Since nobody answer this question, I'll do it for you. I hope you still need it.
To factor x3+7x2-28x-160, I will assume that you know already synthetic division and factoring a quadratic trinomial.
First list all the factors of -160 and put sort them out like this:
±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20 , ±32, ±40, ±80, ±160
There are 24 factors and at most 24 trials that you can do. Choose one number at a time that you can use as a divisor for your synthetic division. I advise you to choose the middle four (±8, ±10, ±16, ±20) because that where you can get lucky in finding the right divisor.
Let's start with -8 and use the coefficients of the polynomials. Using the synthetic division you'll get the following numbers:
-8 _/ 1 7 -28 -160
-8 8 160
-------------------------------
1 -1 -20 0
If the last number is zero, then that's exactly what you need. If the last number is not zero, then you have to pick another factor of -160 until you get the zero as the last number in your synthetic division.
The numbers 1,-1, -20, 0 are the coefficients of the polynomials of one of the factors:
x2 - x - 20 with zero remainder. Therefore:
x3+7x2-28x-160 = (x + 8) (x2 - x - 20). We use (x + 8). It's +8 because it has to be the opposite sign of the divisor in synthetic division.
Then factor x2 - x - 20, by getting the factors of -20 that add up to -1 (-1 is the coefficient of the middle term) we have (x + 4) (x - 5).
x3+7x2-28x-160 = (x + 8) (x + 4) (x - 5)
