All potential rational zeros of a polynomial with only integer
coefficients are numbers of the form ±P/Q where P is a divisor 
of the absolute value of the constant term and Q is a divisor of the 
absolute value of the leading coefficient. 

f(x) = x³-5x²-9x+45

The constant term is 45
The leading term is x³
The leading coefficient is 1.


In this case since the leading coefficient is 1, which has only
the divisor 1, we only need to consider as potential rational zeros
± the divisors of 45, which are

±1, ±3, ±5, ±9, ±15, and ±45.

We try 1

1 | 1 -5  -9  45
  |    1  -4 -13
    1 -4 -13  32

The remainder is 32, not 0, so 1 is not a zero of f(x)

We try -1

-1 | 1 -5  -9  45
   |   -1   6   4
     1 -6  -4  48

The remainder is 48, not 0, so -1 is not a zero of f(x)

We try 3

 3 | 1 -5  -9  45
   |    3  -6 -45
     1 -2 -15   0

The remainder is 0, so 3 is a zero of f(x).

Since the synthetic division was actually a division
of f(x) by x-3, we can use the numbers on the bottom
line of the synthetic division to factor f(x) as

f(x) = (x-3)(x²-2x-15)

We only need to find the zeros of x²-2x-15 to find
the other zeros, rational or otherwise of f(x).

We are able to do further factoring:

f(x) = (x-3)(x+3)(x-5)

Therefore all zeros of f(x) are found by setting each of
these expressions = 0 and solving:

    x-3 = 0;   x+3 = 0;   x-5 = 0
      x = 3      x = -3     x = 5

Thus there are three zeros, 3,-3, and -5.

In we graph of f(x) we see that these three zeros are the 
x-intercepts:



Edwin