how find all the zeroes of the polynomial Find all the zeroes of the polynomial
$f(x) = 2x^7  - 17x^6  -45x^5  +390x^4  + 28x^3  + 1832x^2  +960x$
this is my try
$f(x)= 2x^7  - 16x^6-x^6-42x^5-3x^5+390x^4  + 28x^3  + 1820x^2+12x^2  +960x$
$f(x)=2x^6 (x - 8)-x^5 (x+42)-3x^4 (x-130)+28x^2 (x + 1820)+12x(x+80)$
 A: For problems like this, the rational root theorem.  First, note that you can distribute out $x$, showing that $0$ is a root. Then the remaining rational roots will be (factor of 960)/(factor of 2)  As you find each one, divide it out.  You will come down to a quadratic that has complex roots at the end.
A: It would be nice to know exactly what the context and expectations for this problem are.  As another answer and a comment point out, you can check for integer or rational solutions after narrowing down the list of candidates using the rational root theorem.  Unfortunately, the factors of $960$ are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 32, 40, 48, 60, 64, 80, 96, 120, 160, 192, 240, 320, 480, and 960.  You need to check $\pm$ each of these, as well as each of them over $2$.  Well, you don't have to check things like $4/2$ twice, at least.
The point is that you might very well be working with a graphing utility but not a full blown computer algebra system.  (If you are working with a computer algebra system, then you can just plug it into something like WolframAlpha.)  If you look at a graph, though, then you can see the roots:

It sure looks like the roots are at $-5$, $-0.5$, $0$, $6$, and $8$.  It's easy to plug these numbers in and check, as well.  It sure doesn't look like there are any more roots and none of them appear to be multiple roots.  So there must be an irreducible quadratic that divides the polynomial.
