Question

Is there a way to find the number of roots of a polynomial without finding the roots?

Answer 1

yes Descartes rule of signs http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs

Answer 2

Number of Roots = Degree Of Polynomial...

Answer 3

There will be as many roots of the polynomial as there is the highest power of that polynomial...

Answer 4

Then, do I need to use Descartes Rule of Signs or just look at the highest degree?

Answer 5

Degree means highest power just look for the highest power...

Answer 6

Great! You've saved the day! Thanks!

Answer 7

Welcome dear...

Answer 8

you are getting answers to two different questions

Answer 9

One is to find simply how many roots are there...
Other is to find out of those roots which are positive negative and complex that is explained by Descartes Rule...

Answer 10

one question is how many zeros (roots) does a polynomial of degree \(n\) has and the answer is it has \(n\) roots counting multiplicity

HOWEVER these roots need not be real numbers!
if the question is how many REAL roots a polynomial has, Descartes rule of sign can help answer that question, but does not in fact answer it

Answer 11

Will you please explain what is the multiplicity what is its concept @satellite73 ??

Answer 12

if you have a specific question we could work throught it
@waterineyes for the polynomial \((x-2)^3(x+1)\) the root 2 is a zero with multiplicity 3

Answer 13

Here in this example how do you calculated 3 as its multiplicity??

Answer 14

Suppose I have:

\[(x-1)^5(x-3)^2(x-2)(x+5) = 0\]

Here explain the concept of multiplicity @satellite73

Answer 15

the zeros of your polynomial are 1, 3, 2 and -5
the multiplicity of 1 is 5, the multiplicity of 3 is 2 and 2 and -5 have multiplicty 1

Answer 16

It is related to powers..
Am I right??

For \((x+5)^{10}\)

Root : -5 with, Multiplicity = 10 right @satellite73