SOLUTION: Hello, can you state the possible number of imaginary zeros for these two functions? 1. f(x) = ­-x³ ­-x² + 14x - 24 2. f(x) = 2x³ ­- x² + 16x - 5

Algebra ->  Test -> SOLUTION: Hello, can you state the possible number of imaginary zeros for these two functions? 1. f(x) = ­-x³ ­-x² + 14x - 24 2. f(x) = 2x³ ­- x² + 16x - 5      Log On


   



Question 1037022: Hello, can you state the possible number of imaginary zeros for these two functions?
1. f(x) = ­-x³ ­-x² + 14x - 24
2. f(x) = 2x³ ­- x² + 16x - 5

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
For polynomial functions with real coefficients, imaginary zeros come in pairs, so a polynomial function with real coefficients can have 0,2,4, 6,...imaginary zeros.
However, a polynomial with degree 3 cannot have more than 3 zeros,
so these functions can have highlight%280%29 or highlight%282%29 imaginary zeros.
Polynomials of odd degree must have at least 1 real zero.
They could have 1, 3, 5, all the way up to their degree.

NOTE:
Each of those functions happens to have exactly 1 real zero.
Some calculus plus calculations (or a graphing calculator) would tell you that.
Since they are cubic (degree=3) polynomials, they must have a total of 3 zeros,
so the other 3-1=2 zeros are imaginary.
You are probably not expected to be able to reach that conclusion, though.