The number of positive real roots 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 roots 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)=-4x^4 - 2x^3 - 12x^2-1x - 23
We can see above that there are zero sign changes. Therefore, there are 0 positive real zeros.
Negative Real Zeros
Now consider f(- x).
f(- x)=-4(- x)^4 - 2(- x)^3 - 12(- x)^2-(- x)-23
⇕
f(- x)=- 4x^4 + 2x^3-12x^2+ 1x-23
We can see that there is four sign changes, (-) to ( +), ( +) to (-), (-) to ( +), and ( +) to (-). Therefore, there is either 4, 2, or 0 negative real zeros.
Imaginary Zeros
To calculate the possible number of imaginary roots, we first need to find the total number of complex roots (both real and imaginary). According to the Fundamental Theorem of Algebra, the number of complex roots is given by the degree of the polynomial.
-4x^4 - 2x^3 - 12x^2 - x -23
In this case, there are 4 complex roots. The difference between this number and the number of real roots — positive and negative — is the number of imaginary roots. 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
4
0+4=4
4-4=0
4
0+2=2
4-2=2
4
0+0=
4- =4
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 roots. Let's recall the Conjugate Root Theorem.
If the complex numbera + bi is a zero of a
polynomial in one variable with real coefficients,
then the complex conjugate a - bi is also a zero
of that polynomial.
Therefore, the number of imaginary roots is always even. Since our polynomial has four zeros, the possible number of imaginary roots is 0 , 2 or 4.
