What are all possible zeros and actual zeros of the function f(x)=x3-9x2+27x-28

Answer 1

By the Rational Root Theorem±

±1, ±2, ±4, ±7, ±14, ±28

Real/Rational Root: 4

Complex Roots: (5 ± i√3) / 2

Answer 2

Possible zeros are (factors of p)/(factors of q)

p is the constant term

q is the leading coefficient

since q is 1, we really only need to worry about the factors of p for this problem.

±1, ±2, ±4, ±7, ±14, ±28

You can either test these or graph the equation to see that the only real zero is 4. Doing synthetic division, The remaining polynomial is:

x²-5x+7 This will need to be solved using the quadratic formula.

a=1, b=-5, c=7

[-(-5)±√(-5)2-4(1)(7)]/[2(1)] Simplify

(5±i√3)/2 are the remaining zeros in addition to 4 found earlier