If f(2) = 0, what are all the factors of the function f(x) = x^3 – 2x^2 – 68x + 120 using the remainder theorem?

To determine all the factors of the function f(x) = x^3 – 2x^2 – 68x + 120, we first need to confirm the value of f(2) to ensure that 2 is indeed a root of the polynomial. According to the remainder theorem, if f(2) = 0, then (x – 2) is a factor of f(x).

Now, let’s calculate f(2):

f(2) = (2)^3 - 2(2)^2 - 68(2) + 120
     = 8 - 8 - 136 + 120
     = 8 - 8 - 136 + 120
     = 0

Since f(2) equals 0, we can conclude that (x – 2) is indeed a factor of f(x). Next, we’ll perform polynomial long division or synthetic division to divide f(x) by (x – 2).

            x^2  +  0x  - 60
         _______________________
(x - 2) | x^3 - 2x^2 - 68x + 120
          - (x^3 - 2x^2)
          _______________________
                   0 - 68x + 120
                   - (0 - 60x + 120)
              _______________________
                            -8x + 120
                            - (-8x + 16)
                     _______________________
                                   104

The quotient from the division is x^2 – 60, which means:

f(x) = (x – 2)(x^2 – 60)

Next, we can factor x^2 – 60 further:

x^2 – 60 = (x – 0)(x + 0)

Putting it all together, the complete factorization of f(x) is:

f(x) = (x – 2)(x – 0)(x + 0)

Thus, the factors of the function are (x – 2), (x – √60), and (x + √60).