Here is a "Math is fun" quote: "When we divide by a polynomial of degree $1$ (such as "$x-3$") the remainder will have degree $0$ (in other words a constant, like "$4$")."

I'm hoping someone could give me intuitive proof of why that is.

  • $\begingroup$ oops i meant the remainder $\endgroup$ – Carefullcars May 19 '15 at 15:46
  • $\begingroup$ Strictly speaking, the possible remainder $0$ has degree $-\infty$ by convention. $\endgroup$ – Travis May 19 '15 at 15:47
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    $\begingroup$ Think informally about the "long division" process. As long as what is left has degree $\ge 1$, the process continues. $\endgroup$ – André Nicolas May 19 '15 at 15:51
  • $\begingroup$ This may not be the case for polynomials over a ring instead of a field. For example, in $\mathbb Z[X]$, whenever you have $X^2-1=Q(X)\cdot(2X-3)+R(X)$, with $Q,R\in\mathbb Z[X]$, you will have $\deg R>0$. $\endgroup$ – Hagen von Eitzen May 19 '15 at 15:58
  • $\begingroup$ It is true over any commutative ring as long as the lead coeff of the divisor is a unit (invertible). If not, then the division is possible if one scales the dividend by a power of the lead coef, see the nonmonic polynomial division algorithm. $\endgroup$ – Bill Dubuque May 19 '15 at 16:07

When you divide by a number, the remainder should always be less than that number - otherwise, you could "put in" one more:

$27 : 5 = 5\:\mathrm{rem}\:2$, not $4\:\mathrm{rem}\:7$.

Similarly, when you divide by a polynomial, the remainder should always be less (in degree) than your divisor polynom:

$\frac{x^2+3x+1}{x+1}=x+2-\frac{1}{x+1}$, not $x+1+\frac{x}{x+1}$.

When you divide by a polynomial, you eliminate everything of a degree greater than or equal to the degree of you divisor, and leave only what remains (which is less than this degree).


When we do Polynomial Long Division, we are actually using Euclidean division algorithm. We are finding $q$ and $r$ such that $$a=bq+r$$ and $$\deg(r)<\deg(b)$$By definition, it is stated that we are only interested in solutions where the divisor has a greater degree, even though other solutions do exist.

Therefore, if $\deg(b)=1$, we have $\deg(r)=0$.


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