Question 1209698: Let p(x) be a polynomial with integer coefficients. Suppose p(4) = 5 and p(-2) = 3. If p(x) has an integer root r, then find all possible values of r.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! 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 $\boxed{-1}$
Answer by ikleyn(52903)  (Show Source):
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