i need help with math please, functions?

Question

function: f(x) = x^3 - x^2 - x - 2

1. list all possible rational roots

2. find all real rational roots

3. find any irrational roots (exact values where possible)

4. find imaginary roots of function (exact values where possible)

5. How many complex zeros does your function have? classify each zero as real, rational, irrational, or imaginary.

6. write your function as a linear factorization

7. describe the intervals of increase and decrease

8. find local max and min

9. find where f(x) < 0, f(x) > 0, and f(x) = 0

Answer 1

1. checking all possible factors of -2, you have 1, -1, 2, -2. By substituting in the equation, you find that 2 is a root of the equation or (x - 2) is a factor. 2 is the only rational root of your equation.

2. This leaves you with x^2 + x + 1 as the remaining factor of your expression. Using the binomial theorem to solve for your remaining values of x when f(x) = 0, you find you have two imaginary roots, which are not rational roots. Therefore, 2 is the only real, rational root of your equation.

3. Likewise, there are no irrational roots

4. When you use the binomial theorem for your remaining roots, you have [-1 +/- (1-4)^-2]/2 =[ -1 +/- isqrt(3)]/2. These roots are both imaginary.

5. ??

6. (x - 2) (x - [(1 + i sqrt 3]/2)(x -[1 - isqrt3]/2)

7. ???

8. ???

9. ???

Answer 2

DesCarte's law of signs says there can be only one real root. By inspection, that real root is 2.
The Fundamental Theorem of Agebra says the function has three roots so the other two roots are complex.

x^3 - x^2 - x - 2 = (x-2)(x^2+x+1)
Use the quadratic formula to find the complex roots:
x = [-1 +/- sqrt(-3)]/2
x = - .5 +/- i.5sqrt(3) <-- 2 complex roots

y' = 3x^2-2x -1=0
(3x+1)(x-1) =0
x = -1/3 , 1
Local max at x = -1/3 and local min at x= 1
f(x) increases (- infinity, -1/3)
f(x) decreases (-1/3, 1)
f(x) increases (1, infinity)
f(x) < 0 when x <2
f(x) >0 when x > 2
f(x) = 0 whe x =2

I think that covered everything.