<Question>
Let $p(x)$ be a polynomial of degree 5 such that
\[p(n) = \frac{n}{n^2 - 1}\]for $n = 2,$ 3, 4, $\dots,$ 7.  Find $p(8).$
</Question>

<think>
<Plan_1>
Okay, let's try to figure out how to solve this problem. So, we have a polynomial p(x) of degree 5, and it's given that p(n) = n/(n² - 1) for n = 2, 3, 4, ...., 7. We need to find p(8). Hmm, okay. Let me start by recalling some concepts about polynomials and interpolation.

<prompt_1.1> Recall the concept of polynomial interpolation and how it applies to this problem. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, I remember that if a polynomial of degree k passes through k+1 points, then it's uniquely determined. Here, the polynomial is degree 5, and we have 6 points (from n=2 to n=7). So that makes sense—it should uniquely determine the polynomial. But evaluating it at n=8 isn't one of the given points, so we need some way to extrapolate or use the given points to find p(8).

But how exactly? Let's think. Since p(x) is a degree 5 polynomial, and we know its values at 6 distinct points, maybe we can construct the polynomial using interpolation formulas, like Lagrange or Newton. But constructing the entire polynomial might be tedious, especially since we have to compute it for 6 points. Is there a smarter way?

Alternatively, maybe we can consider the difference between p(x) and the rational function n/(n² - 1). Wait, but p(x) is a polynomial, and n/(n² - 1) is a rational function. Maybe we can express p(x) in terms of a rational function plus some multiple of a polynomial that has roots at n=2,3,4,5,6,7. Hmm, that might be a way. Let me explore that.

Let's denote the rational function as f(n) = n/(n² - 1). So, for each n=2,3,....,7, we have p(n) = f(n). So, p(x) - f(x) has roots at x=2,3,....,7. But p(x) is a polynomial, and f(x) is a rational function. To combine them, maybe we can write f(x) as a polynomial divided by another polynomial. Let's see: f(x) = x / (x² - 1) = x / [(x - 1)(x + 1)]. So, the denominator is (x - 1)(x + 1). Therefore, if we consider p(x) - x / [(x - 1)(x + 1)] has roots at x=2,3,....,7. But since p(x) is a polynomial, the left-hand side would need to be a polynomial minus a rational function, which isn't a polynomial unless the denominators cancel out. So perhaps we need to multiply both sides by (x - 1)(x + 1) to eliminate the denominator.

Wait, let's try that. Let me define a new polynomial Q(x) such that:

Q(x) = (x - 1)(x + 1)p(x) - x
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Then, for x = 2,3,....,7, we have Q(n) = (n - 1)(n + 1)p(n) - n = (n - 1)(n + 1)(n / (n² - 1)) - n = (n - 1)(n + 1)(n / ((n - 1)(n + 1))) - n = n - n = 0. So Q(n) = 0 for n = 2,3,....,7. Therefore, Q(x) has roots at x=2,3,....,7, so it must be divisible by (x - 2)(x - 3)....(x - 7).

<prompt_2.1> Determine the degree of the polynomial Q(x) and the implications of its roots. </prompt_2.1>
-----

But let me verify this step-by-step to ensure I didn't make any errors.

First, confirming the construction of Q(x):

<prompt_2.2> Verify the construction of Q(x) and the calculation of its degree. </prompt_2.2>
-----

But let me verify the degree calculation again.

<prompt_2.3> Re-verify the degree calculation and confirm the final answer. </prompt_2.3>

Wait, but let me check if there's another way to approach this problem, just to confirm.

<prompt_2.4> Explore an alternative approach to confirm the result of p(8). </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
Therefore, Q(x) = k(x - 2)(x - 3)....(x - 7), where k is some constant.

But Q(x) is a polynomial. Let's check its degree. The original polynomial p(x) is degree 5, so (x - 1)(x + 1)p(x) is (degree 2)(degree 5) = degree 7. Then subtracting x (degree 1) gives Q(x) as degree 7. However, the product (x - 2)(x - 3)....(x - 7) is a degree 6 polynomial. Therefore, Q(x) is a degree 7 polynomial on the left, but the right-hand side is a degree 6 polynomial. The only way this can happen is if the coefficient of x^7 in Q(x) is zero. Let's see:

Wait, let's compute the degree of Q(x). The leading term of (x - 1)(x + 1)p(x) would be x^2 times the leading term of p(x), which is a x^5, so the leading term is a x^7. Then subtracting x gives a leading term of a x^7. On the other side, the product (x - 2)(x - 3)....(x - 7) is a degree 6 polynomial, so its leading term is x^6. Therefore, the only way Q(x) can be equal to both a degree 7 polynomial and a degree 6 polynomial is if the coefficient a is zero, which would make Q(x) degree 6. But that contradicts unless Q(x) is indeed degree 6. Wait, maybe I made a mistake here.

Wait, Q(x) is defined as (x - 1)(x + 1)p(x) - x. Since p(x) is degree 5, (x - 1)(x + 1)p(x) is degree 7, so Q(x) is degree 7. But we also have Q(x) = k(x - 2)(x - 3)....(x - 7), which is degree 6. The only way a degree 7 polynomial equals a degree 6 polynomial is if the coefficient of x^7 is zero. Therefore, the coefficient of x^7 in Q(x) must be zero. Let's calculate that coefficient.

The leading term of (x - 1)(x + 1)p(x) is x^2 * (a x^5) = a x^7. Then, subtracting x gives Q(x) = a x^7 + ... - x. Therefore, the coefficient of x^7 in Q(x) is a. But since Q(x) is also equal to k(x - 2)(x - 3)....(x - 7), which has a leading term of k x^6. Therefore, the coefficient of x^7 in Q(x) must be zero, so a = 0. But that would mean the leading term of p(x) is zero, which can't be because p(x) is a degree 5 polynomial. Wait, this seems like a contradiction. Therefore, my initial approach might be flawed.

Hmm, maybe I need to reconsider. Let me think again. The problem states that p(x) is a polynomial of degree 5, so p(x) is degree 5, and f(x) = x/(x² - 1) is a rational function. So, when I set Q(x) = (x - 1)(x + 1)p(x) - x, then Q(x) is indeed a polynomial of degree 7, since (x - 1)(x + 1)p(x) is degree 7. However, Q(x) is supposed to be equal to k(x - 2)(x - 3)....(x - 7), which is degree 6. Therefore, the only way this can happen is if Q(x) is the zero polynomial. But that would imply that (x - 1)(x + 1)p(x) = x for all x, which would mean p(x) = x / [(x - 1)(x + 1)]. But p(x) is supposed to be a polynomial of degree 5, and x / [(x - 1)(x + 1)] is a rational function, not a polynomial. Therefore, this suggests that my initial approach is incorrect. Hmm. So perhaps I need a different strategy.
</execution_2.1>

<execution_2.2>
Q(x) is defined as (x - 1)(x + 1)p(x) - x. Since p(x) is a degree 5 polynomial, multiplying by (x - 1)(x + 1) gives a degree 7 polynomial. Then subtracting x, which is degree 1, so Q(x) is degree 7. Then, since Q(n) = 0 for n=2,3,....,7, that's 6 roots. Therefore, Q(x) can be written as (x - 2)(x - 3)....(x - 7) multiplied by some linear factor, because 6 roots and degree 6, but Q(x) is degree 7. Wait, hold on, that might be a problem. Wait, Q(x) is degree 7, and (x - 2)(x - 3)...(x - 7) is degree 6. Therefore, Q(x) = (x - 2)(x - 3)....(x - 7)(ax + b), where ax + b is a linear polynomial. So we need to determine a and b.
</execution_2.2>

<execution_2.3>
We said that Q(x) is equal to (x -1)(x +1)p(x) - x. Since p(x) is degree 5, (x -1)(x +1)p(x) would be a polynomial of degree 5 + 2 = 7. Then subtracting x (degree 1) gives Q(x) as degree 7. Then, since Q(x) has roots at 2,3,4,5,6,7, which are six roots, but Q(x) is degree 7, so Q(x) must be (x -1)(x +1)p(x) - x = k(x -2)(x -3)...(x -7), where k is a constant. Wait, but hold on. Wait, (x -2)(x -3)...(x -7) is a degree 6 polynomial. However, Q(x) is degree 7. Therefore, the right-hand side is k*(x -2)...(x -7), which is degree 6, but Q(x) is degree 7. Therefore, this suggests that perhaps my initial construction is missing something. Because Q(x) is degree 7, but the product (x -2)...(x -7) is degree 6. Therefore, unless there's a mistake in the degree analysis.

Wait, let's re-examine. Let's see:

Q(x) = (x -1)(x +1)p(x) - x. If p(x) is degree 5, then (x -1)(x +1)p(x) is degree 5 + 2 = 7. Then subtract x, which is degree 1. So Q(x) is indeed degree 7. Then, Q(n) = 0 for n=2,3,....,7, which are six roots. But a degree 7 polynomial can have up to 7 roots, unless it's the zero polynomial. So, if Q(x) has six roots, but it's degree 7, then there must be another root somewhere. But according to the construction, Q(x) is (x -1)(x +1)p(x) - x. Let's check x=1 and x=-1.

If x=1, then Q(1) = (1 -1)(1 +1)p(1) -1 = 0 -1 = -1 ≠ 0. Similarly, x=-1: Q(-1) = (-1 -1)(-1 +1)p(-1) - (-1) = (-2)(0)p(-1) +1 = 0 +1 =1 ≠0. Therefore, x=1 and x=-1 are not roots of Q(x). Therefore, the roots of Q(x) are only at x=2,3,....,7, which are six roots. But Q(x) is degree 7, so there must be a seventh root. Wait, but unless there's a mistake here.

Alternatively, perhaps there's a miscalculation in the degree. Let's see:

If p(x) is degree 5, then (x -1)(x +1)p(x) is degree 5 + 2 = 7. Then subtract x, which is degree 1. So the leading term is the same as the leading term of (x -1)(x +1)p(x). Therefore, Q(x) is indeed degree 7. Then, unless there's a multiple root or something else. Hmm. But according to the problem statement, p(n) is defined for n=2,3,....,7, so those are six points. So Q(x) has six roots. But Q(x) is degree 7, so there must be another root. But according to the equation Q(x) = (x -1)(x +1)p(x) - x, unless there's another value of x where this expression is zero. Let's see. Suppose x is not 2,3,....,7. Then, unless (x -1)(x +1)p(x) = x. But since p(x) is a degree 5 polynomial, unless x is chosen such that this equality holds. However, since p(x) is defined for all x, except maybe where the denominator is zero, but p(x) is a polynomial, so it's defined everywhere. Therefore, perhaps the seventh root is at x=1 or x=-1? But when we checked x=1 and x=-1, Q(1) = -1 and Q(-1)=1, so they are not roots. Therefore, maybe there's an error in the degree analysis.
</execution_2.3>

<execution_2.4>
Alternatively, sometimes with interpolation problems, especially when the polynomial degree is one less than the number of points, there's a trick using finite differences or constructing a related polynomial. Let me see.

Alternatively, consider that p(x) is a degree 5 polynomial such that p(n) = n/(n² -1) for n=2,3,....,7. So, if we define a polynomial R(x) = (x -1)(x +1)p(x) -x. Then R(x) is a polynomial, since (x -1)(x +1) cancels the denominator in p(x). Then, R(n) =0 for n=2,3,....,7, as before. So R(x) is a polynomial of degree 7 (since p(x) is degree 5, multiplied by quadratic, so 5+2=7) minus degree 1, but actually, when we multiply (x² -1)p(x), which is degree 5+2=7, and then subtract x, which is degree 1, so R(x) is indeed degree 7. But we know that R(x) has roots at 2,3,....,7, so R(x) = k(x -2)(x -3)....(x -7). Then, to find k, we can plug in x=1 or x=-1, since those are the other factors in (x -1)(x +1). Let's try x=1:

R(1) = (1 -1)(1 +1)p(1) -1 = 0 -1 = -1

But also, R(1) = k(1 -2)(1 -3)....(1 -7) = k(-1)(-2)....(-6) = k(-1)^6 * 6! = k*720

So, -1 = 720k => k = -1/720, which matches earlier. Similarly, check x=-1:

R(-1) = ((-1) -1)((-1) +1)p(-1) - (-1) = (-2)(0)p(-1) +1 = 0 +1 =1

R(-1) = k(-1 -2)(-1 -3)....(-1 -7) = k(-3)(-4)....(-8) = k*(-3)(-4)(-5)(-6)(-7)(-8)

Calculate the product: (-3)(-4)=12; 12*(-5)= -60; -60*(-6)=360; 360*(-7)= -2520; -2520*(-8)=20160

So R(-1)=k*20160=1 => k=1/20160. Wait, but earlier we had k=-1/720. Hmm, contradiction? Wait, but R(-1) should also equal k times that product. Wait, but if we use x=1, we found k=-1/720. If we use x=-1, we have 1 = k*20160, so k=1/20160. But that's a problem because k can't be both -1/720 and 1/20160. Wait, that suggests a mistake in reasoning.

Wait, no. Wait, actually, R(x) is defined as (x² -1)p(x) -x. So when we plug x=1 or x=-1, we don't necessarily get information about the roots. But when we constructed R(x) as having roots at 2,3,....,7, that's correct, but the leading coefficient is determined by the original polynomial.

Wait, but if R(x) is a degree 7 polynomial with leading term from (x² -1)p(x). Since p(x) is degree 5, the leading term of p(x) is, say, a x^5. Then, (x² -1)p(x) would have leading term a x^7, and then subtracting x gives leading term a x^7. Therefore, the leading coefficient of R(x) is a. On the other hand, when we write R(x) = k(x -2)(x -3)....(x -7), the leading term is k x^6. Wait, but that's a problem because R(x) is degree 7, but the right-hand side is degree 6. Wait, that can't be. So there's a mistake here. Wait, no, (x -2)(x -3)....(x -7) is degree 6, but R(x) is degree 7. Therefore, my earlier assumption that R(x) is equal to k(x -2)...(x -7) is incorrect. Instead, R(x) must be equal to (x -2)(x -3)....(x -7) multiplied by another linear factor, since R(x) is degree 7. So, R(x) = (x -2)(x -3)....(x -7)(ax + b), where ax + b is a linear polynomial. Then, to find a and b, we can use values of x that are not roots, such as x=1 and x=-1. Let's try that.

So, R(x) = (x -1)(x +1)p(x) -x = (x -2)(x -3)....(x -7)(ax + b)

We can use x=1:

R(1) = (1 -1)(1 +1)p(1) -1 = 0 -1 = -1

On the other side:

R(1) = (1 -2)(1 -3)....(1 -7)(a*1 + b) = (-1)(-2)(-3)(-4)(-5)(-6)(a + b)

Compute the product: (-1)(-2)=2; 2*(-3)=-6; -6*(-4)=24; 24*(-5)=-120; -120*(-6)=720. So 720*(a + b) = -1 => a + b = -1/720

Similarly, use x=-1:

R(-1) = ((-1) -1)((-1) +1)p(-1) - (-1) = (-2)(0)p(-1) +1 = 0 +1 =1

On the other side:

R(-1) = (-1 -2)(-1 -3)....(-1 -7)(a*(-1) + b) = (-3)(-4)(-5)(-6)(-7)(-8)(-a + b)

Compute the product: (-3)(-4)=12; 12*(-5)=-60; -60*(-6)=360; 360*(-7)=-2520; -2520*(-8)=20160. So 20160*(-a + b) =1 => -a + b =1/20160

Now, we have a system of equations:

1. a + b = -1/720

2. -a + b = 1/20160

Let's solve this system. Adding both equations:

( a + b ) + ( -a + b ) = (-1/720) + (1/20160)

2b = (-1/720) + (1/20160)

Convert to common denominator. Let's see, 720 and 20160. 20160 ÷ 720 = 28. So, 720*28=20160.

So, -1/720 = -28/20160

Thus, 2b = (-28 +1)/20160 = (-27)/20160

Therefore, b = (-27)/(2*20160) = (-27)/40320 = (-9)/13440 = (-3)/4480

Then, from equation 1: a + (-3/4480) = -1/720

So, a = -1/720 + 3/4480

Convert to common denominator. 720 and 4480. Let's see, 720 factors into 2^4 * 3^2 *5, and 4480 is 2^7 *5 *7. So LCM is 2^7 *3^2 *5 *7 = 128 *9 *5 *7 = 128*315=40320.

So, -1/720 = -56/40320

3/4480 = 27/40320

Therefore, a = (-56 +27)/40320 = (-29)/40320

So, a = -29/40320 and b = -3/4480

Therefore, R(x) = (x -2)(x -3)....(x -7)(-29/40320 x -3/4480)

Hmm, this seems complicated, but perhaps we can factor out some terms. Let me see.

Alternatively, perhaps we can write R(x) as (x -2)(x -3)....(x -7)(c x + d). But regardless, now we need to find p(8). To find p(8), we can use the definition of R(x):

R(x) = (x² -1)p(x) -x

So, R(8) = (8² -1)p(8) -8 = (64 -1)p(8) -8 = 63 p(8) -8

But also, R(8) = (8 -2)(8 -3)....(8 -7)(a*8 + b) = (6)(5)(4)(3)(2)(1)(a*8 + b) = 6! * (8a + b) = 720*(8a + b)

We already found a and b:

8a + b = 8*(-29/40320) + (-3/4480) = (-232/40320) - (3/4480)

Convert 3/4480 to 27/40320 (since 4480*9=40320), so:

= (-232 -27)/40320 = (-259)/40320

Thus, R(8) = 720*(-259/40320) = (720/40320)*(-259) = (1/56)*(-259) = -259/56

Therefore, 63 p(8) -8 = -259/56

Solving for p(8):

63 p(8) = 8 -259/56

Convert 8 to 448/56:

63 p(8) = 448/56 -259/56 = (448 -259)/56 = 189/56

Therefore, p(8) = (189/56)/63 = (189)/(56*63)

Simplify 189 and 63: 189 ÷63=3, so:

= 3/(56)

So, p(8)=3/56. Wait, that's the same result as before! So even though we had to go through a more complicated process by considering the linear term, we still end up with p(8)=3/56. Therefore, this confirms that the answer is indeed 3/56.

But let me just recap the steps to ensure there's no oversight.
</execution_2.4>
</Execution_2>

<Plan_3>
Based on execution_2:
1. Define R(x) = (x² -1)p(x) -x, which is a degree 7 polynomial.
2. R(x) has roots at n=2,3,....,7, so R(x) = (x -2)(x -3)....(x -7)(ax + b).
3. Use x=1 and x=-1 to create equations for a and b.
4. Solve for a and b.
5. Evaluate R(8) using both expressions to solve for p(8).

This seems thorough. Therefore, the answer should be 3/56.
</Plan_3>

<Execution_3>
No prompt to execute.
</Execution_3>

<Final_answer>
Therefore, the value of $p(8)$ is $\boxed{\dfrac{3}{56}}$.
</Final_answer>
</think>