Bezout's Theorem

Chang Woo-JIn

Hello~ Nice to meet everyone Smile

I wonder what Bezout's Theorem is.

If someone know about it, Please post it..

**Posted:** Thu Feb 12, 2004 4:04 am

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Valentin Vornicu

Okay. Bezout theorem's is a simple but very used Theorem. It states that for a polynomial f(x) we have that f(x) = (x-a)g(x) where g(x) is another polynomial if and only if f(a)=0.

**Posted:** Thu Feb 12, 2004 4:19 am

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Arne

Euhm, in Belgium we call the following theorem 'Bezout's theorem':

If a, b are integers then there exist integers m, n such that am + bn = gcd(a, b).

**Posted:** Thu Feb 12, 2004 11:38 am

3X.lich

This is the first time I have heard of 'Bezout's Theorem' and yet the two theorems are already known to me. First one is the factor theorem and the other one is the theorem of Euclid's Algorithm of finding gcd's.

**Posted:** Thu Feb 12, 2004 11:42 am

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belenos · Riemann Hypothesis Offline Joined: 12 Sep 2003 Posts: 463

http://mathworld.wolfram.com/BezoutsIdentity.html
and
http://mathworld.wolfram.com/BezoutsTheorem.html

Actually it is Claude Gaspard Bachet de M�ziriac who discovered this theorem and not Etienne B�zout. But it's not that important of course Mr. Green

**Posted:** Thu Feb 12, 2004 11:48 am

Chang Woo-JIn

^^ I found what I wanted

In projective field, generally, the number of intersections of two same-

powered polynomials is equal to the square power of them.

This theorem can be used in any field...

(Sorry for my broken English..)

**Posted:** Sat Feb 14, 2004 7:52 am

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mashimaro

me too,sorry for my English

**Posted:** Wed Dec 29, 2004 6:34 am

Devendra Patel

I never knew its called Bezauts Theorem,
We use it extensively to factorise polynomials of the order 3 and above and also to find the other factors, we call it the factor theorem

**Posted:** Tue May 10, 2005 10:27 pm

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gkodumudi

**Posted:** Thu Jun 16, 2005 6:06 pm

bomb · Riemann Hypothesis Offline Joined: 29 Sep 2004 Posts: 363

Bezout's theorem goes by many names. It's also known as factor/ remainder thorem for polynoms.

Bomb

**Posted:** Fri Jun 24, 2005 9:09 pm

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leibniz

I think that bezout's theorem exist also in arithmetic Question

**Posted:** Sun Jul 03, 2005 7:54 am

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gotztahbeazn

isn't bezouts theorem kinda like the division of polynomials thing?

where you have $f(x) = q(x)g(x) + r(x)$

in this case you are dividing $f(x)$ by some $q(x) = x-a$

since it states that the theorem holds for when $f(a) = 0$ , that just means that $f(x)$ is divisible without remainders by $x-a$ , meaning $r(x) =0$

so we just get $f(x) = (x-a)g(x)$

yea i ono that may have seemed obvious but that's just what i was thinking about in my head

**Posted:** Sat Aug 27, 2005 12:13 pm

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nuahonthuongdau

We should distinguish the Bezout theorem about Polinimial to the Bezout theo rem for Number Theory. Two theorems have many useful to solve and understand many problems.

**Posted:** Thu Dec 29, 2005 6:58 pm

leepakhin

Isn't that the Factor Theorem?! Shocked

**Posted:** Fri Dec 30, 2005 7:51 am

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$\int\int_\Omega\,dA=\int\int_\Sigma \sqrt{E(u,v)G(u,v)-F(u,v)^{2}}\,du\,dv$
$P_{G}(\sigma_{1}, \sigma_{2}, \cdots, \sigma_{n})=\frac{1}{|G|}\sum_{\tau\in G}ind(\tau)$

mathmanman

Read the above posts Razz

**Posted:** Fri Dec 30, 2005 7:59 am

leepakhin

From the above posts, I know that there are two versions of Bezout's Theorem: one in algebra and one in number theory. For the algebra one, I feel shocked because it is just the same as Factor Theorem.

**Posted:** Fri Dec 30, 2005 8:09 am

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$\int\int_\Omega\,dA=\int\int_\Sigma \sqrt{E(u,v)G(u,v)-F(u,v)^{2}}\,du\,dv$
$P_{G}(\sigma_{1}, \sigma_{2}, \cdots, \sigma_{n})=\frac{1}{|G|}\sum_{\tau\in G}ind(\tau)$

mathmanman

Yes, and this has already been said Smile

(about the factor theorem)

**Posted:** Fri Dec 30, 2005 8:25 am

Nukular

The following "Bezout's theorem" appears in algebraic geometry:

In complex projective space, a variety of degree $n$ intersects a variety of degree $m$ in exactly $mn$ points, when the varieties are in general position. More fundamentally, the zero set of polynomial of degreem $m$ intersects the zero set of a polynomial of degree $n$ in exactly $mn$ points.

**Posted:** Thu Jan 26, 2006 3:57 pm

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jmerry

And "points" isn't exactly right, either. We count them with multiplicity, and the precise definition of multiplicity is algebraic.

**Posted:** Thu Jan 26, 2006 4:11 pm

Nukular

yes, I wrote "general position" when I meant "generic," trying to get rid of those nasty multiplicities Wink

.

**Posted:** Thu Jan 26, 2006 5:15 pm

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