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In one my question, a guy told me about the long divison of polynomials and said that-

"Given any polynomials $f$ and $g$, there exist polynomials $q$ (the quotient) and $r$ (remainder) such that $$ f=q⋅g+r $$

and the degree of $r$ is strictly smaller than the degree of $g$."

I understand that in divison the reminder is smaller than the divisor but is it necessary for the remainder to be smaller in degree than the divisor?

I mean for example if the divisor was $2x^2$ then is it possible for the remainder to be $x^2$ or does it have to be in the degree of $1$ or smaller?

Is it possible for the remainder to be the same degree as the divisor but still be smaller than it.

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  • $\begingroup$ What do you mean by "smaller"? The only reasonable way to order polynomials here is by degree. $\endgroup$ – florence Jul 6 '16 at 18:06
  • $\begingroup$ If you allow rational coefficients, yes, smaller in degree. $\endgroup$ – Will Jagy Jul 6 '16 at 18:18
  • $\begingroup$ @WillJagy Well, you could have coefficients like π/e for a really simple example, that works too, but things get really messy then(considering the simplicity of the problem) $\endgroup$ – Ariana Jul 6 '16 at 18:34
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The degree is all that it is possible to specify. Numerically, the situation is entirely unknown, and the coefficients may well be much larger in the remainder. Each step in a polynomial long division will multiply the divisor to eliminate the highest degree of the running remainder, and such a step only becomes impossible (and the division terminates) when the degree is below that of the divisor polynomial.

An example of non-integer coefficients

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No. Consider integers. Let's divide 25 by 4. We could say the answer is 5, with a remainder of 5, since 4 • 5 + 5 = 25. But 4 also divides the remainder, so a better answer is 6, with a remainder of 1. Similarly, in polynomial division, the remainder has to be smaller in degree than both dividend and divisor because it wasn't, you could still divide more.

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  • $\begingroup$ but 6 and 1 are both polynomials in degree of 0 too right?Then why does 1 have the same degree as 6? $\endgroup$ – MartianCactus Jul 6 '16 at 18:27
  • $\begingroup$ There is a difference between dividing polynomials and dividing integers. In polynomials divison ( with rational coefficients ) the remainder degree must be smaller than divisor degree but in integers ( in fact natural numbers )division ( the example above ) the remainder must be smaller in value $\endgroup$ – AmerYR Jul 6 '16 at 18:37
  • $\begingroup$ @Adi In this answer, we are assuming fractions do not exist, in polynomial long division, we allow fraction, but negative powers are not 'allowed' in polynomials(and polynomial long division) $\endgroup$ – Ariana Jul 6 '16 at 18:38
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A easy way to do this is to explain polynomial division, then it would be obvious why that is the case.

Let p be $4x^3+6x+15$ and q be $2x^2+x+1$

To divide p by q, first we try to multiply q by $cx^n$ and then subtract it from p such that the resulting polynomial has a smaller degree.

$2x^2 \times 2x=4x^3$(you could just divide the largest power of both peoples polynomials)

$2x \times q=4x^3+2x^2+2x=g_1$

$p-g_1=-2x^2+4x+15=r_1$

This is like normal long division, where you divide the frontmost digit, in polynomial long division, you 'eliminate' the highest power first.

From the first step, it should be obvious $p=2x\times q+r_1$

Well, for polynomial long division, you do it until you can't multiply q by $cx^n$ to subtract r, then that will be your remainder

$-1q=-2x^2-x-1$

$r_1-q=5x+16=r$

So now, we get $p=2x\times q+-1\times q+(5x+16)$

Simplifying it, we get $p=(2x-1)q+(5x+16)$

Hopefully this can clear any doubts about polynomial long division, if you have any doubts, feel free to ask in the comments

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