<Question>
Let $a,$ $b,$ and $c$ be distinct real numbers.  Find the degree of the polynomial
\[p(x) = \frac{(x - b)(x - c)}{(a - b)(a - c)} + \frac{(x - a)(x - c)}{(b - a)(b - c)} + \frac{(x - a)(x - b)}{(c - a)(c - b)}.\]
</Question>

<think>
<Plan_1>
Okay, so I need to find the degree of the polynomial p(x) given by that expression. Let me first write it down 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, all right. The question says that a, b, and c are distinct real numbers, so none of the denominators are zero, which is good because otherwise, the expression would be undefined. So each term here is a rational function, but when we add them up, we need to check if the result is a polynomial and, if so, what its degree is.

First thought: Each numerator is a quadratic polynomial in x, right? Because each term is (x - something)(x - something else), which expands to x² minus some linear term plus a constant. The denominators are constants because they only involve a, b, c, which are fixed numbers. So each fraction is a quadratic polynomial divided by a constant.

<prompt_1.1> Analyze the degree of each term in the polynomial expression. </prompt_1.1>
-----

But just to make sure, maybe there's another way to think about it. Maybe using Lagrange interpolation? Because the form of the expression looks a lot like the Lagrange basis polynomials for three points.

<prompt_1.2> Consider the possibility of using Lagrange interpolation to determine the degree of the polynomial. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
Therefore, individually, each term is a quadratic. But when you add three quadratics together, the result could be a quadratic, linear, or even constant function, depending on how the terms cancel out. However, the question is asking for the degree of the polynomial, so we need to figure out if there's an x² term, and if so, whether it survives after adding the three terms.
</execution_1.1>

<execution_1.2>
Let me recall: Lagrange's formula for interpolation through three points (a, f(a)), (b, f(b)), (c, f(c)) is given by p(x) = f(a) * [(x - b)(x - c)] / [(a - b)(a - c)] + f(b) * [(x - a)(x - c)] / [(b - a)(b - c)] + f(c) * [(x - a)(x - b)] / [(c - a)(c - b)]. 

Wait, that's exactly the form of p(x) here! But in the standard Lagrange interpolation formula, the coefficients f(a), f(b), f(c) are the function values at those points. However, in our problem, the coefficients are 1 in each term. So if we compare, that would mean that f(a) = f(b) = f(c) = 1. So p(x) would be the interpolating polynomial that passes through the points (a, 1), (b, 1), (c, 1). Since all three points have the same y-value, 1, the interpolating polynomial would be the constant polynomial 1. 

But hold on, the Lagrange interpolating polynomial for three points with the same y-value is indeed a constant polynomial. But according to our earlier expansion, p(x) is a quadratic polynomial. That seems contradictory. So there's a conflict here. On one hand, expanding each term gives a quadratic, but if it's supposed to be the Lagrange interpolating polynomial for three points all at y=1, then it should be the constant polynomial 1. 

But that would mean that the quadratic terms cancel out. So the conclusion is that p(x) is the constant polynomial 1, hence degree 0.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let's check the coefficients of x² in each term. Let me denote each term as T1, T2, T3 for simplicity:
T1 = [(x - b)(x - c)] / [(a - b)(a - c)] T2 = [(x - a)(x - c)] / [(b - a)(b - c)] T3 = [(x - a)(x - b)] / [(c - a)(c - b)]

So expanding each numerator:

<prompt_2.1> Expand each term to identify the coefficient of \(x^2\) and determine the degree of the polynomial. </prompt_2.1>
-----

Based on execution_1.2:
But wait, maybe I made a mistake in the Lagrange interpolation part.

<prompt_2.2> Re-evaluate the Lagrange interpolation approach to confirm the degree of the polynomial. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
For T1: (x - b)(x - c) = x² - (b + c)x + bc. So the coefficient of x² is 1. Similarly, T2: (x - a)(x - c) = x² - (a + c)x + ac, so coefficient of x² is 1. T3: (x - a)(x - b) = x² - (a + b)x + ab, coefficient of x² is 1. 

Therefore, each term T1, T2, T3 has a leading term of (1 / denominator) x². So when we add them up, the x² terms will add up as [1/( (a - b)(a - c) ) + 1/( (b - a)(b - c) ) + 1/( (c - a)(c - b) ) ] x². 

Therefore, the coefficient of x² in p(x) is the sum of those three fractions. If that sum is zero, then the x² terms cancel out, and the degree would be less than 2. Otherwise, it would be 2.
</execution_2.1>

<execution_2.2>
Let me think again: If we have three points, and we construct the Lagrange interpolating polynomial, it should be a quadratic polynomial (degree at most 2) that passes through those three points. However, in this case, all three points are (a, 1), (b, 1), (c, 1). So the polynomial that passes through these three points is indeed the constant polynomial 1, because all the y-values are 1. 

But Lagrange interpolation through three points should give a quadratic, but if all the y-values are the same, then the quadratic and linear terms must cancel out, leaving just the constant term. So in this case, the interpolating polynomial is indeed the constant polynomial 1. 

Therefore, p(x) is equal to 1 for all x, which is a constant function, hence degree 0.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So let's compute that coefficient. Let me denote the coefficient as K:

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

Let's compute K step by step.

First, note that (b - a) = -(a - b), and (c - b) = -(b - c). Similarly, (c - a) = -(a - c). So perhaps we can rewrite the denominators to have a common structure.

<prompt_3.1> Simplify the expression for the coefficient \(K\) by rewriting the denominators. </prompt_3.1>
-----

Based on execution_2.2:
But wait, but in our case, p(x) is written as the sum of those three Lagrange basis polynomials multiplied by 1. So the interpolating polynomial for the points (a,1), (b,1), (c,1) is indeed 1, which is a constant.

<prompt_3.2> Reaffirm the conclusion that the polynomial is constant and determine its degree. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Let's rewrite each denominator:

Denominator of T1: (a - b)(a - c) = (a - b)(a - c)

Denominator of T2: (b - a)(b - c) = (-1)(a - b)(-1)(c - b) = (a - b)(c - b). Wait, let me check that again:

Wait, (b - a) = -(a - b), and (b - c) = -(c - b). So (b - a)(b - c) = [-(a - b)][-(c - b)] = (a - b)(c - b). So denominator of T2 is (a - b)(c - b).

Similarly, denominator of T3: (c - a)(c - b) = (-1)(a - c)(-1)(b - c) = (a - c)(b - c). Wait:

(c - a) = -(a - c), (c - b) = -(b - c), so product is (-1)^2 (a - c)(b - c) = (a - c)(b - c). So denominator of T3 is (a - c)(b - c).

Therefore, K can be written as:

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

Hmm, perhaps we can factor out 1/(a - b) from the first two terms? Let's see:

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

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

Wait, maybe another approach. Let's get a common denominator for all three terms. The denominators are (a - b)(a - c), (a - b)(c - b), and (a - c)(b - c). Let's note that (c - b) = -(b - c), so the second denominator is (a - b)(-1)(b - c) = - (a - b)(b - c). Similarly, the third denominator is (a - c)(b - c) = (-1)(c - a)(b - c).

But perhaps the common denominator would be (a - b)(a - c)(b - c). Let's try expressing each term with this common denominator.

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

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

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

Therefore, combining all three terms over the common denominator:

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

Let me compute that:

First term: (b - c)

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

Third term: (a - b)

Adding them up:

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

Simplify term by term:

b - c - a + c + a - b

Let's rearrange:

b - b + (-c + c) + (-a + a) = 0 + 0 + 0 = 0

Wait, that sums to zero? So the numerator is zero, which means K = 0 / [common denominator] = 0.
</execution_3.1>

<execution_3.2>
Therefore, all the x² and x terms must cancel out, leaving 1. Therefore, p(x) = 1, so it's a constant polynomial, degree 0. 

But let me verify this by expanding one of the terms. Let's take T1:

[(x - b)(x - c)] / [(a - b)(a - c)] = [x² - (b + c)x + bc] / [ (a - b)(a - c) ]

Similarly for T2 and T3. Then, when we add them up, the x² terms will be:

[1 / ( (a - b)(a - c) ) + 1 / ( (b - a)(b - c) ) + 1 / ( (c - a)(c - b) ) ] x²

If that coefficient K is zero, then there's no x² term. Let's compute K.

Compute K:

First, let's write all denominators in terms of (a - b), (a - c), etc.:

Denominator of T1: (a - b)(a - c)

Denominator of T2: (b - a)(b - c) = (-1)(a - b)(-1)(c - b) = (a - b)(c - b)

Wait, let's check:

(b - a) = -(a - b)

(b - c) = -(c - b)

So (b - a)(b - c) = [-(a - b)][-(c - b)] = (a - b)(c - b)

Similarly, denominator of T3: (c - a)(c - b) = [-(a - c)][-(b - c)] = (a - c)(b - c)

Wait, let's verify:

(c - a) = -(a - c)

(c - b) = -(b - c)

So (c - a)(c - b) = [-(a - c)][-(b - c)] = (a - c)(b - c)

So denominators:

T1: (a - b)(a - c)

T2: (a - b)(c - b)

T3: (a - c)(b - c)

Hmm, perhaps we can factor out the negatives. Let's see.

Alternatively, maybe we can factor the denominators as follows:

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

D2 = (b - a)(b - c) = (-1)(a - b)(-1)(c - b) = (a - b)(c - b)

D3 = (c - a)(c - b) = (-1)(a - c)(-1)(b - c) = (a - c)(b - c)

So K = 1/D1 + 1/D2 + 1/D3 = 1/[ (a - b)(a - c) ] + 1/[ (a - b)(c - b) ] + 1/[ (a - c)(b - c) ]

Let me factor out 1/(a - b) from the first two terms:

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

Compute the term inside the brackets:

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

Simplify numerator:

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

So the term becomes (a - b) / [ (a - c)(c - b) ]

Therefore, the first part of K is:

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

So now, K = 1 / [ (a - c)(c - b) ) ] + 1 / [ (a - c)(b - c) ) ]

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

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

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

So adding them together:

-1 / [ (a - c)(b - c) ) ] + 1 / [ (a - c)(b - c) ) ] = 0

Therefore, K = 0.

Therefore, the coefficient of x² is zero.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
That suggests that the x² terms cancel out.
-----

Based on execution_3.2:
Thus, the highest degree term is x. Wait, but if K is zero, then there is no x² term. Then, we need to check the coefficient of x.

<prompt_4.1> Verify the cancellation of the \(x^2\) term and determine the degree of the polynomial. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Let's denote the coefficient of x as L. Let's compute L:

Each term T1, T2, T3 has a linear term. Let's expand each numerator:

T1: (x - b)(x - c) = x² - (b + c)x + bc. So coefficient of x is -(b + c)

T2: (x - a)(x - c) = x² - (a + c)x + ac. Coefficient of x is -(a + c)

T3: (x - a)(x - b) = x² - (a + b)x + ab. Coefficient of x is -(a + b)

Therefore, when we add the three terms, the coefficient of x will be:

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

Hmm, this looks more complicated. Let's see if this sum is zero or not.

But since the coefficient of x² is zero, as we found earlier, maybe the coefficient of x is also zero? Let's check.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
First, let's write all denominators in terms of positive factors:

<prompt_5.1> Simplify the expression for the coefficient \(L\) by rewriting the denominators. </prompt_5.1>
-----

Based on execution_3.2:
To confirm, let's pick specific values for a, b, c and compute p(x) numerically. Let's choose a=1, b=2, c=3.

<prompt_5.2> Numerically verify the degree of the polynomial by evaluating \(p(x)\) for specific values of \(a, b, c\). </prompt_5.2>
</Plan_5>

<Execution_5>
<execution_5.1>
Denominator of T1: (a - b)(a - c)

Denominator of T2: (b - a)(b - c) = (-1)(a - b)(-1)(c - b) = (a - b)(c - b)

Denominator of T3: (c - a)(c - b) = (-1)(a - c)(-1)(b - c) = (a - c)(b - c)

So denominators:

T1: (a - b)(a - c)

T2: (a - b)(c - b)

T3: (a - c)(b - c)

Note that (c - b) = -(b - c) and (b - c) = -(c - b). So we can write all denominators in terms of (a - b), (a - c), and (b - c) with appropriate signs.

But perhaps instead of dealing with signs, let's factor out the negatives. Let's note that:

Denominator of T2: (a - b)(c - b) = (a - b)(-(b - c)) = - (a - b)(b - c)

Similarly, denominator of T3: (a - c)(b - c) = (a - c)(b - c)

So let's rewrite the coefficients:

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

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

Third term: - (a + b) / [ (a - c)(b - c) ]

So L becomes:

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

Let's factor out 1/(a - b) from the first two terms and 1/(a - c)(b - c) from the third term:

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

Wait, perhaps another approach. Let's get a common denominator for all three terms. The denominators are:

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

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

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

So the common denominator is (a - b)(a - c)(b - c). Let's express each term with this denominator.

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

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

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

Therefore, combining all three terms:

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

Let's compute each part step by step.

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

So adding all three parts together:

(-b² + c²) + (a² - c²) + (-a² + b²)

Let's simplify term by term:

- b² + c² + a² - c² - a² + b²

Combine like terms:

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

So the numerator is zero. Therefore, L = 0 / [common denominator] = 0.
</execution_5.1>

<execution_5.2>
Then:

T1 = [(x - 2)(x - 3)] / [(1 - 2)(1 - 3)] = (x² - 5x + 6) / [(-1)(-2)] = (x² -5x +6)/2

T2 = [(x -1)(x -3)] / [(2 -1)(2 -3)] = (x² -4x +3) / [(1)(-1)] = (x² -4x +3)/(-1) = -x² +4x -3

T3 = [(x -1)(x -2)] / [(3 -1)(3 -2)] = (x² -3x +2)/(2*1) = (x² -3x +2)/2

Now, sum them up:

T1 + T2 + T3 = (x² -5x +6)/2 + (-x² +4x -3) + (x² -3x +2)/2

Combine the terms:

First, combine the x² terms:

(1/2)x² - x² + (1/2)x² = (1/2 -1 +1/2)x² = (0)x²

Then the x terms:

(-5/2)x +4x + (-3/2)x = [ (-5/2) +4 + (-3/2) ]x = [ (-5/2 -3/2) +4 ]x = [ -4 +4 ]x = 0x

Then the constant terms:

6/2 -3 +2/2 = 3 -3 +1 =1

So total p(x) = 0x² +0x +1 =1. Therefore, p(x)=1, which is a constant polynomial. Hence, degree 0.

Therefore, even through computation, it's confirmed that p(x) is indeed the constant polynomial 1, so degree is 0.
</execution_5.2>
</Execution_5>

<Plan_6>
Based on execution_5.2:
But just to confirm once more, let's try another set of values.

<prompt_6.1> Verify the result further with another set of values for \(a, b, c\). </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Let me choose a=0, b=1, c=2.

Compute each term:

T1 = (x -1)(x -2)/[(0 -1)(0 -2)] = (x² -3x +2)/[(-1)(-2)] = (x² -3x +2)/2

T2 = (x -0)(x -2)/[(1 -0)(1 -2)] = (x(x -2))/(1*(-1)) = -x(x -2) = -x² +2x

T3 = (x -0)(x -1)/[(2 -0)(2 -1)] = x(x -1)/(2*1) = (x² -x)/2

Sum them up:

T1 + T2 + T3 = (x² -3x +2)/2 + (-x² +2x) + (x² -x)/2

Combine x² terms:

(1/2 -1 +1/2)x² = 0x²

x terms:

(-3/2 +2 -1/2)x = (-3/2 +1)x = (-1/2)x

Constants:

2/2 =1

Wait, but according to previous logic, it should be 1. But here, we have (-1/2)x +1. Wait, that contradicts the previous conclusion. Wait, did I compute correctly?

Wait, let's recompute:

Compute T1: (x² -3x +2)/2

T2: -x² +2x

T3: (x² -x)/2

Adding them:

T1 + T2 + T3 = (x²/2 -3x/2 +1) + (-x² +2x) + (x²/2 -x/2)

Combine x² terms:

(1/2)x² - x² + (1/2)x² = (1/2 -1 +1/2)x² = 0x²

x terms:

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

Constants:

1 from T1.

So total p(x) = 0x² +0x +1 =1. So indeed, even with a=0, b=1, c=2, p(x)=1. So previous calculation must have an error.

Wait, in my calculation above, I must have miscalculated. Let me check again:

T1: (x² -3x +2)/2

T2: -x² +2x

T3: (x² -x)/2

Adding the x terms:

From T1: -3x/2

From T2: +2x

From T3: -x/2

Total x terms: (-3/2 +2 -1/2)x = (-3/2 -1/2 +2)x = (-2 +2)x =0x

Constants:

From T1: 2/2=1

So yes, total is 1. So previous miscalculation was my error. So indeed, even with a=0, b=1, c=2, p(x)=1. 

Therefore, this confirms that p(x) is indeed the constant polynomial 1, regardless of the distinct values of a, b, c. Therefore, the degree is 0.

But let me just check another case where maybe the cancellation isn't so obvious. Let's take a= -1, b=0, c=1.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
Compute each term:

T1 = (x -0)(x -1)/[(-1 -0)(-1 -1)] = x(x -1)/[(-1)(-2)] = x(x -1)/2

T2 = (x -(-1))(x -1)/[(0 -(-1))(0 -1)] = (x +1)(x -1)/[(1)(-1)] = (x² -1)/(-1) = -x² +1

T3 = (x -(-1))(x -0)/[(1 -(-1))(1 -0)] = (x +1)x/[2*1] = (x² +x)/2

Sum them up:

T1 + T2 + T3 = [x(x -1)/2] + (-x² +1) + [(x² +x)/2]

Expand T1: (x² -x)/2

So total:

(x² -x)/2 -x² +1 + (x² +x)/2

Combine x² terms:

(1/2)x² -x² + (1/2)x² = (1/2 -1 +1/2)x² =0x²

x terms:

(-x)/2 +x/2 = ( (-1 +1)/2 )x =0x

Constants:

1

Therefore, again p(x)=1. So regardless of the values, as long as a, b, c are distinct, the polynomial p(x) reduces to 1. Therefore, the degree is 0.
**Final Answer**
The degree of the polynomial is \boxed{0}.
</Plan_7>

<Execution_7>
No prompt to execute.
</Execution_7>

<Final_answer>
Therefore, the degree of the polynomial $p(x)$ is $\boxed{0}$.
</Final_answer>
</think>