How do i find the roots of this polynomial?

  • Thread starter Thread starter iScience
  • Start date Start date
  • Tags Tags
    Polynomial Roots
AI Thread Summary
To find the roots of the polynomial x^3 - 5x - 6 = 0, various methods can be employed. Trial and error can help identify potential solutions, while graphing the equation can provide visual estimates of the roots. The rational root theorem suggests a limited set of rational candidates, but in this case, none of them yield actual roots, indicating that the roots are likely irrational. More complex methods, such as using the cubic formula or approximation techniques like Newton's method, may be necessary to find the roots. Ultimately, the discussion highlights the challenges of finding roots for this specific cubic equation.
iScience
Messages
466
Reaction score
5
x^3-5x-6=0

i've tried the p/q calculations in accordance with the rational roots theorem but I've yet to find the answers...
 
Physics news on Phys.org
You can either:
1. Use trial and error to find a solution.
2. Graph the equation in order to estimate a solution.
3. Use the formula for the roots of a cubic equation (most complicated approach).

See: http://en.wikipedia.org/wiki/Cubic_function
 
4. Use an approximation scheme such as regula falsi, secant method, or Newton's method.

The rational root theorem says yields a small number of candidate rational roots. Not a single one of those candidates will yield a root. That doesn't mean there aren't any roots. It just means the root/roots aren't rational numbers.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top