Answer:
We will use Descarte's Rule to solve the following question:
It says that the number of sign changes in the function f(x) tells the maximum number of positive roots that could exist.
Similarly the number of sign changes in the function f(-x) tells the maximum number of negative roots that could exist.
Now let us consider a example as:
We know that the total number of zeros of a polynomial function is always equal to the degree of the polynomial.
This is a polynomial function of degree 3 hence it has total 3 zeros.
Now f(x) has total 2 sign changes first from + to - and then from - to +.
Hence atmost 2 positive real zeros are possible.
Also f(-x)=-x^3-x^2+5
This function has only one sign change i.e. from - to +.
Hence atmost 1 negative real roots are possible.
Also on solving the cubic equation we got that we have one real zeros and 2 complex zeros.
In table we could write as:
total zeros: 3 (-1.4334 , 1.2167-1.4170 i , 1.2167+1.4170 i)
No. of positive : None
real zero
No. of negative : 1 (-1.4334)
real zero
Complex zero : 2 (1.2167-1.4170 i and 1.2167+1.4170 i)
This is a polynomial function of degree 4 hence it has total 4 zeros.
Now f(x) has total 1 sign changes first from + to - .
Hence atmost 1 positive real zeros are possible.
Also f(-x)= x^4-x^3-1
This function has only one sign change i.e. from + to -.
Hence atmost 1 negative real roots are possible.
Also on solving the cubic equation we got that we have two real zeros and 2 complex zeros.
In table we could write as:
total zeros: 4 (-1.3803, 0.81917 , -0.21945-0.91447 i ,
, -0.21945+0.91447 i)
No. of positive : 1 (0.81917)
real zero
No. of negative : 1 (-1.3803)
real zero
Complex zero : 2 (-0.21945-0.91447 i , -0.21945+0.91447 i)
