Answer:
1. The possible roots are ±1, ±2, ±3, ±6. These are the factors of the trailing constant (6) divided by the factors of the leading coefficient (1). When the leading coefficient is 1, the possible roots are the factors of the constant
2. The polynomial is easier to evaluate when it is written in Horner form:
f(x) = ((x -4)x +1)x +6
To show that 2 is a zero, we want to find f(2):
f(2) = ((2 -4)2 +1)2 +6 = (-4 +1)2 +6 = -6 +6 = 0
f(2) = 0, so 2 is one of the zeros of this function
3. Using synthetic division (attached) or polynomial long division, we can divide the given polynomial by (x-2) to find the remaining factors. This division gives (x^2 -2x -3), which can be factored as (x -3)(x +1), so the three actual roots are ...
x = 2 (from above), x = 3, x = -1 (from our factorization)
4. In factored form, the polynomial can be written ...
f(x) = (x +1)(x -2)(x -3)
The first factor was found from the fact that 2 was given as a zero of the function. For any zero "a", a factor of the polynomial is (x-a).
The remaining factors were found by factoring the quadratic trinomial that resulted from the division of f(x) by x-2. That trinomial is x^2 -2x -3.
There are a number of methods that can be used to factor x^2 -2x -3. Again, the rational root theorem can help. It suggests that ±1 and ±3 are possible roots.
