We can know the answer to this by letting p(x) be equal to zero and by substituting the values of x to the function.

To know what to substitute, we equate all factors to zero since even if only one of these factors is zero, the product of all factors would immediately be zero.
1.

2.

3.

4.

We then try to substitute each value of x to the original polynomial and see if it will equal to zero.
1.

2.

3.

4.

As we can see, all but number 1 equated to zero therefore numbers 2, 3, and 4 are all linear factors of p(x).
ANSWER: Numbers 2, 3, and 4.