In polynomial long division, what happens when the lead terms' coefficients don't divide each other?

Asked Nov 15 '20 at 20:32

Active Nov 15 '20 at 20:43

Viewed 36 times

I'm currently tasked with acquiring polynomial long division. So far, all examples make sense. However, what happens in a situation like $(3x^3 + 2x^2 - 3) \div (2x^2 + 1)$?
To my knowledge, iterating once would yield as an interim step:
$(3x^3 + 2x^2 - 3) - (2x^3 + x) = x^2 + 2x^2 - x - 3$
But now, we've advanced no further in reducing the polynomial!

The only solution I can see is using non-integer coefficients. Is this how a situation like this is meant to be dealt with?

(Sorry for formatting this like it's not long division. I'm talking about the long division of polynomials.)

edited Nov 15 '20 at 20:43

asked Nov 15 '20 at 20:32

Mark J. Isaac Neu

1

Yes, this is the way. – Tito Eliatron Nov 15 '20 at 20:34
3

Yes, you have to perform the division in $\mathbf Q[X]$. – Bernard Nov 15 '20 at 20:39
So $\frac{3}{2}x + \text{ the rest}$? – Mark J. Isaac Neu Nov 15 '20 at 20:40
$$ \left( 3 x^{3} + 2 x^{2} - 3 \right) = \left( 2 x^{2} + 1 \right) \cdot \color{magenta}{ \left( \frac{ 3 x + 2 }{ 2 } \right) } + \left( \frac{ - 3 x - 8 }{ 2 } \right) $$ – Will Jagy Nov 15 '20 at 20:41
I mean long polynomial division – Mark J. Isaac Neu Nov 15 '20 at 20:43
The ring $\Bbb Z[X]$ doesn’t have Euclidean division, but $\Bbb Q[X]$ does. That’s the long and short of it. – Lubin Nov 15 '20 at 21:50

In polynomial long division, what happens when the lead terms' coefficients don't divide each other?

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