The number of positive real zeros of f 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 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 givenpolynomial function g(x).
g(x)=x5−2x3−x2+6⇕g(x)=1x5−2x3−1x2+6
We can see above that there are two sign changes, (+) to (−) and (−) to (+). Therefore, there are either 2 or 0 positive real zeros.
Negative Real Zeros
Now consider g(-x).
g(-x)=(-x)5−2(-x)3−(-x)2+6⇕g(-x)=-1x5+2x3−1x2+6
We can see above that there are three sign changes, (−) to (+) two times and (+) to (−). Therefore, there are either 3 or 1 negative real zeros.
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.
x5−2x3−x2+6
In this case, there are 5 complex zeros. The difference between this number and the number of real roots — 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
5
2+1=3
5−3=2
5
0+1=1
5−1=4
5
2+3=5
5−5=0
5
0+3=3
5−3=2
The given polynomial has either 0,2 or 4 imaginary zeros.
Extra
Another Method
There is another way of finding the possible number of imaginary zeros of given function. Let's recall the Conjugate Root Theorem.
Conjugate Root Theorem
If the complex number a+bi is a root of a polynomial in one variable with real coefficients, then the complex conjugate a−bi is also a root of that polynomial.
Thus, the number of imaginary roots is always even. Since our polynomial function has five zeros, the possible number of imaginary roots are 0,2 and 4.
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