Question 1209698
Here's how to find the possible integer roots of p(x):

1. **Use the Integer Root Theorem:** If p(x) is a polynomial with integer coefficients, and if r is an integer root of p(x), then r must be a divisor of the constant term of p(x).  However, we don't know the constant term.

2. **Use the property that if a and b are integers, then a-b divides p(a) - p(b):**
Since p(x) has integer coefficients, for any integers a and b, a - b must divide p(a) - p(b).

3. **Apply this to our problem:**
We have p(4) = 5 and p(-2) = 3.  Therefore,
* 4 - (-2) = 6 must divide p(4) - p(-2) = 5 - 3 = 2.
The divisors of 2 are ±1 and ±2.  However, 6 must divide 2, which is impossible.  There must be an error in the problem statement.

Let's assume the question meant that the coefficients are integers, and that *if* p(x) has an integer root r, then we want to know all possible values of r.

We know that for any integers a and b, $a-b$ divides $p(a)-p(b)$.

Let r be an integer root of p(x).  Then p(r) = 0.

* Case 1: a = 4, b = r.  Then 4 - r divides p(4) - p(r) = 5 - 0 = 5.
The divisors of 5 are ±1 and ±5.
    * 4 - r = 1 => r = 3
    * 4 - r = -1 => r = 5
    * 4 - r = 5 => r = -1
    * 4 - r = -5 => r = 9

* Case 2: a = -2, b = r. Then -2 - r divides p(-2) - p(r) = 3 - 0 = 3.
The divisors of 3 are ±1 and ±3.
    * -2 - r = 1 => r = -3
    * -2 - r = -1 => r = -1
    * -2 - r = 3 => r = -5
    * -2 - r = -3 => r = 1

The possible values of r are the integers that appear in both lists: -1.

Therefore, the only possible integer root is -1.

Final Answer: The final answer is $\boxed{-1}$