Given a polynomial function P(x) = a_nx^n + a_(n-1)x^(n-1)+ ... + a_1x+a_0 where a_0, a_1, ... , a_(n-1), a_n are real coefficients:
The number of positive real zeros of P(x) is the same as the number of changes in sign of the coefficient of the terms, or is less than this by an even number.
Then number of negative real zeros of P(x) is the same as the number of changes in sign of the coefficients of the terms of P(- x), or is less than this by an even number.
Therefore, we can determine the possible number of real zeros by examining the number of sign changes in the polynomial function.
As there were 3 sign changes, the possible number of positive real zeros is either 3 or 3-2= 1. We would do something similar for the negative roots, but this time we will examine the polynomial f(- x). Let's proceed by finding f(- x) first.
f(- x)=(- x)^4-2(- x)^3+6(- x)^2+5(- x)-12
⇕
f(- x)=x^4+2x^3+6x^2-5x-12
Now we can count the number of sign changes in the polynomial function.
As there is just 1 sign change, the only possibility is 1 negative zero. Let's put everything together. As the polynomial is degree 4 we will have 4 zeros according to the Fundamental Theorem of Algebra. There are two possibilities.
Having 1 positive and 1 negative zeros and 2 imaginary zeros.
