<Question>
Suppose that all four of the numbers \[3 - 2\sqrt{2}, \; -3-2\sqrt{2}, \; 1+\sqrt{7}, \; 1-\sqrt{7}\]are roots of the same nonzero polynomial with rational coefficients. What is the smallest possible degree of the polynomial?
</Question>

<Plan_1>
Okay, let's see. The problem states that all four numbers: 3 - 2√2, -3 - 2√2, 1 + √7, and 1 - √7 are roots of the same nonzero polynomial with rational coefficients. We need to find the smallest possible degree of such a polynomial.

Hmm, right. Since the polynomial has rational coefficients, I remember that there's something about irrational roots needing their conjugates to also be roots. Maybe that's the key here. Let me think through this step by step.

First, let's recall that if a polynomial has rational coefficients, then any irrational root must come with its conjugate. This is due to the Conjugate Root Theorem, which says that if a polynomial has rational coefficients and a + b√c is a root, where a, b are rational and c is a square-free integer, then a - b√c must also be a root. So, for each of these irrational roots, we need to check if their conjugates are already included in the list or if we need to add them.

Looking at the given roots:

1. 3 - 2√2: Its conjugate would be 3 + 2√2. Since this isn't listed as a root, the polynomial must include this conjugate as well if we're to have rational coefficients.

<prompt_1.1> Identify the conjugates of the given roots and determine if they are already included or need to be added. </prompt_1.1>
-----

Let me check numerically.

Compute 3 - 2√2 approximately.

<prompt_1.2> Perform numerical verification to check if the roots are distinct. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
2. -3 - 2√2: Its conjugate would be -3 + 2√2. This also isn't in the list, so we need to add this.

3. 1 + √7: Its conjugate is 1 - √7, which is already given. So that pair is okay.

4. 1 - √7: Same as above, already accounted for.

So, for the first two roots (3 - 2√2 and -3 - 2√2), we need their conjugates (3 + 2√2 and -3 + 2√2) to be included as roots. The other two roots (1 + √7 and 1 - √7) are already a conjugate pair, so they don't require additional roots.

Therefore, the minimal polynomial must have all these roots: the original four plus the two conjugates we just identified. That would make a total of six roots.

But wait, hold on. Let me count again.

The original roots are four, but two of them (1 + √7 and 1 - √7) are conjugates, so they form a pair. The other two roots (3 - 2√2 and -3 - 2√2) each need their own conjugates. So:

- For 3 - 2√2, we need 3 + 2√2.
- For -3 - 2√2, we need -3 + 2√2.

Therefore, adding two more roots, making a total of 6 roots. Hence, the minimal polynomial would have degree 6?
</execution_1.1>

<execution_1.2>
√2 is about 1.414, so 2√2 ≈ 2.828. So 3 - 2.828 ≈ 0.172. Similarly, 1 + √7 ≈ 1 + 2.645 ≈ 3.645. These are different. Similarly, -3 - 2√2 ≈ -3 - 2.828 ≈ -5.828, and 1 - √7 ≈ -1.645. All these roots are distinct. So there's no overlap.

Therefore, all six roots are distinct, and each quadratic is distinct and irreducible, so their product is the minimal polynomial of degree 6.

Wait, but let me verify that there isn't a polynomial of lower degree that can contain all four given roots plus the necessary conjugates.

For example, maybe the minimal polynomial for 3 - 2√2 and -3 - 2√2 is of degree 4, as we saw, but then combined with the minimal polynomial for 1 + √7 and 1 - √7, which is degree 2, so total degree 4 + 2 = 6. So that still gives 6.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But wait, let me check if there's any overlap or if some roots can be generated by the same minimal polynomial. Maybe some of these roots are related in a way that their minimal polynomials share factors?

Alternatively, perhaps these roots come from different quadratic factors, each contributing two roots, and then combining them.

Let's try to construct the polynomial step by step.

Starting with 3 - 2√2 and its conjugate 3 + 2√2.

<prompt_2.1> Construct the polynomial step by step using the identified roots. </prompt_2.1>
-----

Based on execution_1.2:
Alternatively, could there be some algebraic relation between these roots that allows for a lower degree? Let's see.

<prompt_2.2> Explore further algebraic relations and confirm the minimal degree. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
The minimal polynomial for these would be (x - (3 - 2√2))(x - (3 + 2√2)) = (x - 3 + 2√2)(x - 3 - 2√2). Multiplying this out: (x - 3)^2 - (2√2)^2 = x² - 6x + 9 - 8 = x² - 6x + 1. So that's a quadratic with rational coefficients.

Similarly, for -3 - 2√2 and its conjugate -3 + 2√2: The minimal polynomial is (x - (-3 - 2√2))(x - (-3 + 2√2)) = (x + 3 + 2√2)(x + 3 - 2√2) = (x + 3)^2 - (2√2)^2 = x² + 6x + 9 - 8 = x² + 6x + 1. Another quadratic with rational coefficients.

Then, the pair 1 + √7 and 1 - √7: Their minimal polynomial is (x - (1 + √7))(x - (1 - √7)) = (x - 1)^2 - (√7)^2 = x² - 2x + 1 - 7 = x² - 2x - 6. Also a quadratic.

So, if we have these three quadratics, each irreducible over the rationals, then the minimal polynomial would be their product, which is degree 2 * 2 * 2 = 8.

Wait, but that contradicts the previous thought of 6 roots. Wait, why?

Because each quadratic corresponds to two roots, so three quadratics give 6 roots, which is a degree 6 polynomial. Wait, but 2*2*2 is 8. Wait, no, the product of three quadratics is degree 6. Each quadratic is degree 2, so 2+2+2=6. Yeah, right, multiplying three quadratics gives a degree 6 polynomial.
</execution_2.1>

<execution_2.2>
For instance, suppose that 3 - 2√2 and -3 - 2√2 are related in some way. Let's check if they satisfy a common quadratic equation.

Take 3 - 2√2 and -3 - 2√2. Let's call them a and b. If we can find a quadratic with rational coefficients that has both a and b as roots, then maybe we can combine them. Let's check if such a quadratic exists.

If a = 3 - 2√2 and b = -3 - 2√2, then sum of roots a + b = (3 - 2√2) + (-3 - 2√2) = -4√2. Product of roots a * b = (3 - 2√2)(-3 - 2√2) = -9 -6√2 +6√2 + (2√2)^2 = -9 + 8 = -1. Wait, so the product is -1, which is rational, but the sum is -4√2, which is irrational. Therefore, the minimal polynomial for a and b would have coefficients that are not rational because the sum is irrational. Therefore, we cannot have a quadratic with rational coefficients that has both a and b as roots. Therefore, they must be in separate quadratics, each with their own conjugates. Hence, the minimal polynomial must have factors corresponding to each conjugate pair, leading to degree 6.

Therefore, the minimal degree is 6.

So, the answer should be 6. But let me double-check.

Wait, the problem states that "all four of the numbers [... ] are roots of the same nonzero polynomial with rational coefficients". It doesn't say that these are the only roots. So the minimal polynomial would be the least common multiple of the minimal polynomials of each of these roots. Since each of these roots requires their conjugates, the minimal polynomial must include all conjugates. Therefore, the minimal polynomial is indeed the product of the three quadratics, which is degree 6. Therefore, the answer is 6.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So the minimal polynomial would be (x² - 6x + 1)(x² + 6x + 1)(x² - 2x - 6). But then, is this irreducible over Q?

<prompt_3.1> Finalize the minimal polynomial and confirm its degree. </prompt_3.1>
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Based on execution_2.2:
But wait, just to be absolutely thorough, let me confirm that none of these quadratics can be factored into a product of lower-degree polynomials with rational coefficients. For example, the quadratic x² - 6x + 1: discriminant is 36 - 4 = 32, which is not a perfect square, so it's irreducible over Q. Similarly, x² + 6x + 1 has discriminant 36 - 4 = 32, same thing.

<prompt_3.2> Check for irreducibility of quadratics and confirm the solution. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Wait, each of the quadratics is irreducible over Q because their roots are irrational and they have rational coefficients, so they can't be factored further into linear terms with rational coefficients. Therefore, multiplying them together gives a polynomial of degree 6, which is the minimal polynomial containing all these roots. Hence, the minimal degree would be 6.

But hold on, let me check if there are any dependencies or if some of these quadratics can be combined or if there's a higher degree factor that includes some of these roots. For example, could the roots 3 - 2√2 and -3 - 2√2 be related in some way that allows them to be part of a higher degree minimal polynomial?

Wait, 3 - 2√2 and -3 - 2√2. Let me see. Suppose we have a polynomial that has both 3 - 2√2 and -3 - 2√2 as roots. If we consider their conjugates as well, 3 + 2√2 and -3 + 2√2, maybe there's a quartic polynomial that can be formed from these four roots. Let's check.

If we take (x - (3 - 2√2))(x - (3 + 2√2))(x - (-3 - 2√2))(x - (-3 + 2√2)). Let's compute this step by step.

First, pair them into the two quadratics we did before: (x² - 6x + 1) and (x² + 6x + 1). Multiplying these two together: (x² - 6x + 1)(x² + 6x + 1) = (x² + 1)^2 - (6x)^2 = x^4 + 2x² + 1 - 36x² = x^4 - 34x² + 1. So that's a quartic. Then, multiplying by the third quadratic (x² - 2x - 6) gives the degree 6 polynomial as before. But is there a way to have a lower degree polynomial that includes all these roots?

Alternatively, is there any overlap between the roots of these quadratics and the other quadratic? For example, could any of the roots from the first two quadratics be related to the roots of the third quadratic?
</execution_3.1>

<execution_3.2>
The quadratic x² - 2x -6: discriminant 4 + 24 = 28, which is not a perfect square. So all three quadratics are irreducible over Q. Therefore, their product is indeed the minimal polynomial, which has degree 6.

Therefore, the smallest possible degree is 6.

**Final Answer**

The smallest possible degree of the polynomial is \boxed{6}.
</execution_3.2>
</Execution_3>

<Final_answer>
Therefore, the smallest possible degree of the polynomial is $\boxed{6}$.
</Final_answer>