State the possible number of imaginary zeros of f(x) = 2x^3 - 4x^2 + 8x - 3.?

Answer 1

0 or 2
You just have to count the sign changes to find the number of possible positive real roots. In this case 3 sign changes, so there could be 3 or 1 positive real root.
Replace all x's with (-x)'s and count up the number of sign changes to find negative real roots: 2(-x)^3 - 4(-x)^2 + 8(-x) - 3 = -2x^3 - 4x^2 - 8x - 3
In this case 0, so there are negative real roots.
Since polynomial is of degree 3 (the largest exponent), there must be 3 roots.
If there are 3 positive real roots, there are 0 negative and 0 imaginary roots.
If there is 1 positive real root, 0 negative roots, there would have to be 2 imaginary roots.
If you want to know what they are and how to find them, you need to use the rational root theorem

Answer 2

A polynomial has exactly its degree-many zeros counting multiplicity. If you have a polynomial with real coefficients (which you do), then the complex zeros come in pairs, i.e. if z is a zero then its complex conjugate is also a zero. So we see that the equation you have has either 0 or 2 complex zeros. If you definetely want to know imaginary zeros you will have to work a little harder. Either use the cubic formula to solve the equation or try using what you know about the coefficients of the polynomial and the zeros to come up with an argument. Hint: zeros of this polynomial add up to 2 and if there are two imaginary roots, their real parts are both zero!

Answer 3

A cubic equation constantly has the two 3 real zeros or a million real 0 and a complicated conjugate pair. So its the two 0 or 2. For thsi equation it has zeros while x= 0.455 and 0.7725±a million.6433i