State the possible number of imaginary zeros of f(x)= 2x^3 - 4x^2 + 8x - 3.
0,1,3?
For any polynomial of odd degree, large negative values of x yield large negative values for y.
Same for positive values. So, somewhere along the way, y must change from negative to positive, meaning there is at least one real root.
Now, it's quite possible that there is only on real root, leaving possibly 2 complex roots.
So, there are 0,2 possible complex roots.
Note I said complex, not imaginary. Imaginary numbers have no real part, so they look like 0+bi. Most complex roots have a nonzero real part, so they look like a+bi.
USEFUL FACT: For a polynomial with real coefficients (the vast majority of the examples you will encounter) the complex roots come in conjugate pairs, meaning there are always an even number of them. This fact will help you solve a lot of the problems posed.
Thank you
To determine the possible number of imaginary zeros of a polynomial, we need to apply the Fundamental Theorem of Algebra. According to this theorem, the number of complex zeros (which includes imaginary zeros) is equal to the degree of the polynomial.
In this case, the polynomial f(x) = 2x^3 - 4x^2 + 8x - 3 is a cubic polynomial, which means it has a degree of 3. Therefore, the possible number of imaginary zeros is 3.
So the answer is 3.