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.

you have orange juice on wednesday. (idk why this is the answer)

step-by-step explanation:

answer: im gonna go with milli

step-by-step explanation: