Explanation:
Okay, let's see... Charlie is riding his unicycle, and I need to find his speed in inches per second. The problem says the wheel has a radius of 9 inches and makes 2 revolutions every 3 seconds. Hmm, speed... so I think I need to figure out how far the unicycle moves in one second.

First, maybe I should remember the formula for the circumference of a circle because the distance the wheel covers in one revolution is equal to its circumference. The formula for circumference is C = 2πr, right? Where r is the radius. So here, the radius is 9 inches. Let me calculate that first.

So, circumference C = 2 * π * 9. Let me compute that... 2 times 9 is 18, so it's 18π inches. So each full revolution, the wheel covers 18π inches.

But the wheel isn't making just one revolution; it's making 2 revolutions every 3 seconds. So I need to figure out how much distance that is per second. Maybe first find the distance per revolution and then multiply by the number of revolutions per second?

Yes, that makes sense. So if the wheel makes 2 revolutions in 3 seconds, then the number of revolutions per second is 2 divided by 3. Let me write that down: revolutions per second = 2/3 rev/sec.

Then, the distance per second would be the circumference multiplied by the number of revolutions per second. So that's 18π inches/rev * (2/3) rev/sec. Wait, the revolutions unit should cancel out, leaving inches per second, which is what we want. Let me compute that.

So 18π multiplied by 2/3. Let me do the math. 18 divided by 3 is 6, and 6 times 2 is 12. So that's 12π inches per second? Wait, is that right?

Let me check again. If the circumference is 18π, and in 3 seconds, the wheel makes 2 revolutions, so total distance in 3 seconds is 2 * 18π = 36π inches. Then, to find the speed per second, divide that by 3 seconds. So 36π divided by 3 is 12π inches per second. Yeah, same result. So that seems consistent.

But maybe I should convert π to a numerical value? The problem says to put the answer in inches per second, but they might want it in terms of π or as a decimal? Wait, the question just says "what is the unicycle's speed in inches per second?" without specifying, so maybe both are acceptable, but maybe it's better to leave it in terms of π unless told otherwise. Let me check the problem again.

The original problem says: "what is the unicycle's speed in inches per second?" No specific instructions, so maybe either is okay, but since π is exact, perhaps they want the answer in terms of π. However, maybe the answer expects a numerical value? Hmm. But the problem didn't specify, so maybe I should present both? Wait, no, let me check standard practice. In math problems, unless specified, often leaving it in terms of π is acceptable. But maybe the problem expects a numerical value. Wait, the radius is given as 9 inches, which is exact, and revolutions per time is exact. So 12π is exact. However, maybe they want a decimal? But since π is irrational, it's better to leave it as 12π. But maybe I should check again.

Wait, let me confirm my calculations once more. Radius is 9 inches. Circumference is 2πr = 18π. 2 revolutions per 3 seconds. So distance per second is (2 revolutions / 3 seconds) * 18π inches per revolution. Multiply those: 2*18π / 3 = 36π / 3 = 12π inches per second. Yes, that's correct. So if I write 12π inches per second, that should be the answer. But maybe the problem expects a decimal? Hmm. The problem is presented with exact numbers, so 12π is exact. If I convert π to approximately 3.1416, then 12*3.1416 is about 37.699 inches per second. But unless told to approximate, exact value is better.

But let me check the problem's source or if there are any clues. The problem is presented in a general way, so maybe they want the exact answer. So I think 12π inches per second is the correct answer. Wait, let me check once again. Let me think about units.

So revolutions per second is 2/3 rev/sec. Multiply by circumference (inches per revolution) gives inches per second. Yep, that's correct. So 2/3 * 18π = 12π. Correct.

Alternatively, maybe there's another way to approach it. Let's think about the angular velocity. But that might complicate. Angular velocity is 2 revolutions per 3 seconds, which is 2/3 rev/sec. Then, linear speed v = rω, but ω needs to be in radians per second. So 2 revolutions per 3 seconds is 2*(2π radians)/3 seconds = 4π/3 radians per second. Then, linear speed v = rω = 9 inches * 4π/3 rad/s = (36π)/3 = 12π inches per second. Same answer! So that's another way to get there. So that confirms it again.

Therefore, the speed is 12π inches per second. Therefore, the answer should be 12π inches per second. So I think that's the answer. Wait, but let me check once more if there's any mistake in calculations.

Alternatively, think about how much distance is covered in 3 seconds. 2 revolutions. Each revolution is 18π inches. So total distance 2*18π = 36π inches in 3 seconds. Therefore, speed is 36π / 3 = 12π inches per second. Yep, same. So all methods give me 12π. So that's solid.

So I think that's the answer. So I can write that in a box as \boxed{12\pi}.

**Final Answer**
The unicycle's speed is \boxed{12\pi} inches per second.

To determine the speed of Charlie's unicycle, we break the problem into two essential components: **distance traveled per revolution** and **number of revolutions per second**.

---

### 1. **Calculating the Circumference of the Wheel**

The unicycle's wheel has a **radius of 9 inches**, and the **distance a wheel covers in one full revolution** is equal to its **circumference**, given by the formula:

$$
C = 2\pi r
$$

Substituting the given radius:

$$
C = 2\pi \times 9 = 18\pi \text{ inches}
$$

So, each full revolution covers **18π inches**.

---

### 2. **Finding the Number of Revolutions per Second**

The problem states that the unicycle’s wheel makes **2 revolutions every 3 seconds**. Therefore, the **revolutions per second (RPS)** is:

$$
\text{RPS} = \frac{2}{3} \text{ revolutions per second}
$$

---

### 3. **Calculating the Speed in Inches per Second**

Now, to find the **speed** in inches per second, we multiply the distance per revolution (i.e., the circumference) by the number of revolutions per second:

$$
\text{Speed} = 18\pi \times \frac{2}{3} = \frac{36\pi}{3} = 12\pi \text{ inches per second}
$$

---

### Final Answer

$$
\boxed{12\pi}
$$

The unicycle's speed is **12π inches per second**.

[Assistant]
Explanation:
Okay, let's see... Charlie is riding his unicycle, and I need to find his speed in inches per second. The problem says the wheel has a radius of 9 inches and makes 2 revolutions every 3 seconds. Hmm, speed... so I think I need to figure out how far the unicycle moves in one second.

First, maybe I should remember the formula for the circumference of a circle because the distance the wheel covers in one revolution is equal to its circumference. The formula for circumference is C = 2πr, right? Where r is the radius. So here, the radius is 9 inches. Let me calculate that first.

So, circumference C = 2 * π * 9. Let me compute that... 2 times 9 is 18, so it's 18π inches. So each full revolution, the wheel covers 18π inches.

But the wheel isn't making just one revolution; it's making 2 revolutions every 3 seconds. So I need to figure out how much distance that is per second. Maybe first find the distance per revolution and then multiply by the number of revolutions per second?

Yes, that makes sense. So if the wheel makes 2 revolutions in 3 seconds, then the number of revolutions per second is 2 divided by 3. Let me write that down: revolutions per second = 2/3 rev/sec.

Then, the distance per second would be the circumference multiplied by the number of revolutions per second. So that's 18π inches/rev * (2/3) rev/sec. Wait, the revolutions unit should cancel out, leaving inches per second, which is what we want. Let me compute that.

So 18π multiplied by 2/3. Let me do the math. 18 divided by 3 is 6, and 6 times 2 is 12. So that's 12π inches per second? Wait, is that right?

Let me check again. If the circumference is 18π, and in 3 seconds, the wheel makes 2 revolutions, so total distance in 3 seconds is 2 * 18π = 36π inches. Then, to find the speed per second, divide that by 3 seconds. So 36π divided by 3 is 12π inches per second. Yeah, same result. So that seems consistent.

But maybe I should convert π to a numerical value? The problem says to put the answer in inches per second, but they might want it in terms of π or as a decimal? Wait, the question just says "what is the unicycle's speed in inches per second?" without specifying, so maybe both are acceptable, but maybe it's better to leave it in terms of π unless told otherwise. Let me check the problem again.

The original problem says: "what is the unicycle's speed in inches per second?" No specific instructions, so maybe either is okay, but since π is exact, perhaps they want the answer in terms of π. However, maybe the answer expects a numerical value? Hmm. But the problem didn't specify, so maybe I should present both? Wait, no, let me check standard practice. In math problems, unless specified, often leaving it in terms of π is acceptable. But maybe the problem expects a numerical value. Wait, the radius is given as 9 inches, which is exact, and revolutions per time is exact. So 12π is exact. However, maybe they want a decimal? But since π is irrational, it's better to leave it as 12π. But maybe I should check again.

Wait, let me confirm my calculations once more. Radius is 9 inches. Circumference is 2πr = 18π. 2 revolutions per 3 seconds. So distance per second is (2 revolutions / 3 seconds) * 18π inches per revolution. Multiply those: 2*18π / 3 = 36π / 3 = 12π inches per second. Yes, that's correct. So if I write 12π inches per second, that should be the answer. But maybe the problem expects a decimal? Hmm. The problem is presented with exact numbers, so 12π is exact. If I convert π to approximately 3.1416, then 12*3.1416 is about 37.699 inches per second. But unless told to approximate, exact value is better.

But let me check the problem's source or if there are any clues. The problem is presented in a general way, so maybe they want the exact answer. So I think 12π inches per second is the correct answer. Wait, let me check once again. Let me think about units.

So revolutions per second is 2/3 rev/sec. Multiply by circumference (inches per revolution) gives inches per second. Yep, that's correct. So 2/3 * 18π = 12π. Correct.

Alternatively, maybe there's another way to approach it. Let's think about the angular velocity. But that might complicate. Angular velocity is 2 revolutions per 3 seconds, which is 2/3 rev/sec. Then, linear speed v = rω, but ω needs to be in radians per second. So 2 revolutions per 3 seconds is 2*(2π radians)/3 seconds = 4π/3 radians per second. Then, linear speed v = rω = 9 inches * 4π/3 rad/s = (36π)/3 = 12π inches per second. Same answer! So that's another way to get there. So that confirms it again.

Therefore, the speed is 12π inches per second. Therefore, the answer should be 12π inches per second. So I think that's the answer. Wait, but let me check once more if there's any mistake in calculations.

Alternatively, think about how much distance is covered in 3 seconds. 2 revolutions. Each revolution is 18π inches. So total distance 2*18π = 36π inches in 3 seconds. Therefore, speed is 36π / 3 = 12π inches per second. Yep, same. So all methods give me 12π. So that's solid.

So I think that's the answer. So I can write that in a box as \boxed{12\pi}.

**Final Answer**
The unicycle's speed is \boxed{12\pi} inches per second.

To determine the speed of Charlie's unicycle, we break the problem into two essential components: **distance traveled per revolution** and **number of revolutions per second**.

---

### 1. **Calculating the Circumference of the Wheel**

The unicycle's wheel has a **radius of 9 inches**, and the **distance a wheel covers in one full revolution** is equal to its **circumference**, given by the formula:

$$
C = 2\pi r
$$

Substituting the given radius:

$$
C = 2\pi \times 9 = 18\pi \text{ inches}
$$

So, each full revolution covers **18π inches**.

---

### 2. **Finding the Number of Revolutions per Second**

The problem states that the unicycle’s wheel makes **2 revolutions every 3 seconds**. Therefore, the **revolutions per second (RPS)** is:

$$
\text{RPS} = \frac{2}{3} \text{ revolutions per second}
$$

---

### 3. **Calculating the Speed in Inches per Second**

Now, to find the **speed** in inches per second, we multiply the distance per revolution (i.e., the circumference) by the number of revolutions per second:

$$
\text{Speed} = 18\pi \times \frac{2}{3} = \frac{36\pi}{3} = 12\pi \text{ inches per second}
$$

---

### Final Answer

$$
\boxed{12\pi}
$$

The unicycle's speed is **12π inches per second**.

Solution: We need to know how many inches are in one revolution of the unicycle's wheel. In other words, we need to know the circumference of the wheel. Since the radius is $9$ inches, the circumference is $18\pi$ inches. So, if there are $18\pi$ inches in one revolution, and the wheel makes $2$ revolutions every $3$ seconds, the unicycle's speed is $18\pi\cdot\frac{2}{3} = \boxed{12\pi}$ inches per second.


Explanation:
Okay, let me try to figure out the degree of this polynomial p(x). The problem says that a, b, and c are distinct real numbers, so I don't have to worry about division by zero or anything like that. The polynomial is given by the sum of three fractions, each of which is a quadratic numerator divided by a constant denominator. Let me write it out again to make sure I have it right:

p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)]

Hmm, so each term in the sum is a quadratic polynomial divided by some constant. So each term individually is a quadratic, but when added together, maybe they simplify to something of lower degree? Or maybe not? The question is asking for the degree of the resulting polynomial p(x).

First, let me recall that if you have a polynomial expressed as a sum of other polynomials, the degree of the sum is the maximum degree of the individual polynomials, unless they cancel out terms. So if each term is quadratic, then if they don't cancel out, the degree would be 2. However, maybe they do cancel out? Let me check.

Alternatively, maybe there's a smarter way to figure this out. The expression reminds me of Lagrange interpolation polynomials. Let me recall: the Lagrange basis polynomials are of the form L_i(x) = product_{j ≠ i} (x - x_j)/(x_i - x_j). So for three points, each basis polynomial is quadratic. Then, if you have three points, the interpolating polynomial is a combination of these basis polynomials.

Wait, the given polynomial p(x) looks exactly like the Lagrange interpolation polynomial for the function f(x) = x², evaluated at points a, b, c? Let me think. Because if you have three points, the Lagrange interpolating polynomial would be the sum of f(a)L_a(x) + f(b)L_b(x) + f(c)L_c(x). So if f(x) is quadratic, then the interpolating polynomial would be quadratic. But since we are interpolating a quadratic function with three points, the interpolating polynomial is exactly that function. So p(x) should be equal to x². Therefore, the degree is 2.

But wait, let me verify that. Let me check with specific numbers. Let me choose specific values for a, b, c and see what p(x) evaluates to.

Let me take a = 0, b = 1, c = 2. Let me compute each term.

First term: [(x - 1)(x - 2)] / [(0 - 1)(0 - 2)] = [(x² - 3x + 2)] / [(-1)(-2)] = (x² - 3x + 2)/2.

Second term: [(x - 0)(x - 2)] / [(1 - 0)(1 - 2)] = [x(x - 2)] / [1 * (-1)] = [x² - 2x]/(-1) = -x² + 2x.

Third term: [(x - 0)(x - 1)] / [(2 - 0)(2 - 1)] = [x(x - 1)] / [2 * 1] = (x² - x)/2.

Now add them all together:

First term: (x² - 3x + 2)/2

Second term: -x² + 2x

Third term: (x² - x)/2

Let me combine them:

Let me convert all terms to have denominator 2:

First term: (x² - 3x + 2)/2

Second term: (-2x² + 4x)/2

Third term: (x² - x)/2

Adding them up:

[ (x² - 3x + 2) + (-2x² + 4x) + (x² - x) ] / 2

Compute numerator:

x² - 3x + 2 - 2x² + 4x + x² - x

Combine like terms:

x² - 2x² + x² = 0

-3x + 4x - x = 0x

+2

So total numerator is 0 + 0 + 2 = 2

Therefore, p(x) = 2 / 2 = 1. Wait, that's a constant function? But that's strange. Wait, but if we are interpolating f(x) = x² at points a=0, b=1, c=2. But if the interpolating polynomial is a quadratic, then when we evaluate it at x=0,1,2, it should equal f(x). Let me check.

For example, if p(x) is supposed to be x², then p(0) should be 0² = 0, but according to the calculation, p(0) would be 1? Wait, there's a mistake here. Wait, when I plugged in a=0, b=1, c=2, and computed p(x), I ended up with p(x) = 1, which is a constant function. But that contradicts the Lagrange interpolation idea. Wait, why?

Wait, maybe I made a mistake in the calculation. Let me check again.

First term: [(x - 1)(x - 2)] / [(0 - 1)(0 - 2)] = [(x² - 3x + 2)] / [ ( -1)( -2 ) ] = (x² - 3x + 2)/2. Correct.

Second term: [(x - 0)(x - 2)] / [(1 - 0)(1 - 2)] = [x(x - 2)] / [1*(-1)] = (x² - 2x)/(-1) = -x² + 2x. Correct.

Third term: [(x - 0)(x - 1)] / [(2 - 0)(2 - 1)] = [x(x - 1)]/[2*1] = (x² - x)/2. Correct.

Then adding all three terms:

First term + second term + third term:

(x² - 3x + 2)/2 + (-x² + 2x) + (x² - x)/2

Convert all to halves:

First term: (x² - 3x + 2)/2

Second term: (-2x² + 4x)/2

Third term: (x² - x)/2

Sum numerators:

(x² -3x +2) + (-2x² +4x) + (x² -x) = (x² -2x² +x²) + (-3x +4x -x) + 2 = 0x² + 0x +2 = 2

Therefore, 2/2 = 1. So p(x) is 1? But that's a constant function. But if we are interpolating f(x) = x², then the interpolating polynomial should pass through (0, 0), (1,1), (2,4). Let me check p(0):

p(0) = [ (0 - 1)(0 - 2) ] / [ (0 - 1)(0 - 2) ] + [ (0 - 0)(0 - 2) ] / [ (1 - 0)(1 - 2) ] + [ (0 - 0)(0 - 1) ] / [ (2 - 0)(2 - 1) ]

Compute each term:

First term: [ (-1)(-2) ] / [ (-1)(-2) ] = 2 / 2 = 1

Second term: [ 0 * (-2) ] / [ 1 * (-1) ] = 0 / (-1) = 0

Third term: [ 0 * (-1) ] / [ 2 * 1 ] = 0 / 2 = 0

Therefore, p(0) = 1 + 0 + 0 = 1. But f(0) is 0. That's a problem. Wait, so why is the interpolating polynomial giving 1 instead of 0?

Wait, maybe I confused something. Wait, the interpolating polynomial for f(x) = x² at points x=0,1,2 would be a quadratic polynomial that satisfies p(0) = 0, p(1)=1, p(2)=4. However, the given p(x) as calculated here is the constant function 1. Which does not satisfy p(0)=0. Therefore, my assumption that p(x) is the Lagrange interpolating polynomial for f(x)=x² must be wrong.

Alternatively, perhaps the given expression p(x) is the Lagrange interpolating polynomial for some function? Wait, maybe I need to think again.

Wait, the standard Lagrange interpolation polynomial for a function f is L(x) = sum_{i} f(x_i) * L_i(x), where L_i(x) are the basis polynomials. In our case, the given p(x) is sum_{i} [ (x - x_j)(x - x_k) / (x_i - x_j)(x_i - x_k) ] where the indices are cyclic. So if we take, for example, for i = a, then it's [(x - b)(x - c)] / [(a - b)(a - c)], etc. So if we think of this as sum_{i} [ (x - j)(x - k) / (i - j)(i - k) ] where i, j, k are the three points a, b, c.

But if we take the Lagrange basis polynomials for points a, b, c, then each L_i(x) is [(x - j)(x - k)] / [(i - j)(i - k)], where j and k are the other two points. Therefore, the given p(x) is sum_{i} L_i(x). But the sum of the Lagrange basis polynomials is 1, since for any x, sum L_i(x) = 1, because it's the interpolating polynomial for the constant function 1. So that would mean that p(x) is the constant function 1. But that contradicts with the earlier thought that if f(x) was quadratic. However, in the example I did, with a=0, b=1, c=2, p(x) turned out to be 1. So maybe the answer is a constant function, so degree 0? But that seems odd. But let me check with another example.

Wait, let me check with another set of points. Let me take a=0, b=1, c=-1. Let me compute p(x) again.

First term: [(x - 1)(x - (-1))] / [(0 - 1)(0 - (-1))] = [(x - 1)(x + 1)] / [(-1)(1)] = (x² - 1)/(-1) = -x² + 1.

Second term: [(x - 0)(x - (-1))] / [(1 - 0)(1 - (-1))] = [x(x + 1)] / [1*2] = (x² + x)/2.

Third term: [(x - 0)(x - 1)] / [(-1 - 0)(-1 - 1)] = [x(x - 1)] / [(-1)(-2)] = (x² - x)/2.

Adding them all together:

First term: -x² + 1.

Second term: (x² + x)/2.

Third term: (x² - x)/2.

Adding them:

Let me write them all with denominator 2:

First term: (-2x² + 2)/2.

Second term: (x² + x)/2.

Third term: (x² - x)/2.

Total numerator: (-2x² + 2) + (x² + x) + (x² - x) = (-2x² + x² + x²) + (x - x) + 2 = 0x² + 0x + 2 = 2. So p(x) = 2/2 = 1. Again, constant function 1. So regardless of the points, this seems to be the case. Therefore, the polynomial p(x) is actually the constant polynomial 1. Therefore, its degree is 0? But then why is the question asking for the degree? Maybe my initial thought was wrong, but the example shows that p(x) is constant.

But let me check with another function. Suppose instead of summing L_i(x), if I sum f(x_i) * L_i(x). For example, if I take f(x) = 1, then the interpolating polynomial is 1, which is consistent. If I take f(x) = x, then the interpolating polynomial would be... let me compute it. For points a, b, c. Wait, but the problem here is that p(x) is the sum of L_i(x), which is 1.

Therefore, this suggests that p(x) is always 1, regardless of a, b, c? Wait, but if that's the case, then the degree is 0. However, the problem says "Let a, b, and c be distinct real numbers. Find the degree of the polynomial..." So maybe the answer is 0. But that seems counterintuitive. Let me check once more with different a, b, c. Maybe I made a mistake in my previous calculation.

Wait, let me take a = 1, b = 2, c = 3. Let me compute p(x):

First term: [(x - 2)(x - 3)] / [(1 - 2)(1 - 3)] = [(x² -5x +6)] / [(-1)(-2)] = (x² -5x +6)/2

Second term: [(x - 1)(x - 3)] / [(2 - 1)(2 - 3)] = [(x² -4x +3)] / [1*(-1)] = -(x² -4x +3) = -x² +4x -3

Third term: [(x -1)(x -2)] / [(3 -1)(3 -2)] = [(x² -3x +2)] / [2*1] = (x² -3x +2)/2

Adding all three terms:

First term: (x² -5x +6)/2

Second term: (-x² +4x -3)

Third term: (x² -3x +2)/2

Convert all to halves:

First term: (x² -5x +6)/2

Second term: (-2x² +8x -6)/2

Third term: (x² -3x +2)/2

Now adding numerators:

(x² -5x +6) + (-2x² +8x -6) + (x² -3x +2)

Compute term by term:

x² -2x² +x² = 0

-5x +8x -3x = 0x

6 -6 +2 = 2

Total numerator is 2. So p(x) = 2/2 = 1. Again, constant function. So regardless of the points, this seems to be 1. Therefore, this suggests that p(x) is identically equal to 1. Therefore, degree 0.

But the problem says "Find the degree of the polynomial p(x)". But why is the problem structured this way? Maybe there's a mistake here? Or maybe my calculation is wrong? Let me check another perspective.

Alternatively, maybe p(x) is designed to be the interpolating polynomial for the function f(x) = 1, which is a constant function. Therefore, since we are interpolating a constant function with three points, the interpolating polynomial is just the constant function, so degree 0. Hence, the answer is 0. But the problem is presented as if it's more complicated, so maybe I need to think again.

Alternatively, perhaps I made a mistake in thinking that the given expression is the sum of Lagrange basis polynomials. Let me check again.

The Lagrange basis polynomial L_a(x) is [(x - b)(x - c)] / [(a - b)(a - c)], right? Yes. Similarly for L_b(x) and L_c(x). So the sum L_a(x) + L_b(x) + L_c(x) is equal to 1 for all x, because if you take any x, the sum of the Lagrange basis polynomials is 1. Because they are the basis for the space of polynomials of degree at most 2, and their sum is the constant polynomial 1. Therefore, p(x) is equal to 1. Therefore, it is a constant polynomial, so degree 0. Hence, the answer should be 0.

But then why does the problem say "Find the degree of the polynomial p(x)"? Maybe the original problem had different coefficients? Wait, let me check again the problem statement:

"Let a, b, and c be distinct real numbers. Find the degree of the polynomial

p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)]."

Yes, that's exactly what I've been working with. So according to my calculations, p(x) is identically 1. Therefore, degree 0. However, I need to be cautious here. Maybe there's a mistake in my reasoning?

Wait, but another way to think: suppose that p(x) is equal to 1. Then, for example, when I plug in x = a, we should get 1. Let me check that. Let me take x = a. Then:

First term: [(a - b)(a - c)] / [(a - b)(a - c)] = 1

Second term: [(a - a)(a - c)] / [(b - a)(b - c)] = 0 / [...] = 0

Third term: [(a - a)(a - b)] / [(c - a)(c - b)] = 0 / [...] = 0.

Therefore, p(a) = 1 + 0 + 0 = 1. Similarly, p(b) = 1, p(c) = 1. Therefore, p(x) is a polynomial that takes the value 1 at three distinct points a, b, c. If p(x) is a constant polynomial, then yes, that's consistent. However, if p(x) were a non-constant polynomial, it could take the value 1 at three points, but since we have already established that p(x) is 1 (as per the examples), then it is constant. Therefore, degree 0.

But why does the problem present it as a polynomial? Maybe there is something wrong here? Wait, if all the denominators are constants, then each term is a quadratic polynomial divided by a constant, so each term is a quadratic polynomial. Then adding them up, they could result in a quadratic polynomial, but in our case, they sum up to a constant. Therefore, the degree is 0.

But let me check once again with another example. Let me take a = 1, b = 2, c = 3. Let me compute p(x) at x = 0. Wait, in the previous example, p(x) was 1. Let me verify:

p(0) = [(0 - 2)(0 - 3)] / [(1 - 2)(1 - 3)] + [(0 - 1)(0 - 3)] / [(2 - 1)(2 - 3)] + [(0 - 1)(0 - 2)] / [(3 - 1)(3 - 2)]

Calculating each term:

First term: [(-2)(-3)] / [(-1)(-2)] = 6 / 2 = 3

Second term: [(-1)(-3)] / [(1)(-1)] = 3 / (-1) = -3

Third term: [(-1)(-2)] / [(2)(1)] = 2 / 2 = 1

So p(0) = 3 - 3 + 1 = 1. Correct.

Similarly, let's compute p(1):

First term: [(1 - 2)(1 - 3)] / [(1 - 2)(1 - 3)] = [(-1)(-2)] / [(-1)(-2)] = 2 / 2 = 1

Second term: [(1 - 1)(1 - 3)] / [(2 - 1)(2 - 3)] = [0 * (-2)] / [1 * (-1)] = 0

Third term: [(1 - 1)(1 - 2)] / [(3 - 1)(3 - 2)] = 0 / ... = 0

So p(1) = 1 + 0 + 0 = 1. Correct.

Same with p(2):

First term: [(2 - 2)(2 - 3)] / [(1 - 2)(1 - 3)] = 0 / ... = 0

Second term: [(2 - 1)(2 - 3)] / [(2 - 1)(2 - 3)] = [1 * (-1)] / [1 * (-1)] = 1

Third term: [(2 - 1)(2 - 2)] / [(3 - 1)(3 - 2)] = 0 / ... = 0

Thus, p(2) = 0 + 1 + 0 = 1. Correct. Similarly for p(3). Therefore, p(x) is 1 at x = a, b, c, and also at other points like 0, 1, 2, etc. Therefore, since it's a polynomial that takes the value 1 at infinitely many points (in fact, all points), it must be the constant polynomial 1. Hence, its degree is 0. Therefore, the answer should be 0.

But then why is the problem given as such? Maybe I misread it? Let me check again. The original problem is:

Find the degree of the polynomial p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)].

Yes, exactly as before. But maybe the problem is from a source where the answer is different? Or maybe I made a mistake in assuming that the sum of Lagrange basis polynomials is 1. Let me check that again.

In general, the sum of the Lagrange basis polynomials for a set of points is 1. Because if you consider the polynomial Q(x) = L_a(x) + L_b(x) + L_c(x), then Q(x) is a polynomial of degree at most 2. And for any x = a, b, c, Q(x) = 1. Since a polynomial of degree at most 2 that takes the value 1 at three distinct points must be the constant polynomial 1. Hence, Q(x) = 1. Therefore, p(x) is 1. Hence, degree 0.

Therefore, the answer should be 0. But maybe the problem is designed to trick you into thinking it's quadratic? But the examples show that it's constant. Therefore, I think the correct answer is 0.

But wait, hold on. Let me think again. Is p(x) really equal to 1? Suppose that I take a = 0, b = 1, c = -1. Then, p(x) = 1 for all x. But suppose that I take a different function. Suppose I take p(x) as defined, and compute its derivative. If p(x) is 1, then its derivative is 0. Let me check the derivative at x = 0. From the earlier example, when a=0, b=1, c=-1, p(x) = 1. So derivative is 0. But if I take the expression for p(x):

In that case, p(x) = -x² + 1 + (x² + x)/2 + (x² -x)/2. Which simplifies to (-x² +1) + (x² +x +x² -x)/2 = (-x² +1) + (2x²)/2 = (-x² +1) + x² = 1. Therefore, yes, it's 1. Therefore, derivative is zero. Therefore, the polynomial is constant. Hence, degree 0.

Therefore, despite the expression looking like a quadratic, when combined, they cancel out to a constant. Therefore, the answer is 0. But why would the problem give such a complicated expression? Maybe to test if you can recognize the Lagrange basis polynomials and realize their sum is 1?

Alternatively, maybe I made some mistake in the problem statement. Let me check again. The problem says:

"Find the degree of the polynomial p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)]."

Yes. So maybe there's a different interpretation. For example, if someone thought that the denominators are (a - b)(a - c) etc., which are constants, so each term is a quadratic over a constant, hence each term is quadratic. But when added together, they result in a constant.

Alternatively, maybe the problem is not about Lagrange interpolation, but about something else. Let me think. Suppose I didn't know about Lagrange interpolation. How else could I approach this?

Let me consider that each term is a quadratic polynomial. Let me denote them as follows:

Let me define f(x) = [(x - b)(x - c)] / [(a - b)(a - c)]

Similarly, g(x) = [(x - a)(x - c)] / [(b - a)(b - c)]

h(x) = [(x - a)(x - b)] / [(c - a)(c - b)]

Then p(x) = f(x) + g(x) + h(x). Now, if I can show that p(x) is constant, then degree is 0.

Alternatively, maybe I can substitute specific values of x to determine p(x). For example, as we saw, p(a) = 1, p(b) = 1, p(c) = 1. If p(x) is a polynomial that equals 1 at three distinct points, and if p(x) is of degree ≤ 2, then since the difference p(x) - 1 is a polynomial of degree ≤ 2 that has three roots, so p(x) - 1 must be identically zero, hence p(x) = 1. Therefore, degree 0.

Therefore, regardless of the values of a, b, c (as long as they are distinct), p(x) is constant. Therefore, the degree is 0. Therefore, the answer is 0.

But I need to make sure that there isn't a case where p(x) is not constant. For example, suppose that a, b, c are such that... Wait, but p(x) - 1 has roots at a, b, c. So, if p(x) is a polynomial of degree ≤ 2, then p(x) - 1 is a degree ≤ 2 polynomial with three roots, hence must be zero. Hence, p(x) = 1. So regardless of a, b, c (as long as they are distinct), p(x) is 1.

Therefore, the degree is 0.

But maybe there's a mistake here? Let me check with another example. Let me pick a = 1, b = 2, c = 3. Let me take x = 4. Then compute p(4):

First term: [(4 - 2)(4 - 3)] / [(1 - 2)(1 - 3)] = (2 * 1)/[(-1)*(-2)] = 2/2 = 1

Second term: [(4 - 1)(4 - 3)] / [(2 - 1)(2 - 3)] = (3 * 1)/[1 * (-1)] = 3/-1 = -3

Third term: [(4 - 1)(4 - 2)] / [(3 - 1)(3 - 2)] = (3 * 2)/[2 * 1] = 6/2 = 3

Adding them: 1 - 3 + 3 = 1. p(4) = 1.

Another point, x = 5:

First term: [(5 - 2)(5 - 3)] / [(1 - 2)(1 - 3)] = (3 * 2)/[(-1)(-2)] = 6/2 = 3

Second term: [(5 - 1)(5 - 3)] / [(2 - 1)(2 - 3)] = (4 * 2)/[1 * (-1)] = 8/-1 = -8

Third term: [(5 - 1)(5 - 2)] / [(3 - 1)(3 - 2)] = (4 * 3)/[2 * 1] = 12/2 = 6

Sum: 3 - 8 + 6 = 1.

So p(5) = 1. Therefore, it is indeed always 1. Therefore, p(x) is the constant polynomial 1. Hence, degree 0. Therefore, the answer is 0.

But then why does the problem ask for the degree? Because maybe the problem is designed to think that each term is quadratic, so the sum is quadratic, but then you realize it cancels to constant? So it's a trick question? Or perhaps there's a misunderstanding.

Alternatively, maybe I made a mistake in assuming that p(x) is equal to 1. Wait, but with multiple examples, and the general argument using the fact that p(x) - 1 has three roots and is degree 2, hence zero, it's valid. Therefore, the answer is 0.

Therefore, I think the correct answer is 0. However, to be thorough, let me think if there's another way to approach this.

Suppose I expand each term:

Let me consider the general expression:

p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)]

Let me denote the denominators as constants for each term:

Let me compute each term as a quadratic in x. Let me expand them:

First term:

[(x - b)(x - c)] = x² - (b + c)x + bc. Divided by (a - b)(a - c). So term1 = [x² - (b + c)x + bc]/D1, where D1 = (a - b)(a - c).

Similarly, term2 = [x² - (a + c)x + ac]/D2, where D2 = (b - a)(b - c).

term3 = [x² - (a + b)x + ab]/D3, where D3 = (c - a)(c - b).

Now, p(x) = term1 + term2 + term3.

Let me write p(x) as:

[ x²/D1 - (b + c)x/D1 + bc/D1 ] + [ x²/D2 - (a + c)x/D2 + ac/D2 ] + [ x²/D3 - (a + b)x/D3 + ab/D3 ]

Combine like terms:

For x² terms: [1/D1 + 1/D2 + 1/D3] * x²

For x terms: [ - (b + c)/D1 - (a + c)/D2 - (a + b)/D3 ] * x

Constant terms: [ bc/D1 + ac/D2 + ab/D3 ]

Therefore, if I can compute these coefficients, and see if they result in zero for the x² and x terms, and a constant.

Let me compute each coefficient.

First, compute the coefficient of x²:

1/D1 + 1/D2 + 1/D3.

Given that D1 = (a - b)(a - c), D2 = (b - a)(b - c) = -(a - b)(b - c), D3 = (c - a)(c - b) = (c - a)(c - b) = ( - (a - c) ) ( - (b - c)) = (a - c)(b - c). Wait:

Let me compute D1, D2, D3:

D1 = (a - b)(a - c)

D2 = (b - a)(b - c) = -(a - b)(b - c)

D3 = (c - a)(c - b) = ( - (a - c) ) ( - (b - c) ) = (a - c)(b - c)

So D1 = (a - b)(a - c)

D2 = -(a - b)(b - c)

D3 = (a - c)(b - c)

Therefore, let me compute 1/D1 + 1/D2 + 1/D3:

= 1/[(a - b)(a - c)] + 1/[ - (a - b)(b - c) ] + 1/[ (a - c)(b - c) ]

Let me factor out 1/(a - b)(a - c)(b - c) or something. Let me get common denominators.

Let me denote:

Let me compute each term:

Term1: 1 / [(a - b)(a - c)]

Term2: -1 / [(a - b)(b - c)]

Term3: 1 / [(a - c)(b - c)]

Let me write them with denominator (a - b)(a - c)(b - c):

Term1: [ (b - c) ] / [ (a - b)(a - c)(b - c) ]

Term2: - [ (a - c) ] / [ (a - b)(a - c)(b - c) ]

Term3: [ (a - b) ] / [ (a - b)(a - c)(b - c) ]

Adding them together:

[ (b - c) - (a - c) + (a - b) ] / [ (a - b)(a - c)(b - c) ]

Compute numerator:

(b - c) - (a - c) + (a - b)

Simplify term by term:

First term: b - c

Second term: -a + c

Third term: +a - b

Combine:

(b - c) + (-a + c) + (a - b)

= b - c - a + c + a - b

Simplify:

b - b cancels, -c + c cancels, -a + a cancels. Total is 0.

Therefore, coefficient of x² is 0.

Now, the coefficient of x:

- [ (b + c)/D1 + (a + c)/D2 + (a + b)/D3 ]

Let me compute this:

Let me write each term:

Term1: - (b + c)/D1 = - (b + c)/[(a - b)(a - c)]

Term2: - (a + c)/D2 = - (a + c)/[ - (a - b)(b - c) ] = (a + c)/[ (a - b)(b - c) ]

Term3: - (a + b)/D3 = - (a + b)/[ (a - c)(b - c) ]

So combining these terms:

Let me write them with common denominators. Let me denote denominators again.

Term1: - (b + c)/[(a - b)(a - c)]

Term2: (a + c)/[(a - b)(b - c)]

Term3: - (a + b)/[(a - c)(b - c)]

Again, let me get a common denominator of (a - b)(a - c)(b - c)

Term1 becomes: - (b + c)(b - c) / [ (a - b)(a - c)(b - c) ]

Term2 becomes: (a + c)(a - c) / [ (a - b)(a - c)(b - c) ]

Term3 becomes: - (a + b)(a - b) / [ (a - b)(a - c)(b - c) ]

So adding them together:

Numerator:

- (b + c)(b - c) + (a + c)(a - c) - (a + b)(a - b)

Let me compute each part:

First part: - (b + c)(b - c) = - (b² - c²) = -b² + c²

Second part: (a + c)(a - c) = a² - c²

Third part: - (a + b)(a - b) = - (a² - b²) = -a² + b²

Adding all three parts together:

(-b² + c²) + (a² - c²) + (-a² + b²) = (-b² + b²) + (c² - c²) + (a² - a²) = 0 + 0 + 0 = 0

Therefore, the coefficient of x is 0.

Now, the constant term:

bc/D1 + ac/D2 + ab/D3

Again, let's compute this:

bc / [(a - b)(a - c)] + ac / [ (b - a)(b - c) ] + ab / [ (c - a)(c - b) ]

Let me write denominators:

First term: (a - b)(a - c)

Second term: (b - a)(b - c) = -(a - b)(b - c)

Third term: (c - a)(c - b) = ( - (a - c) ) ( - (b - c) ) = (a - c)(b - c)

Therefore, rewrite:

Term1: bc / [ (a - b)(a - c) ]

Term2: ac / [ - (a - b)(b - c) ] = -ac / [ (a - b)(b - c) ]

Term3: ab / [ (a - c)(b - c) ]

Again, common denominator (a - b)(a - c)(b - c):

Term1: bc(b - c) / [ (a - b)(a - c)(b - c) ]

Term2: -ac(a - c) / [ (a - b)(a - c)(b - c) ]

Term3: ab(a - b) / [ (a - b)(a - c)(b - c) ]

Adding them together:

Numerator:

bc(b - c) - ac(a - c) + ab(a - b)

Let me expand each term:

First term: bc(b - c) = b²c - bc²

Second term: -ac(a - c) = -a²c + ac²

Third term: ab(a - b) = a²b - ab²

Adding all together:

b²c - bc² - a²c + ac² + a²b - ab²

Let me rearrange terms:

a²b - a²c + b²c - ab² + (- bc² + ac² )

Factor terms:

a²(b - c) + b²(c - a) + c²(a - b)

Wait, let me see:

Wait:

Group terms:

a²b - a²c = a²(b - c)

b²c - ab² = b²(c - a)

- bc² + ac² = c²(a - b)

Therefore, the entire numerator is:

a²(b - c) + b²(c - a) + c²(a - b)

This is a known expression. Let me compute it:

Let me expand:

a²b - a²c + b²c - ab² + ac² - bc²

Now, group terms:

(a²b - ab²) + (b²c - bc²) + (ac² - a²c)

Factor:

ab(a - b) + bc(b - c) + ac(c - a)

Hmm, but maybe there is a way to show that it's zero.

Alternatively, let me factor further. Let me consider:

a²(b - c) + b²(c - a) + c²(a - b)

This is a symmetric expression. Let me check if it's zero. For example, take a=0, b=1, c= -1:

Compute:

0²(1 - (-1)) + 1²(-1 - 0) + (-1)^2(0 - 1) = 0 + 1*(-1) + 1*(-1) = -1 -1 = -2 ≠ 0. Wait, but earlier calculations show that the constant term is 1? Wait, but earlier when I calculated p(x) with specific values, I found that p(x) was 1. However, here, the numerator is -2, and the denominator is (a - b)(a - c)(b - c) = (0 - 1)(0 - (-1))(1 - (-1)) = (-1)(1)(2) = -2. Therefore, the constant term would be (-2)/(-2) = 1. Which matches our previous result. Therefore, the numerator is not zero, but when divided by the denominator, gives 1.

But in general, how to show that:

Numerator: a²(b - c) + b²(c - a) + c²(a - b) = (a - b)(b - c)(c - a)

Wait, let me check with the example above: a=0, b=1, c=-1.

(a - b)(b - c)(c - a) = (0 - 1)(1 - (-1))(-1 - 0) = (-1)(2)(-1) = 2, while the numerator was -2. Therefore, it's - (a - b)(b - c)(c - a). So let me verify.

Let me compute:

Let me take the numerator: a²(b - c) + b²(c - a) + c²(a - b)

Let me factor:

Let me consider that this expression is equal to (a - b)(b - c)(c - a).

Let me test with a=1, b=2, c=3.

Numerator:

1²(2 - 3) + 2²(3 - 1) + 3²(1 - 2) = 1*(-1) + 4*(2) + 9*(-1) = -1 + 8 -9 = -2

Denominator: (1 - 2)(2 - 3)(3 - 1) = (-1)(-1)(2) = 2. Then the constant term is -2 / 2 = -1? Wait, but according to previous examples, when a=1, b=2, c=3, the constant term should be 1. Wait, but in that case, earlier when I computed p(x), it was 1. Wait, there's something wrong here.

Wait, let me compute the constant term for a=1, b=2, c=3:

From earlier, when I computed p(x) for a=1, b=2, c=3, and x=0, I got p(0)=1. But according to the calculation here, the numerator is -2, denominator is (a - b)(a - c)(b - c) = (1 - 2)(1 - 3)(2 - 3) = (-1)(-2)(-1) = -2. Therefore, constant term is (-2)/(-2) = 1. Correct. So the numerator is -2, denominator is -2, so total is 1. Therefore, the general formula is:

Numerator is equal to -(a - b)(b - c)(c - a). Therefore:

Numerator: a²(b - c) + b²(c - a) + c²(a - b) = -(a - b)(b - c)(c - a)

Thus, the constant term is [ -(a - b)(b - c)(c - a) ] / [ (a - b)(a - c)(b - c) ]

Simplify numerator and denominator:

Numerator: - (a - b)(b - c)(c - a)

Denominator: (a - b)(a - c)(b - c)

Note that (c - a) = - (a - c), so:

Numerator: - (a - b)(b - c)( - (a - c)) = (a - b)(b - c)(a - c)

Therefore, constant term becomes:

[ (a - b)(b - c)(a - c) ] / [ (a - b)(a - c)(b - c) ] = 1

Therefore, the constant term is 1.

Therefore, putting it all together, p(x) has:

Coefficient of x²: 0

Coefficient of x: 0

Constant term: 1

Therefore, p(x) is equal to 1. Therefore, degree 0.

Therefore, after expanding everything, it's clear that the polynomial p(x) is the constant polynomial 1. Therefore, its degree is 0.

Therefore, the answer is \boxed{0}.

**Final Answer**
\boxed{0}

To determine the degree of the polynomial

$$
p(x) = \frac{(x - b)(x - c)}{(a - b)(a - c)} + \frac{(x - a)(x - c)}{(b - a)(b - c)} + \frac{(x - a)(x - b)}{(c - a)(c - b)},
$$

we examine the structure of the expression.

---

### **Key Observations**

Each term in the sum is of the form:

$$
\frac{(x - x_j)(x - x_k)}{(x_i - x_j)(x_i - x_k)},
$$

where $ x_i, x_j, x_k $ are the three distinct real numbers $ a, b, c $. This structure is precisely the form of a **Lagrange basis polynomial** for the interpolation of a function on the three nodes $ a, b, c $.

More formally, for each index $ i $, the basis polynomial is defined as:

$$
L_i(x) = \frac{(x - x_j)(x - x_k)}{(x_i - x_j)(x_i - x_k)},
$$

where $ j, k $ are the other two indices not equal to $ i $. Thus, the given polynomial $ p(x) $ is:

$$
p(x) = L_a(x) + L_b(x) + L_c(x).
$$

It is a well-known fact in interpolation theory that the **sum of the Lagrange basis polynomials** for interpolation at three points is the **constant polynomial** 1. This is because the Lagrange polynomials form a basis for the space of polynomials of degree at most 2 (i.e., $ \mathbb{P}_2 $), and their sum, when evaluated at any point (e.g., the same value or any other variable), results in the sum being equal to 1.

---

### **Verification via Expansion**

To further verify this, we can expand each term of $ p(x) $, combine them, and show that the resulting polynomial is constant.

Let’s compute the **coefficient of** $ x^2 $, **coefficient of** $ x $, and the **constant term**, and observe that they all simplify to zero and zero, respectively, and the constant term simplifies to 1. This confirms that the polynomial is:

$$
p(x) = 1,
$$

a **constant function**, therefore a polynomial of **degree 0**.

---

### **Conclusion**

Although each term in the sum is a quadratic polynomial, they combine to form a constant function. Hence, the degree of $ p(x) $ is:

$$
\boxed{0}
$$

[Assistant]
Explanation:
Okay, let me try to figure out the degree of this polynomial p(x). The problem says that a, b, and c are distinct real numbers, so I don't have to worry about division by zero or anything like that. The polynomial is given by the sum of three fractions, each of which is a quadratic numerator divided by a constant denominator. Let me write it out again to make sure I have it right:

p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)]

Hmm, so each term in the sum is a quadratic polynomial divided by some constant. So each term individually is a quadratic, but when added together, maybe they simplify to something of lower degree? Or maybe not? The question is asking for the degree of the resulting polynomial p(x).

First, let me recall that if you have a polynomial expressed as a sum of other polynomials, the degree of the sum is the maximum degree of the individual polynomials, unless they cancel out terms. So if each term is quadratic, then if they don't cancel out, the degree would be 2. However, maybe they do cancel out? Let me check.

Alternatively, maybe there's a smarter way to figure this out. The expression reminds me of Lagrange interpolation polynomials. Let me recall: the Lagrange basis polynomials are of the form L_i(x) = product_{j ≠ i} (x - x_j)/(x_i - x_j). So for three points, each basis polynomial is quadratic. Then, if you have three points, the interpolating polynomial is a combination of these basis polynomials.

Wait, the given polynomial p(x) looks exactly like the Lagrange interpolation polynomial for the function f(x) = x², evaluated at points a, b, c? Let me think. Because if you have three points, the Lagrange interpolating polynomial would be the sum of f(a)L_a(x) + f(b)L_b(x) + f(c)L_c(x). So if f(x) is quadratic, then the interpolating polynomial would be quadratic. But since we are interpolating a quadratic function with three points, the interpolating polynomial is exactly that function. So p(x) should be equal to x². Therefore, the degree is 2.

But wait, let me verify that. Let me check with specific numbers. Let me choose specific values for a, b, c and see what p(x) evaluates to.

Let me take a = 0, b = 1, c = 2. Let me compute each term.

First term: [(x - 1)(x - 2)] / [(0 - 1)(0 - 2)] = [(x² - 3x + 2)] / [(-1)(-2)] = (x² - 3x + 2)/2.

Second term: [(x - 0)(x - 2)] / [(1 - 0)(1 - 2)] = [x(x - 2)] / [1 * (-1)] = [x² - 2x]/(-1) = -x² + 2x.

Third term: [(x - 0)(x - 1)] / [(2 - 0)(2 - 1)] = [x(x - 1)] / [2 * 1] = (x² - x)/2.

Now add them all together:

First term: (x² - 3x + 2)/2

Second term: -x² + 2x

Third term: (x² - x)/2

Let me combine them:

Let me convert all terms to have denominator 2:

First term: (x² - 3x + 2)/2

Second term: (-2x² + 4x)/2

Third term: (x² - x)/2

Adding them up:

[ (x² - 3x + 2) + (-2x² + 4x) + (x² - x) ] / 2

Compute numerator:

x² - 3x + 2 - 2x² + 4x + x² - x

Combine like terms:

x² - 2x² + x² = 0

-3x + 4x - x = 0x

+2

So total numerator is 0 + 0 + 2 = 2

Therefore, p(x) = 2 / 2 = 1. Wait, that's a constant function? But that's strange. Wait, but if we are interpolating f(x) = x² at points a=0, b=1, c=2. But if the interpolating polynomial is a quadratic, then when we evaluate it at x=0,1,2, it should equal f(x). Let me check.

For example, if p(x) is supposed to be x², then p(0) should be 0² = 0, but according to the calculation, p(0) would be 1? Wait, there's a mistake here. Wait, when I plugged in a=0, b=1, c=2, and computed p(x), I ended up with p(x) = 1, which is a constant function. But that contradicts the Lagrange interpolation idea. Wait, why?

Wait, maybe I made a mistake in the calculation. Let me check again.

First term: [(x - 1)(x - 2)] / [(0 - 1)(0 - 2)] = [(x² - 3x + 2)] / [ ( -1)( -2 ) ] = (x² - 3x + 2)/2. Correct.

Second term: [(x - 0)(x - 2)] / [(1 - 0)(1 - 2)] = [x(x - 2)] / [1*(-1)] = (x² - 2x)/(-1) = -x² + 2x. Correct.

Third term: [(x - 0)(x - 1)] / [(2 - 0)(2 - 1)] = [x(x - 1)]/[2*1] = (x² - x)/2. Correct.

Then adding all three terms:

First term + second term + third term:

(x² - 3x + 2)/2 + (-x² + 2x) + (x² - x)/2

Convert all to halves:

First term: (x² - 3x + 2)/2

Second term: (-2x² + 4x)/2

Third term: (x² - x)/2

Sum numerators:

(x² -3x +2) + (-2x² +4x) + (x² -x) = (x² -2x² +x²) + (-3x +4x -x) + 2 = 0x² + 0x +2 = 2

Therefore, 2/2 = 1. So p(x) is 1? But that's a constant function. But if we are interpolating f(x) = x², then the interpolating polynomial should pass through (0, 0), (1,1), (2,4). Let me check p(0):

p(0) = [ (0 - 1)(0 - 2) ] / [ (0 - 1)(0 - 2) ] + [ (0 - 0)(0 - 2) ] / [ (1 - 0)(1 - 2) ] + [ (0 - 0)(0 - 1) ] / [ (2 - 0)(2 - 1) ]

Compute each term:

First term: [ (-1)(-2) ] / [ (-1)(-2) ] = 2 / 2 = 1

Second term: [ 0 * (-2) ] / [ 1 * (-1) ] = 0 / (-1) = 0

Third term: [ 0 * (-1) ] / [ 2 * 1 ] = 0 / 2 = 0

Therefore, p(0) = 1 + 0 + 0 = 1. But f(0) is 0. That's a problem. Wait, so why is the interpolating polynomial giving 1 instead of 0?

Wait, maybe I confused something. Wait, the interpolating polynomial for f(x) = x² at points x=0,1,2 would be a quadratic polynomial that satisfies p(0) = 0, p(1)=1, p(2)=4. However, the given p(x) as calculated here is the constant function 1. Which does not satisfy p(0)=0. Therefore, my assumption that p(x) is the Lagrange interpolating polynomial for f(x)=x² must be wrong.

Alternatively, perhaps the given expression p(x) is the Lagrange interpolating polynomial for some function? Wait, maybe I need to think again.

Wait, the standard Lagrange interpolation polynomial for a function f is L(x) = sum_{i} f(x_i) * L_i(x), where L_i(x) are the basis polynomials. In our case, the given p(x) is sum_{i} [ (x - x_j)(x - x_k) / (x_i - x_j)(x_i - x_k) ] where the indices are cyclic. So if we take, for example, for i = a, then it's [(x - b)(x - c)] / [(a - b)(a - c)], etc. So if we think of this as sum_{i} [ (x - j)(x - k) / (i - j)(i - k) ] where i, j, k are the three points a, b, c.

But if we take the Lagrange basis polynomials for points a, b, c, then each L_i(x) is [(x - j)(x - k)] / [(i - j)(i - k)], where j and k are the other two points. Therefore, the given p(x) is sum_{i} L_i(x). But the sum of the Lagrange basis polynomials is 1, since for any x, sum L_i(x) = 1, because it's the interpolating polynomial for the constant function 1. So that would mean that p(x) is the constant function 1. But that contradicts with the earlier thought that if f(x) was quadratic. However, in the example I did, with a=0, b=1, c=2, p(x) turned out to be 1. So maybe the answer is a constant function, so degree 0? But that seems odd. But let me check with another example.

Wait, let me check with another set of points. Let me take a=0, b=1, c=-1. Let me compute p(x) again.

First term: [(x - 1)(x - (-1))] / [(0 - 1)(0 - (-1))] = [(x - 1)(x + 1)] / [(-1)(1)] = (x² - 1)/(-1) = -x² + 1.

Second term: [(x - 0)(x - (-1))] / [(1 - 0)(1 - (-1))] = [x(x + 1)] / [1*2] = (x² + x)/2.

Third term: [(x - 0)(x - 1)] / [(-1 - 0)(-1 - 1)] = [x(x - 1)] / [(-1)(-2)] = (x² - x)/2.

Adding them all together:

First term: -x² + 1.

Second term: (x² + x)/2.

Third term: (x² - x)/2.

Adding them:

Let me write them all with denominator 2:

First term: (-2x² + 2)/2.

Second term: (x² + x)/2.

Third term: (x² - x)/2.

Total numerator: (-2x² + 2) + (x² + x) + (x² - x) = (-2x² + x² + x²) + (x - x) + 2 = 0x² + 0x + 2 = 2. So p(x) = 2/2 = 1. Again, constant function 1. So regardless of the points, this seems to be the case. Therefore, the polynomial p(x) is actually the constant polynomial 1. Therefore, its degree is 0? But then why is the question asking for the degree? Maybe my initial thought was wrong, but the example shows that p(x) is constant.

But let me check with another function. Suppose instead of summing L_i(x), if I sum f(x_i) * L_i(x). For example, if I take f(x) = 1, then the interpolating polynomial is 1, which is consistent. If I take f(x) = x, then the interpolating polynomial would be... let me compute it. For points a, b, c. Wait, but the problem here is that p(x) is the sum of L_i(x), which is 1.

Therefore, this suggests that p(x) is always 1, regardless of a, b, c? Wait, but if that's the case, then the degree is 0. However, the problem says "Let a, b, and c be distinct real numbers. Find the degree of the polynomial..." So maybe the answer is 0. But that seems counterintuitive. Let me check once more with different a, b, c. Maybe I made a mistake in my previous calculation.

Wait, let me take a = 1, b = 2, c = 3. Let me compute p(x):

First term: [(x - 2)(x - 3)] / [(1 - 2)(1 - 3)] = [(x² -5x +6)] / [(-1)(-2)] = (x² -5x +6)/2

Second term: [(x - 1)(x - 3)] / [(2 - 1)(2 - 3)] = [(x² -4x +3)] / [1*(-1)] = -(x² -4x +3) = -x² +4x -3

Third term: [(x -1)(x -2)] / [(3 -1)(3 -2)] = [(x² -3x +2)] / [2*1] = (x² -3x +2)/2

Adding all three terms:

First term: (x² -5x +6)/2

Second term: (-x² +4x -3)

Third term: (x² -3x +2)/2

Convert all to halves:

First term: (x² -5x +6)/2

Second term: (-2x² +8x -6)/2

Third term: (x² -3x +2)/2

Now adding numerators:

(x² -5x +6) + (-2x² +8x -6) + (x² -3x +2)

Compute term by term:

x² -2x² +x² = 0

-5x +8x -3x = 0x

6 -6 +2 = 2

Total numerator is 2. So p(x) = 2/2 = 1. Again, constant function. So regardless of the points, this seems to be 1. Therefore, this suggests that p(x) is identically equal to 1. Therefore, degree 0.

But the problem says "Find the degree of the polynomial p(x)". But why is the problem structured this way? Maybe there's a mistake here? Or maybe my calculation is wrong? Let me check another perspective.

Alternatively, maybe p(x) is designed to be the interpolating polynomial for the function f(x) = 1, which is a constant function. Therefore, since we are interpolating a constant function with three points, the interpolating polynomial is just the constant function, so degree 0. Hence, the answer is 0. But the problem is presented as if it's more complicated, so maybe I need to think again.

Alternatively, perhaps I made a mistake in thinking that the given expression is the sum of Lagrange basis polynomials. Let me check again.

The Lagrange basis polynomial L_a(x) is [(x - b)(x - c)] / [(a - b)(a - c)], right? Yes. Similarly for L_b(x) and L_c(x). So the sum L_a(x) + L_b(x) + L_c(x) is equal to 1 for all x, because if you take any x, the sum of the Lagrange basis polynomials is 1. Because they are the basis for the space of polynomials of degree at most 2, and their sum is the constant polynomial 1. Therefore, p(x) is equal to 1. Therefore, it is a constant polynomial, so degree 0. Hence, the answer should be 0.

But then why does the problem say "Find the degree of the polynomial p(x)"? Maybe the original problem had different coefficients? Wait, let me check again the problem statement:

"Let a, b, and c be distinct real numbers. Find the degree of the polynomial

p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)]."

Yes, that's exactly what I've been working with. So according to my calculations, p(x) is identically 1. Therefore, degree 0. However, I need to be cautious here. Maybe there's a mistake in my reasoning?

Wait, but another way to think: suppose that p(x) is equal to 1. Then, for example, when I plug in x = a, we should get 1. Let me check that. Let me take x = a. Then:

First term: [(a - b)(a - c)] / [(a - b)(a - c)] = 1

Second term: [(a - a)(a - c)] / [(b - a)(b - c)] = 0 / [...] = 0

Third term: [(a - a)(a - b)] / [(c - a)(c - b)] = 0 / [...] = 0.

Therefore, p(a) = 1 + 0 + 0 = 1. Similarly, p(b) = 1, p(c) = 1. Therefore, p(x) is a polynomial that takes the value 1 at three distinct points a, b, c. If p(x) is a constant polynomial, then yes, that's consistent. However, if p(x) were a non-constant polynomial, it could take the value 1 at three points, but since we have already established that p(x) is 1 (as per the examples), then it is constant. Therefore, degree 0.

But why does the problem present it as a polynomial? Maybe there is something wrong here? Wait, if all the denominators are constants, then each term is a quadratic polynomial divided by a constant, so each term is a quadratic polynomial. Then adding them up, they could result in a quadratic polynomial, but in our case, they sum up to a constant. Therefore, the degree is 0.

But let me check once again with another example. Let me take a = 1, b = 2, c = 3. Let me compute p(x) at x = 0. Wait, in the previous example, p(x) was 1. Let me verify:

p(0) = [(0 - 2)(0 - 3)] / [(1 - 2)(1 - 3)] + [(0 - 1)(0 - 3)] / [(2 - 1)(2 - 3)] + [(0 - 1)(0 - 2)] / [(3 - 1)(3 - 2)]

Calculating each term:

First term: [(-2)(-3)] / [(-1)(-2)] = 6 / 2 = 3

Second term: [(-1)(-3)] / [(1)(-1)] = 3 / (-1) = -3

Third term: [(-1)(-2)] / [(2)(1)] = 2 / 2 = 1

So p(0) = 3 - 3 + 1 = 1. Correct.

Similarly, let's compute p(1):

First term: [(1 - 2)(1 - 3)] / [(1 - 2)(1 - 3)] = [(-1)(-2)] / [(-1)(-2)] = 2 / 2 = 1

Second term: [(1 - 1)(1 - 3)] / [(2 - 1)(2 - 3)] = [0 * (-2)] / [1 * (-1)] = 0

Third term: [(1 - 1)(1 - 2)] / [(3 - 1)(3 - 2)] = 0 / ... = 0

So p(1) = 1 + 0 + 0 = 1. Correct.

Same with p(2):

First term: [(2 - 2)(2 - 3)] / [(1 - 2)(1 - 3)] = 0 / ... = 0

Second term: [(2 - 1)(2 - 3)] / [(2 - 1)(2 - 3)] = [1 * (-1)] / [1 * (-1)] = 1

Third term: [(2 - 1)(2 - 2)] / [(3 - 1)(3 - 2)] = 0 / ... = 0

Thus, p(2) = 0 + 1 + 0 = 1. Correct. Similarly for p(3). Therefore, p(x) is 1 at x = a, b, c, and also at other points like 0, 1, 2, etc. Therefore, since it's a polynomial that takes the value 1 at infinitely many points (in fact, all points), it must be the constant polynomial 1. Hence, its degree is 0. Therefore, the answer should be 0.

But then why is the problem given as such? Maybe I misread it? Let me check again. The original problem is:

Find the degree of the polynomial p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)].

Yes, exactly as before. But maybe the problem is from a source where the answer is different? Or maybe I made a mistake in assuming that the sum of Lagrange basis polynomials is 1. Let me check that again.

In general, the sum of the Lagrange basis polynomials for a set of points is 1. Because if you consider the polynomial Q(x) = L_a(x) + L_b(x) + L_c(x), then Q(x) is a polynomial of degree at most 2. And for any x = a, b, c, Q(x) = 1. Since a polynomial of degree at most 2 that takes the value 1 at three distinct points must be the constant polynomial 1. Hence, Q(x) = 1. Therefore, p(x) is 1. Hence, degree 0.

Therefore, the answer should be 0. But maybe the problem is designed to trick you into thinking it's quadratic? But the examples show that it's constant. Therefore, I think the correct answer is 0.

But wait, hold on. Let me think again. Is p(x) really equal to 1? Suppose that I take a = 0, b = 1, c = -1. Then, p(x) = 1 for all x. But suppose that I take a different function. Suppose I take p(x) as defined, and compute its derivative. If p(x) is 1, then its derivative is 0. Let me check the derivative at x = 0. From the earlier example, when a=0, b=1, c=-1, p(x) = 1. So derivative is 0. But if I take the expression for p(x):

In that case, p(x) = -x² + 1 + (x² + x)/2 + (x² -x)/2. Which simplifies to (-x² +1) + (x² +x +x² -x)/2 = (-x² +1) + (2x²)/2 = (-x² +1) + x² = 1. Therefore, yes, it's 1. Therefore, derivative is zero. Therefore, the polynomial is constant. Hence, degree 0.

Therefore, despite the expression looking like a quadratic, when combined, they cancel out to a constant. Therefore, the answer is 0. But why would the problem give such a complicated expression? Maybe to test if you can recognize the Lagrange basis polynomials and realize their sum is 1?

Alternatively, maybe I made some mistake in the problem statement. Let me check again. The problem says:

"Find the degree of the polynomial p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)]."

Yes. So maybe there's a different interpretation. For example, if someone thought that the denominators are (a - b)(a - c) etc., which are constants, so each term is a quadratic over a constant, hence each term is quadratic. But when added together, they result in a constant.

Alternatively, maybe the problem is not about Lagrange interpolation, but about something else. Let me think. Suppose I didn't know about Lagrange interpolation. How else could I approach this?

Let me consider that each term is a quadratic polynomial. Let me denote them as follows:

Let me define f(x) = [(x - b)(x - c)] / [(a - b)(a - c)]

Similarly, g(x) = [(x - a)(x - c)] / [(b - a)(b - c)]

h(x) = [(x - a)(x - b)] / [(c - a)(c - b)]

Then p(x) = f(x) + g(x) + h(x). Now, if I can show that p(x) is constant, then degree is 0.

Alternatively, maybe I can substitute specific values of x to determine p(x). For example, as we saw, p(a) = 1, p(b) = 1, p(c) = 1. If p(x) is a polynomial that equals 1 at three distinct points, and if p(x) is of degree ≤ 2, then since the difference p(x) - 1 is a polynomial of degree ≤ 2 that has three roots, so p(x) - 1 must be identically zero, hence p(x) = 1. Therefore, degree 0.

Therefore, regardless of the values of a, b, c (as long as they are distinct), p(x) is constant. Therefore, the degree is 0. Therefore, the answer is 0.

But I need to make sure that there isn't a case where p(x) is not constant. For example, suppose that a, b, c are such that... Wait, but p(x) - 1 has roots at a, b, c. So, if p(x) is a polynomial of degree ≤ 2, then p(x) - 1 is a degree ≤ 2 polynomial with three roots, hence must be zero. Hence, p(x) = 1. So regardless of a, b, c (as long as they are distinct), p(x) is 1.

Therefore, the degree is 0.

But maybe there's a mistake here? Let me check with another example. Let me pick a = 1, b = 2, c = 3. Let me take x = 4. Then compute p(4):

First term: [(4 - 2)(4 - 3)] / [(1 - 2)(1 - 3)] = (2 * 1)/[(-1)*(-2)] = 2/2 = 1

Second term: [(4 - 1)(4 - 3)] / [(2 - 1)(2 - 3)] = (3 * 1)/[1 * (-1)] = 3/-1 = -3

Third term: [(4 - 1)(4 - 2)] / [(3 - 1)(3 - 2)] = (3 * 2)/[2 * 1] = 6/2 = 3

Adding them: 1 - 3 + 3 = 1. p(4) = 1.

Another point, x = 5:

First term: [(5 - 2)(5 - 3)] / [(1 - 2)(1 - 3)] = (3 * 2)/[(-1)(-2)] = 6/2 = 3

Second term: [(5 - 1)(5 - 3)] / [(2 - 1)(2 - 3)] = (4 * 2)/[1 * (-1)] = 8/-1 = -8

Third term: [(5 - 1)(5 - 2)] / [(3 - 1)(3 - 2)] = (4 * 3)/[2 * 1] = 12/2 = 6

Sum: 3 - 8 + 6 = 1.

So p(5) = 1. Therefore, it is indeed always 1. Therefore, p(x) is the constant polynomial 1. Hence, degree 0. Therefore, the answer is 0.

But then why does the problem ask for the degree? Because maybe the problem is designed to think that each term is quadratic, so the sum is quadratic, but then you realize it cancels to constant? So it's a trick question? Or perhaps there's a misunderstanding.

Alternatively, maybe I made a mistake in assuming that p(x) is equal to 1. Wait, but with multiple examples, and the general argument using the fact that p(x) - 1 has three roots and is degree 2, hence zero, it's valid. Therefore, the answer is 0.

Therefore, I think the correct answer is 0. However, to be thorough, let me think if there's another way to approach this.

Suppose I expand each term:

Let me consider the general expression:

p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)]

Let me denote the denominators as constants for each term:

Let me compute each term as a quadratic in x. Let me expand them:

First term:

[(x - b)(x - c)] = x² - (b + c)x + bc. Divided by (a - b)(a - c). So term1 = [x² - (b + c)x + bc]/D1, where D1 = (a - b)(a - c).

Similarly, term2 = [x² - (a + c)x + ac]/D2, where D2 = (b - a)(b - c).

term3 = [x² - (a + b)x + ab]/D3, where D3 = (c - a)(c - b).

Now, p(x) = term1 + term2 + term3.

Let me write p(x) as:

[ x²/D1 - (b + c)x/D1 + bc/D1 ] + [ x²/D2 - (a + c)x/D2 + ac/D2 ] + [ x²/D3 - (a + b)x/D3 + ab/D3 ]

Combine like terms:

For x² terms: [1/D1 + 1/D2 + 1/D3] * x²

For x terms: [ - (b + c)/D1 - (a + c)/D2 - (a + b)/D3 ] * x

Constant terms: [ bc/D1 + ac/D2 + ab/D3 ]

Therefore, if I can compute these coefficients, and see if they result in zero for the x² and x terms, and a constant.

Let me compute each coefficient.

First, compute the coefficient of x²:

1/D1 + 1/D2 + 1/D3.

Given that D1 = (a - b)(a - c), D2 = (b - a)(b - c) = -(a - b)(b - c), D3 = (c - a)(c - b) = (c - a)(c - b) = ( - (a - c) ) ( - (b - c)) = (a - c)(b - c). Wait:

Let me compute D1, D2, D3:

D1 = (a - b)(a - c)

D2 = (b - a)(b - c) = -(a - b)(b - c)

D3 = (c - a)(c - b) = ( - (a - c) ) ( - (b - c) ) = (a - c)(b - c)

So D1 = (a - b)(a - c)

D2 = -(a - b)(b - c)

D3 = (a - c)(b - c)

Therefore, let me compute 1/D1 + 1/D2 + 1/D3:

= 1/[(a - b)(a - c)] + 1/[ - (a - b)(b - c) ] + 1/[ (a - c)(b - c) ]

Let me factor out 1/(a - b)(a - c)(b - c) or something. Let me get common denominators.

Let me denote:

Let me compute each term:

Term1: 1 / [(a - b)(a - c)]

Term2: -1 / [(a - b)(b - c)]

Term3: 1 / [(a - c)(b - c)]

Let me write them with denominator (a - b)(a - c)(b - c):

Term1: [ (b - c) ] / [ (a - b)(a - c)(b - c) ]

Term2: - [ (a - c) ] / [ (a - b)(a - c)(b - c) ]

Term3: [ (a - b) ] / [ (a - b)(a - c)(b - c) ]

Adding them together:

[ (b - c) - (a - c) + (a - b) ] / [ (a - b)(a - c)(b - c) ]

Compute numerator:

(b - c) - (a - c) + (a - b)

Simplify term by term:

First term: b - c

Second term: -a + c

Third term: +a - b

Combine:

(b - c) + (-a + c) + (a - b)

= b - c - a + c + a - b

Simplify:

b - b cancels, -c + c cancels, -a + a cancels. Total is 0.

Therefore, coefficient of x² is 0.

Now, the coefficient of x:

- [ (b + c)/D1 + (a + c)/D2 + (a + b)/D3 ]

Let me compute this:

Let me write each term:

Term1: - (b + c)/D1 = - (b + c)/[(a - b)(a - c)]

Term2: - (a + c)/D2 = - (a + c)/[ - (a - b)(b - c) ] = (a + c)/[ (a - b)(b - c) ]

Term3: - (a + b)/D3 = - (a + b)/[ (a - c)(b - c) ]

So combining these terms:

Let me write them with common denominators. Let me denote denominators again.

Term1: - (b + c)/[(a - b)(a - c)]

Term2: (a + c)/[(a - b)(b - c)]

Term3: - (a + b)/[(a - c)(b - c)]

Again, let me get a common denominator of (a - b)(a - c)(b - c)

Term1 becomes: - (b + c)(b - c) / [ (a - b)(a - c)(b - c) ]

Term2 becomes: (a + c)(a - c) / [ (a - b)(a - c)(b - c) ]

Term3 becomes: - (a + b)(a - b) / [ (a - b)(a - c)(b - c) ]

So adding them together:

Numerator:

- (b + c)(b - c) + (a + c)(a - c) - (a + b)(a - b)

Let me compute each part:

First part: - (b + c)(b - c) = - (b² - c²) = -b² + c²

Second part: (a + c)(a - c) = a² - c²

Third part: - (a + b)(a - b) = - (a² - b²) = -a² + b²

Adding all three parts together:

(-b² + c²) + (a² - c²) + (-a² + b²) = (-b² + b²) + (c² - c²) + (a² - a²) = 0 + 0 + 0 = 0

Therefore, the coefficient of x is 0.

Now, the constant term:

bc/D1 + ac/D2 + ab/D3

Again, let's compute this:

bc / [(a - b)(a - c)] + ac / [ (b - a)(b - c) ] + ab / [ (c - a)(c - b) ]

Let me write denominators:

First term: (a - b)(a - c)

Second term: (b - a)(b - c) = -(a - b)(b - c)

Third term: (c - a)(c - b) = ( - (a - c) ) ( - (b - c) ) = (a - c)(b - c)

Therefore, rewrite:

Term1: bc / [ (a - b)(a - c) ]

Term2: ac / [ - (a - b)(b - c) ] = -ac / [ (a - b)(b - c) ]

Term3: ab / [ (a - c)(b - c) ]

Again, common denominator (a - b)(a - c)(b - c):

Term1: bc(b - c) / [ (a - b)(a - c)(b - c) ]

Term2: -ac(a - c) / [ (a - b)(a - c)(b - c) ]

Term3: ab(a - b) / [ (a - b)(a - c)(b - c) ]

Adding them together:

Numerator:

bc(b - c) - ac(a - c) + ab(a - b)

Let me expand each term:

First term: bc(b - c) = b²c - bc²

Second term: -ac(a - c) = -a²c + ac²

Third term: ab(a - b) = a²b - ab²

Adding all together:

b²c - bc² - a²c + ac² + a²b - ab²

Let me rearrange terms:

a²b - a²c + b²c - ab² + (- bc² + ac² )

Factor terms:

a²(b - c) + b²(c - a) + c²(a - b)

Wait, let me see:

Wait:

Group terms:

a²b - a²c = a²(b - c)

b²c - ab² = b²(c - a)

- bc² + ac² = c²(a - b)

Therefore, the entire numerator is:

a²(b - c) + b²(c - a) + c²(a - b)

This is a known expression. Let me compute it:

Let me expand:

a²b - a²c + b²c - ab² + ac² - bc²

Now, group terms:

(a²b - ab²) + (b²c - bc²) + (ac² - a²c)

Factor:

ab(a - b) + bc(b - c) + ac(c - a)

Hmm, but maybe there is a way to show that it's zero.

Alternatively, let me factor further. Let me consider:

a²(b - c) + b²(c - a) + c²(a - b)

This is a symmetric expression. Let me check if it's zero. For example, take a=0, b=1, c= -1:

Compute:

0²(1 - (-1)) + 1²(-1 - 0) + (-1)^2(0 - 1) = 0 + 1*(-1) + 1*(-1) = -1 -1 = -2 ≠ 0. Wait, but earlier calculations show that the constant term is 1? Wait, but earlier when I calculated p(x) with specific values, I found that p(x) was 1. However, here, the numerator is -2, and the denominator is (a - b)(a - c)(b - c) = (0 - 1)(0 - (-1))(1 - (-1)) = (-1)(1)(2) = -2. Therefore, the constant term would be (-2)/(-2) = 1. Which matches our previous result. Therefore, the numerator is not zero, but when divided by the denominator, gives 1.

But in general, how to show that:

Numerator: a²(b - c) + b²(c - a) + c²(a - b) = (a - b)(b - c)(c - a)

Wait, let me check with the example above: a=0, b=1, c=-1.

(a - b)(b - c)(c - a) = (0 - 1)(1 - (-1))(-1 - 0) = (-1)(2)(-1) = 2, while the numerator was -2. Therefore, it's - (a - b)(b - c)(c - a). So let me verify.

Let me compute:

Let me take the numerator: a²(b - c) + b²(c - a) + c²(a - b)

Let me factor:

Let me consider that this expression is equal to (a - b)(b - c)(c - a).

Let me test with a=1, b=2, c=3.

Numerator:

1²(2 - 3) + 2²(3 - 1) + 3²(1 - 2) = 1*(-1) + 4*(2) + 9*(-1) = -1 + 8 -9 = -2

Denominator: (1 - 2)(2 - 3)(3 - 1) = (-1)(-1)(2) = 2. Then the constant term is -2 / 2 = -1? Wait, but according to previous examples, when a=1, b=2, c=3, the constant term should be 1. Wait, but in that case, earlier when I computed p(x), it was 1. Wait, there's something wrong here.

Wait, let me compute the constant term for a=1, b=2, c=3:

From earlier, when I computed p(x) for a=1, b=2, c=3, and x=0, I got p(0)=1. But according to the calculation here, the numerator is -2, denominator is (a - b)(a - c)(b - c) = (1 - 2)(1 - 3)(2 - 3) = (-1)(-2)(-1) = -2. Therefore, constant term is (-2)/(-2) = 1. Correct. So the numerator is -2, denominator is -2, so total is 1. Therefore, the general formula is:

Numerator is equal to -(a - b)(b - c)(c - a). Therefore:

Numerator: a²(b - c) + b²(c - a) + c²(a - b) = -(a - b)(b - c)(c - a)

Thus, the constant term is [ -(a - b)(b - c)(c - a) ] / [ (a - b)(a - c)(b - c) ]

Simplify numerator and denominator:

Numerator: - (a - b)(b - c)(c - a)

Denominator: (a - b)(a - c)(b - c)

Note that (c - a) = - (a - c), so:

Numerator: - (a - b)(b - c)( - (a - c)) = (a - b)(b - c)(a - c)

Therefore, constant term becomes:

[ (a - b)(b - c)(a - c) ] / [ (a - b)(a - c)(b - c) ] = 1

Therefore, the constant term is 1.

Therefore, putting it all together, p(x) has:

Coefficient of x²: 0

Coefficient of x: 0

Constant term: 1

Therefore, p(x) is equal to 1. Therefore, degree 0.

Therefore, after expanding everything, it's clear that the polynomial p(x) is the constant polynomial 1. Therefore, its degree is 0.

Therefore, the answer is \boxed{0}.

**Final Answer**
\boxed{0}

To determine the degree of the polynomial

$$
p(x) = \frac{(x - b)(x - c)}{(a - b)(a - c)} + \frac{(x - a)(x - c)}{(b - a)(b - c)} + \frac{(x - a)(x - b)}{(c - a)(c - b)},
$$

we examine the structure of the expression.

---

### **Key Observations**

Each term in the sum is of the form:

$$
\frac{(x - x_j)(x - x_k)}{(x_i - x_j)(x_i - x_k)},
$$

where $ x_i, x_j, x_k $ are the three distinct real numbers $ a, b, c $. This structure is precisely the form of a **Lagrange basis polynomial** for the interpolation of a function on the three nodes $ a, b, c $.

More formally, for each index $ i $, the basis polynomial is defined as:

$$
L_i(x) = \frac{(x - x_j)(x - x_k)}{(x_i - x_j)(x_i - x_k)},
$$

where $ j, k $ are the other two indices not equal to $ i $. Thus, the given polynomial $ p(x) $ is:

$$
p(x) = L_a(x) + L_b(x) + L_c(x).
$$

It is a well-known fact in interpolation theory that the **sum of the Lagrange basis polynomials** for interpolation at three points is the **constant polynomial** 1. This is because the Lagrange polynomials form a basis for the space of polynomials of degree at most 2 (i.e., $ \mathbb{P}_2 $), and their sum, when evaluated at any point (e.g., the same value or any other variable), results in the sum being equal to 1.

---

### **Verification via Expansion**

To further verify this, we can expand each term of $ p(x) $, combine them, and show that the resulting polynomial is constant.

Let’s compute the **coefficient of** $ x^2 $, **coefficient of** $ x $, and the **constant term**, and observe that they all simplify to zero and zero, respectively, and the constant term simplifies to 1. This confirms that the polynomial is:

$$
p(x) = 1,
$$

a **constant function**, therefore a polynomial of **degree 0**.

---

### **Conclusion**

Although each term in the sum is a quadratic polynomial, they combine to form a constant function. Hence, the degree of $ p(x) $ is:

$$
\boxed{0}
$$

Solution:
Note that $p(x)$ has degree at most 2.  Also, $p(a) = p(b) = p(c) = 1.$  Thus, the polynomials $p(x)$ and 1 agree at three different values, so by the Identity Theorem, they are the same polynomial.  Hence, the degree of $p(x)$ (which is the constant polynomial 1) is $\boxed{0}.$

You can check manually that
\[p(x) = \frac{(x - b)(x - c)}{(a - b)(a - c)} + \frac{(x - a)(x - c)}{(b - a)(b - c)} + \frac{(x - a)(x - b)}{(c - a)(c - b)}\]simplifies to 1.


Explanation:
Okay, so I need to find cos(angle RPS) given that sin(angle RPQ) = 7/25. Let me try to visualize the diagram first. From the Asymptote code, there are points S, P, Q, and R. P is at the origin (0,0), Q is at (2,0), and S is at (-2,0). So, PQ is the segment from P(0,0) to Q(2,0), and PS is from P(0,0) to S(-2,0). So, PQ and PS are along the x-axis, with Q on the right and S on the left of P.

Then there's a point R, which is obtained by rotating the point (1.5, 0) by some angle. The angle mentioned is arcsin(7/25), which is angle RPQ. So, angle RPQ is the angle at point P between points R and Q. So, angle between PR and PQ is arcsin(7/25). Therefore, PR is a line from P(0,0) to R, making an angle with PQ (which is along the x-axis).

Since R is obtained by rotating (1.5, 0) by arcsin(7/25), maybe that rotation is around point P? Because the Asymptote code says rotate(aSin(7/25))*(1.5,0). So, rotating the point (1.5,0) by angle arcsin(7/25) around P, which is at (0,0). So, that would place R somewhere in the plane.

Let me try to find coordinates of R. Let me denote the angle θ = arcsin(7/25). So, sinθ = 7/25. Then, cosθ = sqrt(1 - (7/25)^2) = sqrt(1 - 49/625) = sqrt(576/625) = 24/25. So, cosθ is 24/25.

Since R is the rotation of (1.5, 0) by angle θ around P, which is at (0,0). Rotating a point (x, y) by angle θ around the origin can be done using rotation matrix:

x' = x cosθ - y sinθ

y' = x sinθ + y cosθ

Original point is (1.5, 0). So, x = 1.5, y = 0.

Therefore:

x' = 1.5 * cosθ - 0 * sinθ = 1.5 * (24/25) = (3/2)*(24/25) = (72)/50 = 36/25 = 1.44

y' = 1.5 * sinθ + 0 * cosθ = 1.5*(7/25) = (3/2)*(7/25) = 21/50 = 0.42

Therefore, coordinates of R are (36/25, 21/50). Wait, but 21/50 is 0.42? Yes. So R is at (1.44, 0.42).

But maybe I should keep fractions for exactness. 36/25 is 1.44, and 21/50 is 0.42.

Now, the question is to find cos(angle RPS). Angle RPS is the angle at point P between points R and S. So, points R, P, S. So, angle between PR and PS. Since PS is from P(0,0) to S(-2,0), which is along the negative x-axis, and PR is from P(0,0) to R(36/25, 21/50). So, angle between PR and PS. Since PS is along negative x-axis, and PR is going to the first quadrant. Therefore, the angle between PR and the negative x-axis is angle RPS. So, perhaps we can find the cosine of that angle.

Alternatively, since we know coordinates of R, we can compute the angle between vectors PR and PS. Since PS is the vector from P to S, which is (-2, 0), and PR is the vector from P to R, which is (36/25, 21/50). Then, the cosine of the angle between them is given by the dot product divided by the product of magnitudes.

Yes, that's a good approach. Let me do that.

First, compute vectors PR and PS.

Vector PR is R - P = (36/25, 21/50) - (0,0) = (36/25, 21/50)

Vector PS is S - P = (-2, 0) - (0, 0) = (-2, 0)

Dot product of PR and PS is (36/25)(-2) + (21/50)(0) = -72/25 + 0 = -72/25

The magnitude of PR is sqrt( (36/25)^2 + (21/50)^2 )

Let me compute that:

First, square of 36/25 is (1296)/625

Square of 21/50 is (441)/2500

So, add them: 1296/625 + 441/2500. Let's convert to common denominator:

Multiply numerator and denominator of 1296/625 by 4: 5184/2500 + 441/2500 = (5184 + 441)/2500 = 5625/2500

Simplify: 5625/2500 = 2.25 = 9/4. Wait, 5625 divided by 2500. Let me check:

Divide numerator and denominator by 25: 225/100 = 9/4. Yes! So sqrt(9/4) = 3/2. So magnitude of PR is 3/2.

Wait, that seems interesting. Let me check again:

(36/25)^2 + (21/50)^2

= (36^2 * 4 + 21^2) / 50^2 ?

Wait, no. Wait, 36/25 squared is (36^2)/25^2. Similarly, 21/50 squared is (21^2)/50^2. To add them, common denominator is 50^2 = 2500. So:

36^2 / 25^2 = (36^2 * 4)/50^2. Since 25^2 * 4 = 50^2.

Therefore, 36^2 * 4 + 21^2 over 50^2.

Compute 36^2 = 1296. 1296 * 4 = 5184. 21^2 = 441. Total numerator 5184 + 441 = 5625. Therefore, sqrt(5625 / 2500) = sqrt(5625)/sqrt(2500) = 75/50 = 3/2. Yes! So magnitude of PR is 3/2, which makes sense because the original point was (1.5, 0), which has magnitude 1.5 = 3/2, and rotating it doesn't change its magnitude. So that checks out.

So magnitude of PR is 3/2. The magnitude of PS is sqrt( (-2)^2 + 0^2 ) = sqrt(4) = 2.

Therefore, the cosine of angle RPS is [dot product]/[|PR| |PS|] = (-72/25) / [ (3/2)(2) ] = (-72/25) / 3 = (-72/25) * (1/3) = -24/25.

Wait, hold on. Wait, denominator is (3/2)*2 = 3. So, yes, (-72/25)/3 = (-24)/25. Therefore, cos(angle RPS) = -24/25?

But let me think again. Since angle between PR and PS. But if PR is in the first quadrant, and PS is along negative x-axis, then the angle between them is actually greater than 90 degrees, so cosine should be negative, which matches. So -24/25. But the problem says "What is cos(angle RPS)"? So is that the answer?

Wait, but let me check if I did the dot product correctly. The dot product is (36/25)*(-2) + (21/50)*0 = -72/25. Yes. The magnitude of PR is 3/2, magnitude of PS is 2, so 3/2 * 2 = 3. Then -72/25 divided by 3 is -24/25. So yes, that seems correct.

But let me think of another approach. Maybe using the angle. Since we know angle RPQ is θ, with sinθ = 7/25, and we want angle RPS. Maybe there is a relation between these angles?

Let me see. Let me think geometrically. Point P is between S and Q on the x-axis. R is somewhere above the x-axis. So angle RPQ is the angle between PR and PQ (which is along the positive x-axis). Then angle RPS is the angle between PR and PS (which is along negative x-axis). Therefore, angle RPS = 180° - angle RPQ? Wait, not exactly. Because if angle between PR and PQ is θ, then angle between PR and the negative x-axis would be 180° - θ, but only if PR is in the upper half-plane. But since angle between PR and positive x-axis is θ, then angle between PR and negative x-axis is π - θ. Therefore, cos(angle RPS) = cos(π - θ) = -cosθ. And since we know cosθ is 24/25, so -24/25. Which matches the previous result. Therefore, the answer is -24/25.

But wait, the problem says "cos(angle RPS)". So, according to this, it's -cosθ, which is -24/25. So that seems to align. Therefore, both methods give me -24/25.

Wait, but just to be thorough, let me check with coordinates. The angle RPS is the angle between PR and PS. Vector PR is going to (36/25, 21/50), and vector PS is going to (-2, 0). So, if I imagine this, the angle between PR and PS is the angle between a vector in the first quadrant and a vector pointing to the left along the x-axis. So, if I think of PR making an angle θ above the x-axis, then the angle between PR and the negative x-axis is 180° - θ, so the cosine would be cos(180° - θ) = -cosθ, which is -24/25.

Alternatively, if I calculate the angle of PR with respect to the x-axis. Let me compute tanθ, where θ is angle between PR and x-axis. tanθ = y/x = (21/50)/(36/25) = (21/50)*(25/36) = (21*25)/(50*36) = (21)/(2*36) = 21/72 = 7/24. Wait, so tanθ = 7/24. Which is consistent, since sinθ = 7/25, cosθ = 24/25, so tanθ = 7/24. Yes. So angle between PR and x-axis is θ, with sinθ = 7/25, cosθ = 24/25. Therefore, angle between PR and negative x-axis is π - θ. Then, cos(π - θ) = -cosθ = -24/25. Therefore, same answer.

Therefore, regardless of the method, answer is -24/25. So, I think that is the answer. However, just to make sure I didn't make any miscalculations. Let me check once again.

First, coordinates of R: rotating (1.5, 0) by θ = arcsin(7/25). So, coordinates are (1.5*cosθ, 1.5*sinθ). Wait, no. Wait, rotating a point (x, y) by θ around origin. Since original point is (1.5, 0), so rotating by θ, so:

x' = x cosθ - y sinθ = 1.5 cosθ - 0 = 1.5 cosθ

y' = x sinθ + y cosθ = 1.5 sinθ + 0 = 1.5 sinθ

Since sinθ = 7/25, cosθ = 24/25, so x' = 1.5*(24/25) = 3/2 * 24/25 = (72)/50 = 36/25. And y' = 1.5*(7/25) = 21/50. So, coordinates of R are (36/25, 21/50). So that is correct.

Then vector PR is (36/25, 21/50), vector PS is (-2, 0). Then dot product is (36/25)(-2) + (21/50)(0) = -72/25. The magnitude of PR is sqrt( (36/25)^2 + (21/50)^2 ). As calculated before, that is 3/2. The magnitude of PS is 2. Therefore, cos(angle between them) is (-72/25)/( (3/2)*2 ) = (-72/25)/3 = -24/25. Correct.

Alternatively, thinking in terms of angles: angle between PR and positive x-axis is θ with sinθ = 7/25, so angle between PR and negative x-axis is π - θ. Therefore, cos(π - θ) = -cosθ = -24/25. So same answer.

Therefore, I think the answer is -24/25. But let me check if there is any chance that angle RPS is not π - θ? Let me make sure.

Angle RPQ is the angle at P between R and Q. Since Q is on the positive x-axis, angle between PR and PQ (which is along positive x-axis) is θ. Therefore, angle RPQ = θ. Then angle RPS is angle between PR and PS (which is along negative x-axis). So if angle between PR and positive x-axis is θ, then angle between PR and negative x-axis is 180° - θ. Hence, cos(angle RPS) = cos(180° - θ) = -cosθ. Since cosθ is 24/25, answer is -24/25. So, that seems right.

Alternatively, if someone thought that angle RPS is just θ, but that would be incorrect because PS is pointing in the opposite direction of PQ. So, if angle between PR and PQ is θ, then angle between PR and PS is π - θ, so cosine is -cosθ.

Therefore, I think the answer is -24/25, so boxed answer is \boxed{-\dfrac{24}{25}}.

**Final Answer**
\boxed{-\dfrac{24}{25}}

To solve for $\cos \angle RPS$, we begin with the given information:

- $\sin \angle RPQ = \frac{7}{25}$
- The diagram places point $P$ at the origin $(0, 0)$, with point $Q$ at $(2, 0)$ (to the right), and point $S$ at $(-2, 0)$ (to the left), forming a horizontal axis through $P$.
- Point $R$ is obtained by rotating the point $(1.5, 0)$ around $P$ by $\theta = \arcsin\left(\frac{7}{25}\right)$.

---

### Step 1: Find $\cos \theta$

Given $\sin \theta = \frac{7}{25}$, we compute $\cos \theta$ using the Pythagorean identity:

$$
\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}
$$

---

### Step 2: Determine Coordinates of Point $R$

Point $R$ is the rotation of point $(1.5, 0)$ by angle $\theta$ around the origin. Using the 2D rotation matrix:

$$
\begin{bmatrix}
x' \\
y'
\end{bmatrix}
=
\begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
$$

Substituting $\cos \theta = \frac{24}{25}$, $\sin \theta = \frac{7}{25}$, and $x = 1.5$, $y = 0$:

- $x' = 1.5 \cdot \frac{24}{25} = \frac{36}{25}$
- $y' = 1.5 \cdot \frac{7}{25} = \frac{21}{50}$

So, the coordinates of $R$ are:
$$
R = \left( \frac{36}{25}, \frac{21}{50} \right)
$$

---

### Step 3: Use Vectors to Compute $\cos \angle RPS$

We now compute the cosine of the angle between vectors $\overrightarrow{PR}$ and $\overrightarrow{PS}$.

- Vector $ \overrightarrow{PR} = R - P = \left( \frac{36}{25}, \frac{21}{50} \right) $
- Vector $ \overrightarrow{PS} = S - P = (-2, 0) $

#### Dot Product:

$$
\overrightarrow{PR} \cdot \overrightarrow{PS} = \left( \frac{36}{25} \right)(-2) + \left( \frac{21}{50} \right)(0) = -\frac{72}{25}
$$

#### Magnitudes:

- $|\overrightarrow{PR}| = \sqrt{\left(\frac{36}{25}\right)^2 + \left(\frac{21}{50}\right)^2} = \frac{3}{2}$ (verified geometrically using magnitudes)
- $|\overrightarrow{PS}| = \sqrt{(-2)^2 + 0^2} = 2$

#### Compute $\cos \angle RPS$:

$$
\cos \angle RPS = \frac{\overrightarrow{PR} \cdot \overrightarrow{PS}}{|\overrightarrow{PR}| \cdot |\overrightarrow{PS}|} = \frac{-\frac{72}{25}}{\frac{3}{2} \cdot 2} = \frac{-\frac{72}{25}}{3} = -\frac{24}{25}
$$

---

### Step 4: Alternative Geometric Verification

Note that $\angle RPQ = \theta$ and $\angle RPS = 180^\circ - \theta$ (since $S$ is on the opposite side of the x-axis from $Q$). Using a trigonometric identity:

$$
\cos(180^\circ - \theta) = -\cos \theta = -\frac{24}{25}
$$

This confirms the result from the vector approach.

---

### Final Answer:

$$
\boxed{-\dfrac{24}{25}}
$$

[Assistant]
Explanation:
Okay, so I need to find cos(angle RPS) given that sin(angle RPQ) = 7/25. Let me try to visualize the diagram first. From the Asymptote code, there are points S, P, Q, and R. P is at the origin (0,0), Q is at (2,0), and S is at (-2,0). So, PQ is the segment from P(0,0) to Q(2,0), and PS is from P(0,0) to S(-2,0). So, PQ and PS are along the x-axis, with Q on the right and S on the left of P.

Then there's a point R, which is obtained by rotating the point (1.5, 0) by some angle. The angle mentioned is arcsin(7/25), which is angle RPQ. So, angle RPQ is the angle at point P between points R and Q. So, angle between PR and PQ is arcsin(7/25). Therefore, PR is a line from P(0,0) to R, making an angle with PQ (which is along the x-axis).

Since R is obtained by rotating (1.5, 0) by arcsin(7/25), maybe that rotation is around point P? Because the Asymptote code says rotate(aSin(7/25))*(1.5,0). So, rotating the point (1.5,0) by angle arcsin(7/25) around P, which is at (0,0). So, that would place R somewhere in the plane.

Let me try to find coordinates of R. Let me denote the angle θ = arcsin(7/25). So, sinθ = 7/25. Then, cosθ = sqrt(1 - (7/25)^2) = sqrt(1 - 49/625) = sqrt(576/625) = 24/25. So, cosθ is 24/25.

Since R is the rotation of (1.5, 0) by angle θ around P, which is at (0,0). Rotating a point (x, y) by angle θ around the origin can be done using rotation matrix:

x' = x cosθ - y sinθ

y' = x sinθ + y cosθ

Original point is (1.5, 0). So, x = 1.5, y = 0.

Therefore:

x' = 1.5 * cosθ - 0 * sinθ = 1.5 * (24/25) = (3/2)*(24/25) = (72)/50 = 36/25 = 1.44

y' = 1.5 * sinθ + 0 * cosθ = 1.5*(7/25) = (3/2)*(7/25) = 21/50 = 0.42

Therefore, coordinates of R are (36/25, 21/50). Wait, but 21/50 is 0.42? Yes. So R is at (1.44, 0.42).

But maybe I should keep fractions for exactness. 36/25 is 1.44, and 21/50 is 0.42.

Now, the question is to find cos(angle RPS). Angle RPS is the angle at point P between points R and S. So, points R, P, S. So, angle between PR and PS. Since PS is from P(0,0) to S(-2,0), which is along the negative x-axis, and PR is from P(0,0) to R(36/25, 21/50). So, angle between PR and PS. Since PS is along negative x-axis, and PR is going to the first quadrant. Therefore, the angle between PR and the negative x-axis is angle RPS. So, perhaps we can find the cosine of that angle.

Alternatively, since we know coordinates of R, we can compute the angle between vectors PR and PS. Since PS is the vector from P to S, which is (-2, 0), and PR is the vector from P to R, which is (36/25, 21/50). Then, the cosine of the angle between them is given by the dot product divided by the product of magnitudes.

Yes, that's a good approach. Let me do that.

First, compute vectors PR and PS.

Vector PR is R - P = (36/25, 21/50) - (0,0) = (36/25, 21/50)

Vector PS is S - P = (-2, 0) - (0, 0) = (-2, 0)

Dot product of PR and PS is (36/25)(-2) + (21/50)(0) = -72/25 + 0 = -72/25

The magnitude of PR is sqrt( (36/25)^2 + (21/50)^2 )

Let me compute that:

First, square of 36/25 is (1296)/625

Square of 21/50 is (441)/2500

So, add them: 1296/625 + 441/2500. Let's convert to common denominator:

Multiply numerator and denominator of 1296/625 by 4: 5184/2500 + 441/2500 = (5184 + 441)/2500 = 5625/2500

Simplify: 5625/2500 = 2.25 = 9/4. Wait, 5625 divided by 2500. Let me check:

Divide numerator and denominator by 25: 225/100 = 9/4. Yes! So sqrt(9/4) = 3/2. So magnitude of PR is 3/2.

Wait, that seems interesting. Let me check again:

(36/25)^2 + (21/50)^2

= (36^2 * 4 + 21^2) / 50^2 ?

Wait, no. Wait, 36/25 squared is (36^2)/25^2. Similarly, 21/50 squared is (21^2)/50^2. To add them, common denominator is 50^2 = 2500. So:

36^2 / 25^2 = (36^2 * 4)/50^2. Since 25^2 * 4 = 50^2.

Therefore, 36^2 * 4 + 21^2 over 50^2.

Compute 36^2 = 1296. 1296 * 4 = 5184. 21^2 = 441. Total numerator 5184 + 441 = 5625. Therefore, sqrt(5625 / 2500) = sqrt(5625)/sqrt(2500) = 75/50 = 3/2. Yes! So magnitude of PR is 3/2, which makes sense because the original point was (1.5, 0), which has magnitude 1.5 = 3/2, and rotating it doesn't change its magnitude. So that checks out.

So magnitude of PR is 3/2. The magnitude of PS is sqrt( (-2)^2 + 0^2 ) = sqrt(4) = 2.

Therefore, the cosine of angle RPS is [dot product]/[|PR| |PS|] = (-72/25) / [ (3/2)(2) ] = (-72/25) / 3 = (-72/25) * (1/3) = -24/25.

Wait, hold on. Wait, denominator is (3/2)*2 = 3. So, yes, (-72/25)/3 = (-24)/25. Therefore, cos(angle RPS) = -24/25?

But let me think again. Since angle between PR and PS. But if PR is in the first quadrant, and PS is along negative x-axis, then the angle between them is actually greater than 90 degrees, so cosine should be negative, which matches. So -24/25. But the problem says "What is cos(angle RPS)"? So is that the answer?

Wait, but let me check if I did the dot product correctly. The dot product is (36/25)*(-2) + (21/50)*0 = -72/25. Yes. The magnitude of PR is 3/2, magnitude of PS is 2, so 3/2 * 2 = 3. Then -72/25 divided by 3 is -24/25. So yes, that seems correct.

But let me think of another approach. Maybe using the angle. Since we know angle RPQ is θ, with sinθ = 7/25, and we want angle RPS. Maybe there is a relation between these angles?

Let me see. Let me think geometrically. Point P is between S and Q on the x-axis. R is somewhere above the x-axis. So angle RPQ is the angle between PR and PQ (which is along the positive x-axis). Then angle RPS is the angle between PR and PS (which is along negative x-axis). Therefore, angle RPS = 180° - angle RPQ? Wait, not exactly. Because if angle between PR and PQ is θ, then angle between PR and the negative x-axis would be 180° - θ, but only if PR is in the upper half-plane. But since angle between PR and positive x-axis is θ, then angle between PR and negative x-axis is π - θ. Therefore, cos(angle RPS) = cos(π - θ) = -cosθ. And since we know cosθ is 24/25, so -24/25. Which matches the previous result. Therefore, the answer is -24/25.

But wait, the problem says "cos(angle RPS)". So, according to this, it's -cosθ, which is -24/25. So that seems to align. Therefore, both methods give me -24/25.

Wait, but just to be thorough, let me check with coordinates. The angle RPS is the angle between PR and PS. Vector PR is going to (36/25, 21/50), and vector PS is going to (-2, 0). So, if I imagine this, the angle between PR and PS is the angle between a vector in the first quadrant and a vector pointing to the left along the x-axis. So, if I think of PR making an angle θ above the x-axis, then the angle between PR and the negative x-axis is 180° - θ, so the cosine would be cos(180° - θ) = -cosθ, which is -24/25.

Alternatively, if I calculate the angle of PR with respect to the x-axis. Let me compute tanθ, where θ is angle between PR and x-axis. tanθ = y/x = (21/50)/(36/25) = (21/50)*(25/36) = (21*25)/(50*36) = (21)/(2*36) = 21/72 = 7/24. Wait, so tanθ = 7/24. Which is consistent, since sinθ = 7/25, cosθ = 24/25, so tanθ = 7/24. Yes. So angle between PR and x-axis is θ, with sinθ = 7/25, cosθ = 24/25. Therefore, angle between PR and negative x-axis is π - θ. Then, cos(π - θ) = -cosθ = -24/25. Therefore, same answer.

Therefore, regardless of the method, answer is -24/25. So, I think that is the answer. However, just to make sure I didn't make any miscalculations. Let me check once again.

First, coordinates of R: rotating (1.5, 0) by θ = arcsin(7/25). So, coordinates are (1.5*cosθ, 1.5*sinθ). Wait, no. Wait, rotating a point (x, y) by θ around origin. Since original point is (1.5, 0), so rotating by θ, so:

x' = x cosθ - y sinθ = 1.5 cosθ - 0 = 1.5 cosθ

y' = x sinθ + y cosθ = 1.5 sinθ + 0 = 1.5 sinθ

Since sinθ = 7/25, cosθ = 24/25, so x' = 1.5*(24/25) = 3/2 * 24/25 = (72)/50 = 36/25. And y' = 1.5*(7/25) = 21/50. So, coordinates of R are (36/25, 21/50). So that is correct.

Then vector PR is (36/25, 21/50), vector PS is (-2, 0). Then dot product is (36/25)(-2) + (21/50)(0) = -72/25. The magnitude of PR is sqrt( (36/25)^2 + (21/50)^2 ). As calculated before, that is 3/2. The magnitude of PS is 2. Therefore, cos(angle between them) is (-72/25)/( (3/2)*2 ) = (-72/25)/3 = -24/25. Correct.

Alternatively, thinking in terms of angles: angle between PR and positive x-axis is θ with sinθ = 7/25, so angle between PR and negative x-axis is π - θ. Therefore, cos(π - θ) = -cosθ = -24/25. So same answer.

Therefore, I think the answer is -24/25. But let me check if there is any chance that angle RPS is not π - θ? Let me make sure.

Angle RPQ is the angle at P between R and Q. Since Q is on the positive x-axis, angle between PR and PQ (which is along positive x-axis) is θ. Therefore, angle RPQ = θ. Then angle RPS is angle between PR and PS (which is along negative x-axis). So if angle between PR and positive x-axis is θ, then angle between PR and negative x-axis is 180° - θ. Hence, cos(angle RPS) = cos(180° - θ) = -cosθ. Since cosθ is 24/25, answer is -24/25. So, that seems right.

Alternatively, if someone thought that angle RPS is just θ, but that would be incorrect because PS is pointing in the opposite direction of PQ. So, if angle between PR and PQ is θ, then angle between PR and PS is π - θ, so cosine is -cosθ.

Therefore, I think the answer is -24/25, so boxed answer is \boxed{-\dfrac{24}{25}}.

**Final Answer**
\boxed{-\dfrac{24}{25}}

To solve for $\cos \angle RPS$, we begin with the given information:

- $\sin \angle RPQ = \frac{7}{25}$
- The diagram places point $P$ at the origin $(0, 0)$, with point $Q$ at $(2, 0)$ (to the right), and point $S$ at $(-2, 0)$ (to the left), forming a horizontal axis through $P$.
- Point $R$ is obtained by rotating the point $(1.5, 0)$ around $P$ by $\theta = \arcsin\left(\frac{7}{25}\right)$.

---

### Step 1: Find $\cos \theta$

Given $\sin \theta = \frac{7}{25}$, we compute $\cos \theta$ using the Pythagorean identity:

$$
\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}
$$

---

### Step 2: Determine Coordinates of Point $R$

Point $R$ is the rotation of point $(1.5, 0)$ by angle $\theta$ around the origin. Using the 2D rotation matrix:

$$
\begin{bmatrix}
x' \\
y'
\end{bmatrix}
=
\begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
$$

Substituting $\cos \theta = \frac{24}{25}$, $\sin \theta = \frac{7}{25}$, and $x = 1.5$, $y = 0$:

- $x' = 1.5 \cdot \frac{24}{25} = \frac{36}{25}$
- $y' = 1.5 \cdot \frac{7}{25} = \frac{21}{50}$

So, the coordinates of $R$ are:
$$
R = \left( \frac{36}{25}, \frac{21}{50} \right)
$$

---

### Step 3: Use Vectors to Compute $\cos \angle RPS$

We now compute the cosine of the angle between vectors $\overrightarrow{PR}$ and $\overrightarrow{PS}$.

- Vector $ \overrightarrow{PR} = R - P = \left( \frac{36}{25}, \frac{21}{50} \right) $
- Vector $ \overrightarrow{PS} = S - P = (-2, 0) $

#### Dot Product:

$$
\overrightarrow{PR} \cdot \overrightarrow{PS} = \left( \frac{36}{25} \right)(-2) + \left( \frac{21}{50} \right)(0) = -\frac{72}{25}
$$

#### Magnitudes:

- $|\overrightarrow{PR}| = \sqrt{\left(\frac{36}{25}\right)^2 + \left(\frac{21}{50}\right)^2} = \frac{3}{2}$ (verified geometrically using magnitudes)
- $|\overrightarrow{PS}| = \sqrt{(-2)^2 + 0^2} = 2$

#### Compute $\cos \angle RPS$:

$$
\cos \angle RPS = \frac{\overrightarrow{PR} \cdot \overrightarrow{PS}}{|\overrightarrow{PR}| \cdot |\overrightarrow{PS}|} = \frac{-\frac{72}{25}}{\frac{3}{2} \cdot 2} = \frac{-\frac{72}{25}}{3} = -\frac{24}{25}
$$

---

### Step 4: Alternative Geometric Verification

Note that $\angle RPQ = \theta$ and $\angle RPS = 180^\circ - \theta$ (since $S$ is on the opposite side of the x-axis from $Q$). Using a trigonometric identity:

$$
\cos(180^\circ - \theta) = -\cos \theta = -\frac{24}{25}
$$

This confirms the result from the vector approach.

---

### Final Answer:

$$
\boxed{-\dfrac{24}{25}}
$$

Solution:
For any angle $x$, we have $\cos(180^\circ - x)=-\cos x$, so $\cos \angle RPS = \cos(180^\circ - \angle RPQ) =- \cos\angle RPQ$.

Since $\sin^2 \angle RPQ + \cos^2 \angle RPQ = 1$, we have $\cos^2\angle RPQ = 1 - \left(\frac{7}{25}\right)^2 = \frac{576}{625}$.  Since $\angle RPQ$ is acute, we have $\cos\angle RPQ = \frac{24}{25}$, which gives us $\cos\angle RPS = -\cos\angle RPQ = \boxed{-\frac{24}{25}}$.