 ## How to describe the behavior of the function?

For example,
P(x)=2x^3+x^2+3x+2 would be: „Falls on the left and rises on the right“.
I don’t really get of how to find that out without using the calculator.

Could you please explain that to me?

Thanks!

The key is the leading term. In this case 2x^3 has a HUGE amount of information there. To make it simple, go through the following check list:

1.) Is the power on the leading term even or odd?
If it is even, then the two ends of the function will match – i.e. The graph will either go up on both ends OR go down on both ends

If it is odd, that beginning and ending behavior will be opposite. Up on one end, down on the other.

2.) So, then you need to look at the leading term coefficient (LTC). In this case you have „2“, but more important, is the fact that it is „positive“. The sign +/- of that leading term will tell you that.

Going back to the even powers: a + LTC will have „up-up“ on both ends of the function
A negative LTC has the effect of „reflecting“ or inverting the graph so it will take an „up-up“ and turn it upside down-down!

For odd powers, watch the ending behavior of the function. (furthest to the right as x gets larger)
If the LTC is positive, then the function will go up (increase) as x gets larger. So, on the left it will go down.
If the LTC is negative, again it will invert this relationship and the graph will look like „up-down“ instead.

If you get confused, break the idea down to the 2 simple graphs of „x squared“ for even powers or „x cubed“ for odd powers. The only difference that YOUR equation will have, is that it will have more turns in between those beginning and ending behaviors.

Hope this helps! ## How to factor a complex polynomial?

I learned that to factor things like x^3 – 24x^2 + 192x – 512 you can use a graphing calculator to find a zero, then use synthetic division to find the rest of the factors.
I also learned that you can use the Rational Root Theorem to narrow down the possibilities.

However, is there any way, without guessing and checking, to find factors? All these other options of either calculator or guess and check doesn’t sound too mathematical..

Use the example I stated previously as an example, and thanks.

Every polynomial of odd order will always have at least one real root. For cubic equations there is a closed form solution for the real root — it’s a little messy to compute, but it’s there. Having that real root you then divide it out as you describe to get a quadratic, which is then soluable using the quadratic equation. So… For cubics at least there is a ’no guessing‘ way to get all the roots. ## What are the combined odds of winning a prize; given there are 2 million entrants for each of 3 events?

I’m trying to get the odds of winning a prize given that there are 2 million entrants for each of 3 events, whereas there is a prize given to the winner of each event and that same person could win multiple events. It is assumed that you enter all 3 events and that you cannot enter each event mroe than once. also wondering if you have a better chance of winning any given prize if you enter each event or if you just enter one event.

The chances of winning the prize for any particular event that you enter will be 1 in 2 million or 0.0000005.

The chances of not winning a prize for a particular event is 1 – 1/2,000,000 = 0.9999995.

If you enter all 3 events then your chances of winning no prize is (1 – 1/2,000,000)^3 = 0.9999995^3 = approx. 0.9999985

So if you enter all 3 events, your chances of winning at least 1 prize = 1 – (1 – 1/2,000,000)^3 = approx. 0.0000015.

This number is actually slightly less than 3 times the probability of winning in 1 event. (My calculator can’t display enough digits to show the difference.)

Entering each event does not improve your chances of winning the prize for a particular event. How could it? Suppose the events were (I’m making this up) the California, Oregon and Washington State Lotteries. Each lottery doesn’t ‚know‘ whether you have entered the other ones. ## My abs are getting a little flabby–best way to get a 4 pack?

I don’t want a six pack because that’s kind of odd for a girl, but I want really toned abs. For the past two weeks I’ve been doing 3 sets of 25 crunches along with other exercises for other body parts, but what are some other ab workouts I can do? And please don’t just give me links unless you’ve actually tried those exercises. Thanks in advance!!
I play tennis every day for two hours at least.

Well, it’s cool that you play tennis! That’s an excellent form of cardio…I love tennis too–but can’t find a partner…:0( Anyhow, as for your abs…I lost 40lbs., in addition to getting those sleek abs I’ve always desired by following this plan:

Ab exercises won’t necessarily reduce fat over the abs. We can’t spot reduce fat from different areas of the body with specific exercises, although that doesn’t mean you can’t lose fat from the abdominal area. That can happen if you reduce your overall body fat, although your genes will determine where the fat comes off. The following steps will help you get started:
1.) Regular cardio exercise, do 30 minutes to one hour (preferably) a day of an aerobic exercise to get your heart rate up, so your body can burn fat.
2.) Strength training for your entire body 1-3 non-consecutive days a week. Try reading, “Total Body Strength for Beginners” at: http://exercise.about.com/cs/exbeginners…
3.) A healthy low-calorie diet. This is a very critical part of your program. To make it work you really should:
A. Keep a food log in which you write down everything you eat or drink.
B. Use a food calculator to determine about how many calories you eat each day.
4.) Begin changing bad habits by replacing them, one by one, with healthier choices (fruits and veggies will fill you up and give you nutrients and fiber).
5.) Keep an eye on your portion sizes.
6.) Make small changes every day instead of changing everything at once.
*This is my favorite site for abs…Pics included to guide you:
http://www.fitnessmagazine.com/fitness/c… ## Is there another way to get the rational zeros of a polynomial function?

I get sick of using synthetic method.I’m currently answering a worksheet with 25 items.I know it will take me a long time to do this.I wonder how my classmates did it so fast?When I use synthetic method,I try all the results of P/Q.

How do you think they calculated it so fast manually using the synthetic method?I mean they don’t use graphing calculators.

If you want to test whether st is a zero of ax^3 + bx^2 + cx + d, calculate
as^3/t^3 + bs^2/t^2 + cs/t + d
then calculate
-as^3//t^3 + bs^2/t^2 – cs/t + d
I think that might be quicker that doing synthetic division.

I think the „trick,“ to the extent there is one, is to try to get some information on possible roots before going through all the possibilities that the rational roots test gives you. It appears that you already know that test, so I won’t go into that here.

The first thing to do is to use Descartes‘ Rule of Signs. You count the sign changes on the coefficients. That tells you that the POSSIBLE number of POSITIVE roots is the number of changes, or that it’s less than that by a multiple of 2. Sounds confusing, but really simple once you use it a few times.

E.g., 5x^3 – 3x^2 + 2x + 7 has two sign changes (the coefficients are + – + +, so you start + change to -, change again to + and then stay + ). This polynomial has either 2 or 0 positive real roots. If you find one, there must be another one.

Then check for negative roots. You change signs on the odd terms (in effect substituting -x for x) and count the sign changes again. That tells you something about the possible number of negative roots. In my example, that gives you
– 5x^3 – 3x^2 – 2x + 7.

There is only one sign change. There can be no more than 1 negative root.

This was just a random example, which WolframAlpha tells me has 1 negative irrational root and 2 complex roots, by the way.

Once you have one root, factor it out or use synthetic division to get a quadratic, and then use the quadratic equation to find the other roots–start by just looking at the discriminant b^2 – 4ac, if you are just interested in real or rational roots. If the discriminant is positive and a perfect square, or zero, the roots are rational. If positive and not a perfect square, they are irrational. If negative, they are complex.

Here’s one more thought. You might be able to get a rough graph fairly quickly. E.g., in my example
5x^3 – 3x^2 + 2x + 7
the leading coefficient is positive, so you know that for very large negative x, like -100, this will be negative. And at x=0 this is +7. That, plus the info from Descartes Rule, tells you there MUST BE one negative zero. So I’d start the search by looking at negative possible roots.

You can also pretty quickly get f(-1) and f(1) by adding up
-5 -3 -2 +7 = -3 = f(-1)
and
5 -3 + 2 + 7 = 11 = f(1)

This info tells you that the negative root is between -1 and 0, since f(x) goes from negative to positive in that interval.

Unfortunately, there is always some work involved–there’s not very quick way. Hopefully these tips will help speed things up.