# I have been thinking about this. Lets say we take 7 divided by 3, we know the remainder is 1. However, if we let x=7 and x-4=3, and we take x/(x-4), after performing long division the remainder is 4. Why is it not 1?

Remainder and long division
I have been thinking about this. Lets say we take 7 divided by 3, we know the remainder is 1. However, if we let x=7 and x-4=3, and we take x/(x-4), after performing long division the remainder is 4. Why is it not 1?
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ahem37
Because the quotients are different:
$7=2\cdot 3+1$
$x=1\cdot \left(x-4\right)+4$
When $x=7$ we get
$7=1\cdot 3+4$
which is correct, of course.
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Janet Hart
If someone handed you $x$ and $x-4$, why would you assume that they are $7$ and $3$?
In general, we know that $\frac{x}{x-4}=1+\frac{4}{x-4}$
For $x=7$, it is the same that $\frac{7}{7-4}=1+\frac{4}{7-4}$, just like any arbitrary $x$
However, it just so happens that $1+\frac{4}{7-4}$ can be rewritten as $2+\frac{1}{7-4}$
The polynomial remainder is still $4$