Tuesday, May 6, 2008

Rational Zero Theorem

http://www.purplemath.com/modules/systprob2.htm
Solve decompostition problem

Next

Find the equation of the parabola that passes through the points (–1, 9), (1, 5), and (2, 12).

33 comments:

C.J. said...

you break up the problem and devide the numerator by the factors of the denominator

Robert Southworth said...

Factor the denominator and then im lost

audra said...

you separate the equation and divide the number above the division sign by the factors of the number below the division sign.

coolconor said...

In the hood math is frowned upon

Robert Southworth said...

i get lost at A B C when u need to distribute

Kay said...

you break the problem down as much as you can and just then solve it and yeahh..

jessica w said...

to solve this i would break up the denominator by factoring out the equation. i would seperate the denominator into seperate fractions and put variables on top. then you do something that reminds me of cramers rule. this will give you three equations which you then solve by any means; substitution, graphing (not as acurate) or any other way. solving these equations will give you the numerations from the begining equation.

coolconor said...

conorhotblog.blogspot.com or mybetterisbetterthenyourbetter.blogspot.com

The Better Campbell said...

the distributing seems easy now that u look at the answer

Steve S said...

you gotta seperate the problem, then simplify or something. i dont really get it.

emilya said...

There was an x in the original problem then they just start adding more letters for some dumb reason, then they factor it, i get that, then they multiply it by something and "compare coefficients" (w/e that means) then they do something that gets them more numbers that they claim are the answers.

Max Wiese said...

Mr. Grasso this problem would be better without the suck in it. They lost me at the step:
=(A+B+C)x^2+(B+3C)x+(-A-2B+2C)1. I dont feel like putting any effort into this and I'm sitting next to Tim who wrote a thesis on this thing making me look like a tool.

Tiim W said...
This comment has been removed by the author.
Dan said...

You seperate the problem and then divide the numerator by the factors of the denominator...

ICBGHHOB654 said...

I have the slightest idea about how to solve this problem

tory said...

you simplify the problem by factoring the denominator and labeling the numerator as a,b,c etc. Then you you make an equation using a numerator term that will be factored to the other denominators, but you don't use its that numerators own denominator. Then you solve, and its confusing....

Blog Manager said...
This comment has been removed by the author.
bri valeri said...

you factor the denominator, separate it and then im lost with the rest.

emily j said...

I think I might understand the decomposition problem. You factor and break it up and somehow it all works out. The only problem I really had with it was that the trees in the first problem were wicked cheap. I wish I could get trees that cheap...

shannon said...

well first you need to know the factors of the denominator. you then seperate the equation and divide the numerator by the factors of the denominator.

P SOUTHSIDE said...

i get it now after a little help from nigel
you do A/B*C B/A*C C/A*B
THEN YOU PULL OUT FACTORS NOT AS HARD ONCE YOU GET IT

P.S. CAPS ARE WAY COOLER

justin said...

i can factor the denominator then i get lost and have no clue what to do next.

P SOUTHSIDE said...

DAN I MADE A TRADE WITH JAKE SO THAT 20 BUCKS WILL SOON BE MINE

maria said...

read some of the comments for this post, and that's how you do the problem

Tim W said...

First, you have to factor the denominatior. After this you put A B and C over the factors. Then you have to take the denominators of what is not under the letter to foil it out. Then you take the denominators of not B to use for B. You do that for C but use the denominators of A and B. Now you foil what you have. I do not get what you do on the next step. I can't figure out where the numbers are coming from. You set A+B+C equal to O. Then you take the equation equal with X and set it equal to 5 since that has an X. After you set the quation with the 1 and set it equal to 7 since that is a constant. Then you can solve the systems of equations to get A,B, and C. You put these over the factors.

coolconor said...

First, you have to factor the denominatior. After this you put A B and C over the factors. Then you have to take the denominators of what is not under the letter to foil it out. Then you take the denominators of not B to use for B. You do that for C but use the denominators of A and B. Now you foil what you have. I do not get what you do on the next step. I can't figure out where the numbers are coming from. You set A+B+C equal to O. Then you take the equation equal with X and set it equal to 5 since that has an X. After you set the quation with the 1 and set it equal to 7 since that is a constant. Then you can solve the systems of equations to get A,B, and C. You put these over the factors.

Nigel said...

You factor the denominator and set the factors under A,B and C. Then you multiply by the common denominator and get the numerator equal to "A" times the two other factors and so on. Then you regroup them with their factors. Finally, you set them equal to the factors in the numerator and solve

coolconor said...

OR You factor the denominator and set the factors under A,B and C. Then you multiply by the common denominator and get the numerator equal to "A" times the two other factors and so on. Then you regroup them with their factors. Finally, you set them equal to the factors in the numerator and solve

tory said...
This comment has been removed by the author.
Blog Manager said...
This comment has been removed by the author.
iitwmemasbedi212 said...

First, you have to factor the denominatior. After this you put A B and C over the factors. Then you have to take the denominators of what is not under the letter to foil it out. Then you take the denominators of not B to use for B. You do that for C but use the denominators of A and B. Now you foil what you have. I do not get what you do on the next step. I can't figure out where the numbers are coming from. You set A+B+C equal to O. Then you take the equation equal with X and set it equal to 5 since that has an X. After you set the quation with the 1 and set it equal to 7 since that is a constant. Then you can solve the systems of equations to get A,B, and C. You put these over the factors.

Blog Manager said...

Finding the partial fraction is basically just separating one huge fraction into three smaller fractions. First you are to factor out the denominator as much as you can. Then you are to place variables on top of these factored denominators. After that you create an equation by taking your first numerator and setting it equal to the variables and the opposite denominators. Then you are to solve the equation for each variable.

From Sarah..

P SOUTHSIDE said...

we should watch Miracle in class

It has hockey and that quote i always think since family size doritios. "Cause we're a family"