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October 27th, 2007, 10:03 AM  #1 
Newbie Joined: Oct 2007 Posts: 5 Thanks: 0  Polynomials and Cones
Hi all, I'm new both here and to mathematics. I'm working on factoring 3rd and 4th order polynomials and finding it difficult to understand. What I want to avoid is getting into a habit of learning math by memorizing procedures and forumulae, which is exactly how I was originally taught in highschool. Lately I have started to appreciate math conceptually and imaginatively. By that I mean that I can see *why* a simple thing like a cube can be described by mathematics  or is mathematics? (See my other thread...) That kind of learning is dominant in early grades of learning, but eventually gets abandoned  probably due to difficulty  and math becomes memorizing and solving hundreds of problems. Great preparation for a test but I was not learning anything at all. I have got to Polynomials working through Barron's "Algebra" book and hit a wall. I could see how the 2nd order polynomials broke down into factors but I couldn't not really visualize what was going on. Also, it was unhelpful that the book recommended "guessing" as a way of finding the solutions. I won't always be working with problems which I can guess at, and that leaves me incapacitated. The 3rd and 4th order stuff is just beyond me at this point. I can see the solutions there on the page and follow it but I feel like I'm just memorizing procedures again. I want to understand this stuff more deeply, if that makes sense. What I am trying to do is develop a conceptual framework in my mind of mathematics for each unit. I'm not trying to accumulate abilities, I'm trying to get my imagination into math. Does this make sense? Therefore, can anyone recommend to me a resource which will help me in this direction? Very disappointed with my textbooks and the books at the bookstore... and my university library stuff is either too basic or too advanced for me. There seems to be no literature which can teach mathematical *thinking*. Why is that? As an example of what doesn't work for me, any time the author does little more than show the transformation of an equation by steps all I get is memorization or procedure. Math books for example, which have very little written english, are usually quite useless resources for me in this direction. I think that is because I have no yet learned to think mathematically. Without that imaginative aspect, it is all very "surface" level and uninteresting. But when someone shows me how to imagine what I am seeing in the math I get an "Ah ha!"... I'm hoping there is an author here or elsewhere that might be able to do that for the more advanced stuff. A polynomial equivalent of showing how a line becomes a plane, so to speak. Hope I am making sense. 
October 28th, 2007, 01:21 PM  #2 
Site Founder Joined: Nov 2006 From: France Posts: 824 Thanks: 7 
Very clever point you are discussing now, as it seems that a mathematician reaches its limits once he loses his ability to conceptualize. I myself cannot conceptualize anything in advanced complex analysis, which doesn't prevent me to solve the problems by analogies but I don't really have a logical framework in my mind (for those of you who are somewhat advanced in math, how do you represent yourself an holomorphic function ?). Concerning the 2nd degree polynomials, you can see where the things come from by "completing the square". This is quite a natural approach indeed, as when you have such a polynomial, you may want to write it as a product of two first degree polynomials, and when dealing with squares, things get easier. As for guessing the roots, well, you may want to have a look at the rational root test first. Of course, then, there are a lot of numeric algorithms to find roots but i think this is not the topic of this post. You'll find out that conceptualizing polynomials is pretty easy after a while, but things may become pretty trickier when you start to tackle notions which are generalizations of no reallife ground (like topological spaces), because then you must be able to conceptualize these notions for yourself. 
October 28th, 2007, 02:32 PM  #3 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 
The first thing is this: Don't get discouraged. I've never read a math book and understood it my first time through, unless I already knew the material beforehand. You can't pick up a math book the same way you can a novel (Something I just learned within the last year...) and expect to read through and understand. You have to fight with it a little bit. I've found that when I read math books, I get a lot more out of them, if I start reading, and just try to follow along (not stopping to do exercises, etc.) until I suddenly realize that I'm completely lost, then I go back to the beginning and read through it carefully, taking notes and doing exercises, until I get to where I left off; then I start again from that point. This does a few things: 1) gets you used to the writer's style before you have to go about the daunting task of understanding what the hell he's talking about. 2)Provides a contextual framework. It's easier to understand something if you know both where it's coming form and where it's going to. 3)Gives you a very (very) basic understand of the material before you "actually" learn it. Down to more specific questions: Polynomials are like pieces that go together to make a whole. Think of it like a jigsaw puzzle, or lego blocks, or something to that effect. When you factor, you're taking these pieces apart. You said you can't really "visualize" what's happening with a second order polynomial, so I'll give you an example one. And I'll explain what I'm doing, and (hopefully) why I'm doing it. x^2+7x+12. You have here 2 parts (ax+b) and (cx+d). They're locked together right now, and you want to take them apart. One thing that you should know is that (ax+b)*(cx+d)= acx^2+adx+bcx+bd(1^2). The emphasis is on the middle part, the "x". This middle part is where the two "lock together". The easiest way I've learned to factor second order polynomials is what my teacher referred to as "divide and conquer": you know the middle term must have a coefficient that is ad+bc, so you try to figure out which numbers work. Step one is to multiply the coefficients on the square terms (the "1" part is squared) In general, this gets you a*b*c*d; in our example, it's 1*12=12 Now we want the factor pairs of this number; In general, this gets us all of the following: (ab,cd), (ac,bd), (ad,bc), (a,bcd).... you get the point. In our case, again: (1,12)(2,6),(3,4). Now, the middle term, we know, is (ad+bc)*x. How do we figure out what 'ac' and 'bd' are? As we saw above, one of our factor pairs is (ad,bc). If we just add the two halves of the pair together, we'll have (ad+bc). But which one? Each pair looks like (m,n); we know that for one of them m+n=ad+bc, so just find the one where m+n is the middle term. In our case, the middle coefficient is 7; 3+4 = 7, so we know either ad =3 and bc = 4, or ad =4 and bc = 3. Now we wanted to do that so we could separate ad from bc. So, in general, we now have: acx^2 + adx + bcx + bd. (In our case: x^2 + 4x + 3x + 12)This next part is perhaps the hardest part to understand, while not too difficult to actually do: You can factor the ax out of the first two terms, and the bx out of the last two: ax(cx+d)+bcx+bd > ax(cx+d)+b(cx+d). You know you've done it right if you have ax*(some polynomial) + b*(the same polynomial). If the polynomial within each pair of parenthesis are different, you can't do the last step: ax(anything)+b(the same anything)=(ax+b)(the same anything). Tis is the same concept as x+yx=x(1+y), it's just that now you have (x+n) rather than (x). In our case: x^2+4x+3x+12 > x(x+4)+3(x+4) > (x+3)(x+4) And now it's factored. In principle, third and fourth order polynomials are exactly the same, but in practice they are messier; also, the higher the order, the less common it is for things to factor cleanly. Alas, I haven't worked with anything but trivial higher order polynomials in years, and have lost the talent (and motivation) to teach how. Hopefully, you can now "see" what's happening when factoring a second order polynomial, rather than just following the formula. And remember, don't expect to understand with a single reading. 
October 28th, 2007, 03:48 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,654 Thanks: 2087 
Don't worry about 3rd and 4th order polynomials. If you can find one factor, you can simplify the problem by dividing by that factor. Also, a polynomial of odd degree with real coefficients must have at least one real zero. Anything beyond that is just more trouble than it's worth at the stage you're at.

October 30th, 2007, 02:29 PM  #5 
Newbie Joined: Oct 2007 Posts: 5 Thanks: 0 
Thanks very much for the help. Good advice too, I fear rationalized frustration creeps in sometimes. cknapp the first section of your example was really helpful, thinking of how the factors lock together. This might be a dumb question, but since the eqation is a square why is it a parabola? Is that because it is a plane? I found a picture here http://en.wikipedia.org/wiki/Image:Conic_sections_2.png that seems to confirm that. So does every value for x and y in the slice of that cone section get described by the equation? Where do the a & b's fit in the plane? And just a curiousity  would we ever have first, (as in order of events), the factors of a polynomial before the equation? What would those be? Getting there I hope... I'm not giving up 
October 31st, 2007, 03:18 PM  #6  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Quote:
This may end up being a bit abstract, but we shall see: Functions "take in" a value and "output" another value. In general, you take one set (You've probably only dealt with the reals, and integers) and "map" it to another set (Or the same set.) In the notation you're used to, y is typically a function of x, and both y and x are arbitrary (but related) members of the reals; which is why sometimes you'll see f(x)=2x interchangeable with y=2x. Functions can be visually represented on a "Cartesian Space", that is, the set made by assigning each pair (member from set one, member from set 2) to a "point" in the plane. To give an example, most people have only experience with the Cartesian space RxR, which equates to the set of all points (x,y) where x and y are real numbers. With 2 dimensions, this Cartesian space is a plane. Now, moving back to functions; as far as I would guess, every function you'll be working with in the next few years will be on a cartesian plane between the reals and the reals. Also, y will almost always equal f(x). That is, the x axis represents one set of the reals, the y axis represents the set that the function maps x to. So, onto our example, with parabolas. The simples tone is y=x^2. So at every value of x, the function outputs to x^2. Visually, this is the point (x,x^2). What you see is a parabola. Not sure how well that answered that question. Quote:
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November 12th, 2007, 06:33 PM  #7 
Newbie Joined: Oct 2007 Posts: 5 Thanks: 0 
Hi cknapp, Sorry for the delay responding, Halloween and 3 birthdays made things very busy. Thanks for this. When we look at the parabola on the graph (x,y), are we looking at the plane from the side? I'm used to thinking of planes as being set by a boundary and including all the values of x,y in between those boundaries. So, when I look at the parabola as a plane, I tend to think it must be like that. Here I drew a hasty parabola in MSPAINT and also drew some colored lines to show two different areas I want to ask about. The blue area represents what I interpret as the "plane" of the parabola. I'm not sure if those values for x,y in the blue area are actually a part of it or not. IOW, are the points x,y of a parabola *strictly* the parabola line or are the areas within and outside of that line called something particularly? The red area would be the outside of the cone, the "exclusion zone" marked by the cone boundary described by that parabola. Here is a picture from Wikipedia that does the same thing but with better artistic skills It seems that the parabolic line represents the boundary, however, I'm not sure what to call the coloured areas. Also, if you rotate the parabola about the x axis in 3d space, you can see 3 areas, because the parabola is flat. The area on the "plane" of the parabola is green (my guess), but the red area completely surrounds it. In other words, does the parabola extend infinitely in the "Z" axis? I guess I'm getting away from cones there a bit. Interesting stuff anyhow. Thanks for the help. 
November 13th, 2007, 08:02 AM  #8  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Quote:
Quote:
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If you bounded either the plane or the cone, that is, took only the region on the object that is above [or below] a certain value of x,y, and or z, the intersection between the two would obviously be bounded. but typically, you don't place artificial bounds on a function. Cheers  
November 13th, 2007, 09:19 AM  #9 
Newbie Joined: Nov 2007 From: East West Kerry Posts: 2 Thanks: 0 
For me the best way to solve this problem is with recourse to the factor theorem. If a,b,c,d,.. are roots of the polynomial. the poly factors as (xa)(xb)(xc)... This is easy to visualise if you consider the graph of the poly. At any of the roots, a say, p(a)=0. Hence your factorisation p(x)=(xa)(xb)(xc)... will equal 0 also. Hence to factor a poly plot the poly from [4,4] at 4,3,...,4. When the plot crosses the xaxis, the poly is zero there and hence the cross point, x=a say, is a root. So factor out (xa) with long division of polys  this is intuitive enough. Then work with your new poly; or revert to the normal equations. This works because realistically any poly you will be asked to factor will have an integer root in [4,4]. This will work for the third degree derived poly also. When you get down to a twodegree poly you have to realise that the roots are going to be often be surds and you won't be able to catch the roots by chance. Loooking at the graph of p(x) and understanding the factor theorem will satisfy the conceptional standards I hope. Let p(x) in P_2[x] have factors (Yb/2)(Y+b/2). See where that leads. Now a teaser: (xa)(xb)(xc)...(xz)=? 
November 15th, 2007, 01:44 PM  #10 
Newbie Joined: Oct 2007 Posts: 5 Thanks: 0 
Got it now cknapp, thanks!


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