the synthetic division is being taught in such a way that your daughter actually understands why it works, it sounds like a great exercise. $\endgroup$
    – David K
    Nov 16, 2021 at 14:04
  • $\begingroup$ Thanks, especially dxiv for the link, but that doesn't really answer my question. Don Thousand is closer. One can contrive some possible uses. I'm wondering if, in doing "real" math, for some purpose other than as an example, people actually use this method. $\endgroup$ Nov 16, 2021 at 16:50

1 Answer 1

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well, a concrete task with polynomial division and remainder, it finds the gcd(over the rationals) of two polynomials, then constructs the Bezout relation (in integers $ax +by = \gcd(a,b)$ )

In this case, one of the polynomials is the derivative of the other, there is a nontrivial common factor, so the square of that factor divides the first... the second step uses continued fractions, it turned out one may use continued fractions with polynomials.

$$ \left( x^{3} - 4 x^{2} - 3 x + 18 \right) $$

$$ \left( 3 x^{2} - 8 x - 3 \right) $$

$$ \left( x^{3} - 4 x^{2} - 3 x + 18 \right) = \left( 3 x^{2} - 8 x - 3 \right) \cdot \color{magenta}{ \left( \frac{ 3 x - 4 }{ 9 } \right) } + \left( \frac{ - 50 x + 150 }{ 9 } \right) $$ $$ \left( 3 x^{2} - 8 x - 3 \right) = \left( \frac{ - 50 x + 150 }{ 9 } \right) \cdot \color{magenta}{ \left( \frac{ - 27 x - 9 }{ 50 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( \frac{ 3 x - 4 }{ 9 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 3 x - 4 }{ 9 } \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - 27 x - 9 }{ 50 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 9 x^{2} + 9 x + 54 }{ 50 } \right) }{ \left( \frac{ - 27 x - 9 }{ 50 } \right) } $$ $$ \left( x^{2} - x - 6 \right) \left( \frac{ 9}{50 } \right) - \left( 3 x + 1 \right) \left( \frac{ 3 x - 4 }{ 50 } \right) = \left( -1 \right) $$ $$ \left( x^{3} - 4 x^{2} - 3 x + 18 \right) = \left( x^{2} - x - 6 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 0 \right) $$ $$ \left( 3 x^{2} - 8 x - 3 \right) = \left( 3 x + 1 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x - 3 \right) } $$ $$ \left( x^{3} - 4 x^{2} - 3 x + 18 \right) \left( \frac{ 9}{50 } \right) - \left( 3 x^{2} - 8 x - 3 \right) \left( \frac{ 3 x - 4 }{ 50 } \right) = \left( - x + 3 \right) $$

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