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.
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oops i meant the remainder – Carefullcars May 19 '15 at 15:46
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Strictly speaking, the possible remainder $0$ has degree $-\infty$ by convention. – Travis Willse May 19 '15 at 15:47
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3Think informally about the "long division" process. As long as what is left has degree $\ge 1$, the process continues. – André Nicolas May 19 '15 at 15:51
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1This 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 $\deg R>0$. – Hagen von Eitzen May 19 '15 at 15:58
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1It 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. – Bill Dubuque May 19 '15 at 16:07
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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).
pascalhein
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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 $\deg(r)=0$.
pascalhein
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