The number of positive real zeros of f(x)=0 is either equal to the number of sign changes between consecutive coefficients of f(x) or is less than that by an even number.
The number of negative real zeros of f(x)=0 is either equal to the number of sign changes between consecutive coefficients of f(- x) or is less than that by an even number.
Positive Real Zeros
Consider the given polynomial f(x).
f(x)=1x^6 - 5x^3+ 1x^2- 1x- 6
We can see above that there are three sign changes, (+) to ( -), ( -) to (+), and (+) to ( -). Therefore, there are either 3 or 1 positive real zeros.
Negative Real Zeros
Now consider f(- x).
f(- x)= (- x)^6-5(- x)^3 +(- x)^2 -(- x)-6
⇕
f(- x)=1x^6+5x^3+1x^2+1x- 6
We can see that there is only one sign change, (+) to ( -). Therefore, there is only one negative real zero.
Imaginary Zeros
To calculate the possible number of imaginary zeros, we first need to find the total number of complex zeros (both real and imaginary). According to the Fundamental Theorem of Algebra, the number of complex zeros is given by the degree of the polynomial.
x^6 - 5c^3+x^2-x-6
In this case, there are 6 complex zeros. The difference between this number and the number of real zeros — positive and negative — is the number of imaginary zeros. We already found these numbers above, so let's calculate the possible number of imaginary zeros.
Total Number of Zeros
Number of ± Real Zeros
Number of Imaginary Zeros
6
3+1=4
6-4=2
6
1+1=2
6-2=4
The given polynomial has either 2 or 4 imaginary zeros.
