RickPalomino
New member
- Joined
- Dec 6, 2008
- Messages
- 4
I have a question for the function:
f(x)=6x^4-7x³-x²+67x-105
Using Descartes Rule, there should be 3 or 1 positive zeros, along with 1 negative zero. And I found the possible zeros to be (Didn't separate with commas.):
±1±3±5±7±15±21±35±105±1/2±3/2±5/2±7/2±15/2±21/2±35/2±105/2±1/3±5/3±7/3±35/3±1/6±5/6±7/6±15/6±21/6±35/6±105/6
Using the "Equation Solver" Calculator on this site, I found these zeros:
-7/3, 3/2, 1-2i, 2i-1
Basically what I would like to know is, is there a shortcut to finding the possible real zeros?
The point I'm at in class is testing every number through synthetic division until getting a quadratic answer to factor and find the imaginary zeros.
f(x)=6x^4-7x³-x²+67x-105
Using Descartes Rule, there should be 3 or 1 positive zeros, along with 1 negative zero. And I found the possible zeros to be (Didn't separate with commas.):
±1±3±5±7±15±21±35±105±1/2±3/2±5/2±7/2±15/2±21/2±35/2±105/2±1/3±5/3±7/3±35/3±1/6±5/6±7/6±15/6±21/6±35/6±105/6
Using the "Equation Solver" Calculator on this site, I found these zeros:
-7/3, 3/2, 1-2i, 2i-1
Basically what I would like to know is, is there a shortcut to finding the possible real zeros?
The point I'm at in class is testing every number through synthetic division until getting a quadratic answer to factor and find the imaginary zeros.