Good At Math?? Then Help Me Out!!!

[GRagland]

I tried all the possibilites for #4 and none worked except -3/2. Maybe thats the only rational zero. But there should be 1 pos. and 2 neg. zeros unless i did the teqnique to find the number of zeros wrong. By the way the method to find the number of pos. and neg. zeros is called Descartes rule of signs.

 

[aux]

Originally posted by: GRagland
I tried all the possibilites for #4 and none worked except -3/2. Maybe thats the only rational zero. But there should be 1 pos. and 2 neg. zeros unless i did the teqnique to find the number of zeros wrong. By the way the method to find the number of pos. and neg. zeros is called Descartes rule of signs.

I didn't check which ones are the zeros, but it's possible that there is only one rational zero. The other zeroes may be real but irrational numbers (see the definition above).

 

[aux]

Originally posted by: GRagland

Originally posted by: aux
For #4: let x=(r/s) is a (rational number which is a) zero of your equation, r and s integers with gcd(r,s) = 1, then plug r/s in the equation and multiply the resulting equation by s^3; the result should be:
2r^3 + 2(r^2)*s - 14 r*(s^2) - 21s^3 = 0
then 2(r^2)*s - 14 r*(s^2) - 21s^3 is divisible by s, 0 is divisible by everything inclusing s, therefore 2r^3 should be divisble by s but gcd(r,c) = 1 so 2 should be divisible by s, therefore the only possibile values for s are 1, -1, 2 and -2
also, 2r^3 + 2(r^2)*s - 14 r*(s^2) is divisible by r, 0 is divisible by r so 21s^3 sould be divisible by r, and again gcd(r,c) = 1 therefore 21 is divisible by r, thus the only possible values for r are 1, -1, 3, -3, 7, -7, 21, -21
then combining the possible values for r and s, you get the posibilities for x (there are gonna be quite a few of them), plug them in the equation and check which ones are zeroes indeed (there is a way to reduce the number of possible values that you need to check but it's too long to explain)

man, there are a lot of combinations. there is a way to find out the number of pos. and neg. zeros, by looking at the number of changes in sign. you know of this? i followed this (too long to explain) and got 1 positive zero and 2 negative zeros. i know -3/2 is a zero, so that means there is one more neg. zero. I tried the other possible negative zeros and none of them worked. Does that mean -3/2 has a multiplicity of 2????

In order to find the multiplicity of -3/2 you have to divide your polinomial by (x + 3/2) and see if the resulting polinomial has -3/2 as a zero, if so -- the multiplicity is at least 2, if not -- it is one; again -- there may be another negative zero of the polinomial, but irrational; you only have to find the rational ones

 

[GRagland]

Anyone knwo how to do #6? as x gets bigger it approaches zero, as it gets smaller it approaches but never reaches a specific number. that specific number is the asymptote. I found the verticle asymptote, which is zero. But i dont know how to find the horizontal.

 

[GRagland]

oh, nevermind, it has no horizontal asymptote.