Questions (I'd rather not make changes to the page since it's out of my area of expertise):
Further examples of polynomials (some are monomials which form a special case with only one term):
If somebody wants to integrate my writeup on E2 to here, feel free. The AC Method may be of particular interest. This is primarily just telling how to factor polynomials so there might be a better place (i.e. factoring[?]) to put it. For simplicity, I'll post a partially wikified version here. If you think it's useful, integrate it. Else, just remove it: http://everything2.org/?node_id=895118 (Note: could contain some errors.)
anxn + an-1xn-1 + an-2xn-2. . . a1x + a0
The degree of a polynomial is the highest total of powers[?] of variables[?] (x, y, etc.) of a single term, so in the polynomial 2xy2 + x2 the degree is three (in the first term, x has a power of one). The standard form of a polynomial is when you write it with the degrees descending (x2 + x + 3, not x + x2 + 3)
To factor a polynomial (If you already know how to then skip down to the AC method. You'll like it. A lot.) you first factor out the common factor[?], if there is one, using the distributive property:
Ex 1) 2x^2 + 4x = 2x(x + 2)
With a binomial[?] (two terms, as in Ex 1) that's all. If you have a trinomial[?] (three terms, as in Ex 2) you're just getting started.
You usually have to find two binomials[?] (B1 and B2) whose first terms multiply to the first term of your trinomial, last terms multiply to the last term of the trinomial, and B1's first term times B2's last term plus vise versa equals the middle term (FOIL[?] users: Inside + Outside=Middle)
Ex 3) x2 + 3x + 2 = (x + 1)(x + 2)
If the first term of your trinomial has a coefficient (a) of 1--as shown above--then the first terms of the binomials are x. Otherwise, you have to play around searching for the proper factors to get it right. That's where the following method comes in:
The AC Method
First factor out the common factor[?]. Always, always, always do this.
Ex 4) 6x2 + 2x-4
Now, I know you're thinking, "What if I have a four-term (or more) polynomial?" Easy: Take a few terms, and slap parenthesis around them (Hint, put together terms that have common factors[?] or that look like they'll factor easily.)
Ex 5) 2x3 - 3x2 + 4x - 6
That last example (first and last steps anyway) was taken from College Algebra[?] by Michael Sullivan[?] because I was having a heck of a time making up a good example. (I'm always coming up with prime polynomials in my example and having to modify them so I can factor them. I wish my math teacher had let me do that in my homework.)
Now you need to do some heavy memorising. These are special polynomials and how to factor them. Knowing how to recognise them will help you enormously, both in multiplication and factoring[?]:
Difference of Squares[?]: x2 - a2 = (x - a)(x + a) (Ex 6) x2 - 144 = (x + 12)(x - 12))
Take the coefficients[?] of (x + y)n and look at the nth row of Pascal's Triangle (the "1" at the top is 0th). Cute and useful.
Sorry for the flood. :-)
Algebra is a subject. Calculus, on the other hand, is something of a hodge-podge --- a collection of subjects that the curriculum brings together. Algebra goes far beyond those things that most students see, and is a subject to which careers of some researchers are devoted. The topics that go far beyond calculus, on the other hand, are not called "calculus", but go by other names, such as "analysis" and "topology". Therefore, it makes sense to say "in algebra", but not as much to say "in calculus". Polynomials of course appear in calculus, as do many things from algebra. -- Mike Hardy
The reason I separated calculus and algebra is that in algebra, one has to distinguish between polynomials and polynomial functions, while in calculus one doesn't. This point is now lost, in fact the first sentence seems to suggest that the two concepts are the same, which they are only in sloppy calculus usage. AxelBoldt 23:41 Nov 30, 2002 (UTC)
I think this article would be improved if some knowledgeable person would add a few sentences about the Fuchsian Function solution to the paragraph which discusses roots of nth order polynomials. They are hinted at with the existing phrase "degree 5 eluded researchers for a long time", which suggests that a solution was eventually found, but this solution is not mentioned in the article. A new article on Fuchsian Funtions would also be welcome. kielhorn@portland.quik.com Dec 22, 2002
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