find all the zeros of the function g(x)=x^3-4x^2-x+22

Answer:x = -2, x = 3 − i√8, and x = 3 + i√8Step-by-step explanation:g(x) = x³ − 4x² − x + 22This is a cubic equation, so it must have either 1 or 3 real roots.Using rational root theorem, we can check if any of those real roots are rational.  Possible rational roots are ±1, ±2, ±11, and ±22.g(-1) = 18g(1) = 18g(-2) = 0g(2) = 12g(-11) = 1782g(11) = 858g(-22) = -12540g(22) = 8712We know -2 is a root.  The other two roots are irrational.  To find them, we must find the other factor of g(x).  We can do this using long division, or we can factor using grouping.g(x) = x³ − 4x² − 12x + 11x + 22g(x) = x (x² − 4x − 12) + 11 (x + 2)g(x) = x (x − 6) (x + 2) + 11 (x + 2)g(x) = (x (x − 6) + 11) (x + 2)g(x) = (x² − 6x + 11) (x + 2)x² − 6x + 11 = 0Quadratic formula:x = [ 6 ± √(36 − 4(1)(11)) ] / 2x = (6 ± 2i√8) / 2x = 3 ± i√8The three roots are x = -2, x = 3 − i√8, and x = 3 + i√8.