Can you have an odd number of imaginary zeros for a function?

Question

I know all imaginary zeros come in pairs (complex conjugate). But for this question the answer is there are 3 complex and 1 real : (state how many complex and real zeros for the function) x^3 - x + 3
Please explain.

Answer 1

You are right. Complex zeroes always come in pairs. Also, the number of roots of a polynomial is the same as the degree of the polynomial. x^3 - x + 3 is degree 3 so there are 3 roots total.
It must be a typo. They must mean 2 complex roots and one real

Answer 2

This one will have two complex roots: The function only hits the x-axis once (somewhere between -1 and -2) and then the next critical point is the change of derivative back to positive at x=1/2. From there, it's off to infinity. So there are two imaginary roots and the one real one.

Answer 3

a million. f(x) = 10x^5 + 2x^4 - 75x^3 - 15x^2 + 125x + 25 f(x) = 2x^4(5x + a million) - 15x^2(5x + a million) + 25(5x + a million) f(x) = (5x + a million)(2x^4 - 15x^2 + 25) f(x) = (5x + a million)(2x^4 - 10x^2 - 5x^2 + 25) f(x) = (5x + a million)(2x^2(x^2 - 5) - 5(x^2 - 5)) f(x) = (5x + a million)(2x^2 - 5)(x^2 - 5) The zeroes are: 5x + a million = 0 or 2x^2 - 5 = 0 or x^2 - 5 = 0 5x = -a million or 2x^2 = 5 or x^2 = 5 x = -a million/5 or x^2 = 5/2 or x^2 = 5 x = -a million/5 or x = +/- sqrt(5/2) or x^2 = +/- sqrt(5) x = -0.2 or x =~ +/- a million.581 or x = +/- 2.236 5 complicated roots, consisting of 5 real and nil non-real a million rational root 2. f(x) = 5x^4 - 2x^2 - 3 f(x) = 5x^4 - 5x^2 + 3x^2 - 3 f(x) = 5x^2(x^2 - a million) + 3(x^2 - a million) f(x) = (5x^2 + 3)(x^2 - a million) The roots are: 5x^2 + 3 = 0 or x^2 - a million = 0 5x^2 = -3 or x^2 = a million x^2 = -3/5 or x^2 = a million x = +/- sqrt(-3/5) or x^2 = +/- sqrt(a million) x = +/- sqrt(-0.6) or x^2 = +/- a million 4 complicated roots, consisting of two real and a pair of non-real 2 rational roots 3. 6 complicated roots, consisting of two real roots the real roots are: x^6 - sixty 4 = 0 x^6 = sixty 4 x = +/- sixty 4^(a million/6) x = +/- 2 2 rational roots

Answer 4

If you have a cubic, there are only 3 zeros you can have. If 1 is real (about 1.7) , the other two are not.