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$\frac{7x^2 + 3x + 9} {2x + 3}$

Replace the $x$s with $10$s to turn into normal division problem.

$\frac {739}{23}$

Solve as mixed fraction.

$32 \frac{3}{23}$

Convert to algebraic form, assuming that $x = 10$.

$3x + 2 + \frac{3}{2x + 3}$

Why in the world would we teach the long or synthetic division method when this is significantly more efficient and intuitive?

Jon Person
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    This is a good idea, but try it for an expression like $\frac{11x}{x+1}$, where it falls apart. – B. Mehta May 20 '17 at 15:18
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    I presume it's for some silly reason such as long division works, while your method doesn't. – Angina Seng May 20 '17 at 15:19
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    @B.Mehta It doesn't even work for the OP's original example. – Angina Seng May 20 '17 at 15:20
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    @jonperson Did you even check your work? $$\frac{7x^2+3x+9}{2x+3}=\frac72x-\frac{15}4+\frac{81/4}{2x+3}$$ On the other hand, obviously, $$(2x+3)\left(3x+2+\frac3{2x+3}\right)=6x^2+13x+9$$ –  May 20 '17 at 15:28
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    "Synthetic division" is in fact the standard long division algorithm taught in most US schools, but easier because the base is "$x$" and not the number $10$ so there's never any carrying or borrowing from one column to the next. Related: https://math.stackexchange.com/questions/2185587/what-actually-is-a-polynomial/2185648#2185648 – Ethan Bolker May 20 '17 at 15:29

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What would you do with $\frac{7x^2 + 13x + 9} {2x + 21}$ ? You could try to do it by calculating $\frac{839}{23}$ and using your method, this will give a wrong answer.

You could keep on writing it like this $\frac{7*10^2+13*10+9}{2*10+21}$ with the powers of 10, and use a division algorithm and it will give you a correct answer. But in the end what you did is just replace the symbol $x$ by the symbol '$10$'.

Jannes Braet
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  • Specifically, the problem is that something like $11x$ is not uniquely represented when $x = 10$; it could be either $11x$ or $x^2+x$, while those two expressions are very different for the general $x$. – fractal1729 May 20 '17 at 15:33
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Let $x=100,$ then $${70309\over203} = 346 + {71\over203},$$ and something obviously did not work out...

Yuri Negometyanov
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By its own nature, your method only yields results in the form $$n_dx^d+n_{d-1}x^{d-1}+\cdots+n_0+\frac{c_0+c_1x+\cdots+c_kx^k}{p(x)}$$ for $n_0,\cdots,n_d,c_0,\cdots,c_k$ natural numbers and $p(x)$ a polynomial (in hindsight, not necessarily the original divisor) with natural coefficients. On the other hand, it is apparent that, even if $s(x),t(x)$ have natural coefficients, the coefficients of the quotient $q(x)$ and remainder $r(x)$ in the division $$\frac{t(x)}{s(x)}=q(x)+\frac{r(x)}{s(x)}$$ are in general rational, non-integer, numbers.

In fact, had you used a tad of self-criticism on your own work, you would have noticed that this is the case for $t(x)=7x^2+3x+9$ and $s(x)=2x+3$, thus answering yourself.