I can find 3 conditions that f(x) can be factorize. f(1) = 1 f(0) = 1 f(1) = 1 Therefore, all the possible factor of f(x) with degree 2 : x[sup]2[/sup] – x + 1 x[sup]2[/sup] + x + 1 2x[sup]2[/sup] + 1 Then expand f(x) to get f(x) = ax[sup]4[/sup] + bx[sup]3[/sup] – ax[sup]2[/sup] – bx + 1 Use f(x) to subtract each of the above possible factor. Then I can get three case. And they all should be divisible by the corresponding factor if the possible factors are really factors of f(x). I can get the three solutions that f(x) can be factorized are: 1. a = 1 b = 2 2. a = 3 b = 1 3. a = 4 b = 0 But I think I can still have more…. Maybe I can get some linear factors that are factors of f(x)…. Or it’s all the possible case already? (Since the factors will degree 2 already contained the linear factors?) The number inside[sup][/sup] is the degree.
I am sorry about that it cannot be shown in the right way.
