Find the real and complex zeros of the following function. Please show all of your work. f(x)=x^3-x^2+4x-4

Question

anonymous · Answer

a good place to start is by trying x = 1, which gives you the sum of the coefficients

anonymous · Answer

$$1-1+4-4=0$$ so 1 is a solution. therefore you can factor out the 1 to get 
$$x^3-x^2+4x-4=(x-1)(\text{something})$$
and you can find the 'something" by either dividing or thinking

anonymous · Answer

diah40, look up "completing the square" on google.  Once you re-remember (as we all have to do) how to do this, then finding the zeros will be a total cinch.

For example, x(x-2)(x-3):  zeros are: 0, 2, and 3.

anonymous · Answer

in any case it has to be 
$$x^3-x^2+4x-4=(x-1)(x^2+4)$$ so only real zero is at x = 1, complex zeros at 
$$x=2i$$ and its conjugate 
$$x=-2i$$

anonymous · Answer

Sorry, not completing the square, but factoring a cubic polynomial (or 3rd degree poly) http://www.youtube.com/watch?v=mXdAiJY7YfY