Answer:
We will use Descarte's Rule to solve the following question:
- Work area to determine possible positive real roots:
It says that the number of sign changes in the function f(x) tells the maximum number of positive roots that could exist.
- Work area to determine possible negative real roots:
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:
1) f(x)=x^3-x^2+5
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)
2) f(x)=x^4+x^3-1
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)