find all zeros of the polynomial function:
p(x)= x^3-9x^2+25x-17

Question

find all zeros of the polynomial function:
p(x)= x^3-9x^2+25x-17

hero · Answer

Since 17 is a prime number, it is likely that either 1 or 17 is one of the roots. If p(x) = 0, then we know that the value of x is a root. In this case, let x = 1, 

p(1) = (1)^3 - 9(1)^2 + 25(1) - 17
       = 1 - 9 + 25 - 17
       =  26 - 26
       = 0

Since p(1) = 0,  x = 1 is a root and x - 1 is a factor. So we can factor the polynomial in the following manner
p(x) = x^3-9x^2+25x-17
       = x^3 - x^2 - 8x^2 + 8x + 17x - 17
       = x^2(x - 1) -8x(x - 1) -17(x - 1)
       = (x - 1)(x^2 - 8x - 17)

hero · Answer

You can continue factoring the quadratic to find the rest of the roots.

anonymous · Answer

Wow, there are some tricks there, Hero.  Like spliting 9x^2 into x^2 and 8x^2.  Also 25x into 8x and 17x.  How would this have been done if 17 hadn't been prime?

hero · Answer

Generally the same way. Once you figure out one of the factors, you split in accordance with that particular factor, and the rest of the factorization behaves quite nicely. It's an algebraic alternative to other methods of factoring degree three and above polynomials.