Find all the zeros of the function

g (x) = x^3-4x^2-x+22

x = - 2, x = 3 - i√8, and x = 3 + i√8

Step-by-step explanation:

g (x) = x³ - 4x² - x + 22

This 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) = 18

g (1) = 18

g (-2) = 0

g (2) = 12

g (-11) = 1782

g (11) = 858

g (-22) = - 12540

g (22) = 8712

We 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 + 22

g (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 = 0

Quadratic formula:

x = [ 6 ± √ (36 - 4 (1) (11)) ] / 2

x = (6 ± 2i√8) / 2

x = 3 ± i√8

The three roots are x = - 2, x = 3 - i√8, and x = 3 + i√8.