I'm following the proof of the Factor theorem in here but I don't understand how $\deg(X-r) = 1$.
Proof of Factor Theorem
3
$\begingroup$
abstract-algebra
polynomials
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0The [Wikipedi$a$ article on the degree of a polynomial](http://en.wikipedia.org/wiki/Degree_of_a_polynomial) might help. – 2012-07-13
1 Answers
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In a polynomial $a_nx^n + a_{n-1}x^{n-1} + \dots a_1x + a_0$, if $a_n \not = 0$, then we say the polynomial has degree $n$. The polynomial $ax + b$ is a special case, and is of degree $1$. It has the special name "linear polynomial" too.
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0Thanks, for some reason that I can't fathom out, I was assuming that X in here is a polynomial(probably because I write polys in capital)! I was just subtracting r from that polynomial and thinking that ain't degree $1$, stupid mistake. – 2012-07-13