Talk:Polynomial remainder theorem

Regarding edits of 8th November 06

Firstly, the general form for polynomial long division is:

${\frac {f(x)}{g(x)}}=q(x)+{\frac {r(x)}{g(x)}}$

and not

${\frac {f(x)}{g(x)}}=q(x)+{\frac {r}{g(x)}}$

Consequently, we need to explain how to get from $r(x)$ to $r$ in the second equation, which was already in place before the edits. I have now reverted this.

Secondly, "if the remainder is zero, then the linear divisor is a factor" is correct, and not "if the remainder is zero, then the linear factor is a divisor". Oli Filth 08:50, 8 November 2006 (UTC)

Theorem's Discoverer?

Okay, so how come it doesn't say who was the person that found out the Theorem? Not only it would be useful information, but doesn't the scientist deserve it? I am currently unsure of the English spelling of his name, but if nobody finds out before me, I will put it up. 79.101.183.132 (talk) 16:57, 22 May 2008 (UTC)

I think it is fair to credit the Remainder Theorem to René Descartes. He certainly stated the closely related Factor Theorem explicitly in his La Géometrie in 1637 and he surely understood its relationship to polynomial division. More details can be found at the Theorem of the Day entry. —Preceding unsigned comment added by Charleswallingford (talk • contribs) 12:41, 2 November 2009 (UTC)

Examples

In the second example, in which the theorem is to be proved for the arbitrary second degree polynomal $f(x)=ax^{2}+bx+c$, I added the explanation that ${\frac {ax(x-r)+(b+ar)x+c}{x-r}}$ results from adding and subtracting arx in the numerator then regrouping terms. I believe this kind of elaboration is useful in mathematics articles, for those who understand the basic principle at work--in this case, algebraic manipulation--but still lack the experience to recognize immediately what is being done, and hence may not understand how the proof proceeds from one step to the next.Pithecanthropus4152 (talk) 21:53, 3 July 2015 (UTC)

Generalized Polynomial Remainder Theorem

Sorry, I have no editing experience on wikipedia. I would like to point out that the Remainder Theorem and Factor Theorem can be easily extended to non-linear divisors. Perhaps you could consider the possibility of inserting a simple example on the page, like the following.

Let $\ f=2x^{15}+5x^{3}+x^{2}-3x+3\quad {\text{and}}\quad m=x^{3}-2\$. Substituting $\ 2\$ for $\ x^{3}\$ in $\ f(x)\$ we obtain

$\ 2\cdot 2^{5}+5\cdot 2+x^{2}-3x+3=64+10+x^{2}-3x+3=x^{2}-3x+77$

which is the remainder of $\ f\$ on division by $\ m$. Then $\ x^{3}-2\$ does not divide $\ 2x^{15}+5x^{3}+x^{2}-3x+3\$ . Flaudano (talk) 23:22, 28 March 2019 (UTC)

References

The first sentence of the article says essentially that the polynomial remainder theorem is a special case of Euclidean division of polynomials when the divisor has degree 1. What you propose is simply a classical way of representing Euclidean division as a rewrite rule: If the divisor is written as $x^{n}-q(x),$ where q has degree less than n, then the remainder may be eventually obtained by substituting $q(x)$ for $x^{n}$ in $f(x)$ until all terms have a degree lower than n. This belongs to the article Euclidean division and should be added there if it is not already there. In any case, this can be generalized further to multivariate polynomials, see Multivariate division algorithm. As this way of viewing Euclidean division seems to be lacking in WP, it could be useful to add here as a generalization. D.Lazard (talk) 23:52, 28 March 2019 (UTC)

The statement "... x-r is a divisor of f(x) if and only if ..." can be misleading because a divisor need not be a factor otherwise what would you call the divisor when there is a remainder? Therefore, you should change this to "...x-r is a factor of f(x)...". 2603:7080:AFF0:60:9CB3:F538:6012:B625 (talk) 14:48, 20 August 2022 (UTC)