Explanation:
Okay, so I need to find the least possible value of |a_{n-1}| for a polynomial with integer coefficients, distinct integer roots, leading coefficient 2, and constant term 66. Let me start by recalling some concepts.

First, since the polynomial has integer coefficients and distinct integer roots, it can be factored as f(x) = 2(x - r_1)(x - r_2)...(x - r_n), where each r_i is a distinct integer. The constant term a_0 is the product of the roots multiplied by the leading coefficient. Wait, actually, the constant term is obtained by multiplying all the roots and then multiplying by the leading coefficient. Let me check that.

If f(x) = a_n(x - r_1)(x - r_2)...(x - r_n), then expanding this, the constant term is a_n * (-1)^n * r_1 * r_2 * ... * r_n. So in our case, a_0 = 2 * (-1)^n * (product of roots). Given that a_0 is 66, so:

2 * (-1)^n * (product of roots) = 66

Which implies that the product of the roots is 66 / [2 * (-1)^n] = 33 / (-1)^n. Since the product of the roots is an integer (because all roots are integers), and 33 is an integer, so this is okay. But since the roots are integers, their product must be an integer, which it is. However, the sign depends on n. Let me note that.

But maybe I can think of the product of the roots as ±33. Because 66 divided by 2 is 33, and then multiplied by (-1)^n. So depending on whether n is even or odd, the product of the roots is either 33 or -33. So the product of the roots is ±33. Let me write that down:

product of roots = ±33.

But since the roots are distinct integers, I need to find all possible sets of distinct integers whose product is ±33, and then for each such set, compute the coefficient a_{n-1}, which is related to the sum of the roots multiplied by the leading coefficient. Wait, let me recall Vieta's formula.

For a polynomial f(x) = a_n x^n + a_{n-1}x^{n-1} + ... + a_0, Vieta's formula tells us that:

a_{n-1} = -a_n * (sum of roots)

Wait, no. Wait, Vieta's formula for the coefficient of x^{n-1} is (-1)^{1} * a_n * (sum of roots). Wait, let me check again.

Actually, Vieta's formula for a monic polynomial (leading coefficient 1) is that the coefficient of x^{n-1} is - (sum of roots). But in our case, the leading coefficient is 2, so the polynomial is 2*(x - r_1)...(x - r_n). So expanding this, the coefficient of x^{n-1} would be 2 * (-1) * (sum of roots). Wait, let me think.

Let me take a simple case. Suppose n=2. Then f(x) = 2(x - r1)(x - r2) = 2x^2 - 2(r1 + r2)x + 2r1r2. So here, a_{n-1} = -2(r1 + r2). So in general, for degree n, the coefficient a_{n-1} is -a_n * (sum of roots). Wait, because if you have f(x) = a_n(x - r1)...(x - rn), then expanding, the coefficient of x^{n-1} is -a_n*(r1 + r2 + ... + rn). Therefore, in our case, since a_n = 2, then a_{n-1} = -2*(sum of roots). Therefore, |a_{n-1}| = 2*|sum of roots|. Therefore, our goal is to find the minimal possible value of 2*|sum of roots|, given that the product of the roots is ±33, and all roots are distinct integers.

Therefore, the problem reduces to: find a set of distinct integers whose product is ±33, and such that the absolute value of their sum is as small as possible. Then multiply that minimal |sum| by 2 to get the minimal |a_{n-1}|.

So first, I need to find all possible sets of distinct integers (with possible positive and negative numbers) whose product is ±33, and then find the one with the minimal |sum|.

But first, let me note that 33 factors into primes as 3 * 11. So the positive divisors of 33 are 1, 3, 11, 33. Similarly, the negative divisors are -1, -3, -11, -33. So the product of the roots is either 33 or -33. Let me consider both cases.

But since the roots are distinct integers, we need to consider all possible factorizations of 33 and -33 into distinct integers. Also, note that the number of roots n can vary. For example, if we take the roots as 1, 3, 11, 33, then n=4. But maybe we can have more roots by including 1 and -1, but then the product would be adjusted accordingly.

Wait, but the product of the roots is fixed as ±33. So if we have more roots, we need to include 1s and -1s, but since the roots must be distinct, we can't have repeated roots. So for example, if we want to have more roots, we need to include factors that multiply to 1 or -1, but since they have to be distinct integers, we can include 1 and -1, but then we need to adjust the other factors accordingly.

But let me think step by step.

First, let me consider the case where the product of the roots is 33. Then, the roots are distinct integers whose product is 33. Similarly, for product -33.

So first, let me list all possible sets of distinct integers whose product is 33. Let me start with positive product first.

Case 1: Product of roots is 33.

Possible factorizations into distinct integers:

First, note that 33 can be written as:

- 33 = 33 (single root, but then n=1, but the polynomial would be degree 1, but the leading coefficient is 2, so f(x) = 2x - 2*33? Wait, but the problem says the roots are distinct integers. If n=1, there's only one root, which is 33. But the problem says "roots are distinct integers", but for a degree 1 polynomial, there's only one root, so it's trivially distinct. However, the problem says "distinct integers", but maybe they mean more than one? Wait, the problem says "its roots are distinct integers". So if the polynomial is degree 1, then it has one root, which is distinct. But maybe the problem allows for n=1? However, the question is about a_{n-1}, which for n=1 would be a_0, but given that a_0 is 66. Wait, but in our case, a_0 is 66, which is given. Let me check.

Wait, if the polynomial is degree n, then a_0 is the constant term. For example, if n=1, then f(x) = 2x + a_0. But in that case, the constant term is a_0 = 2*(-r_1). Wait, earlier we had that a_0 = 2*(-1)^n * product of roots. For n=1, that would be 2*(-1)^1 * r_1 = -2r_1. But given that a_0 is 66, so -2r_1 = 66 => r_1 = -33. So the root is -33, and the polynomial is 2x - 66. Then, a_{n-1} is a_0, which is 66? Wait, no. Wait, for n=1, the polynomial is 2x + a_0. Wait, but Vieta's formula says that for f(x) = a_n x^n + ... + a_0, the constant term is a_n * (-1)^n * product of roots. So for n=1, a_0 = 2 * (-1)^1 * r_1 = -2r_1. Therefore, if a_0 is 66, then -2r_1 = 66 => r_1 = -33. Then, the coefficient a_{n-1} is a_0? Wait, no. For n=1, the polynomial is 2x + a_0. So the coefficients are a_1 = 2, a_0 = 66. Therefore, a_{n-1} is a_0? Wait, n=1, so a_{n-1} is a_0. But the problem says "the least possible value of |a_{n-1}|". So in this case, |a_{n-1}| = |a_0| = 66. But maybe there are other polynomials with higher degree where |a_{n-1}| is smaller. So maybe we need to check all possibilities.

But let's not get ahead of ourselves. Let me first consider all possible factorizations of 33 and -33 into distinct integers, and then compute the sum of roots for each case, then find the minimal |sum|, then multiply by 2 to get |a_{n-1}|.

So starting with product 33.

First, possible factorizations into distinct integers:

Option 1: Single root: 33. Then sum is 33. |sum| = 33. Then |a_{n-1}| = 2*33 = 66.

Option 2: Two roots: 3 and 11. Product is 3*11=33. Sum is 3 + 11 = 14. |sum| = 14. Then |a_{n-1}| = 2*14 = 28.

Option 3: Two roots: 1 and 33. Product is 1*33=33. Sum is 1 + 33 = 34. |sum| = 34. |a_{n-1}| = 68. Worse than 28.

Option 4: Two roots: -1 and -33. Product is (-1)*(-33)=33. Sum is -1 + (-33) = -34. |sum| = 34. Same as above.

Option 5: Two roots: -3 and -11. Product is (-3)*(-11)=33. Sum is -3 + (-11) = -14. |sum| = 14. Same as option 2.

So for two roots, the minimal |sum| is 14, leading to |a_{n-1}|=28.

Option 6: Three roots. Let's think. Since 33 factors into 3*11, but to have three distinct integers, we need to include 1 or -1. Let me try:

For example, 1, 3, 11. Product is 1*3*11=33. Sum is 1 + 3 + 11 = 15. |sum| = 15. |a_{n-1}| = 30. Which is worse than 28.

Alternatively, 1, -1, -33. Product is 1*(-1)*(-33)=33. Sum is 1 + (-1) + (-33) = -33. |sum| = 33. Worse.

Or 1, -3, -11. Product is 1*(-3)*(-11)=33. Sum is 1 - 3 - 11 = -13. |sum| = 13. That's better than 15. So |a_{n-1}| = 2*13=26. That's better than 28. So this is a better candidate.

Wait, so that's a three-root case with sum -13. Let me check that again. 1, -3, -11. Product is 1 * (-3) * (-11) = 33. Yes. Sum is 1 - 3 - 11 = -13. So |sum| is 13. So |a_{n-1}| is 26. That's better.

Another three-root case: -1, 3, -11. Product is (-1)*3*(-11)=33. Sum is -1 + 3 -11 = -9. |sum| = 9. That's even better. So |a_{n-1}| = 18. Wait, that's better. Let me check that.

Wait, roots are -1, 3, -11. Product is (-1)*3*(-11) = 33. Sum is (-1) + 3 + (-11) = -9. So |sum| is 9. So |a_{n-1}| is 2*9=18. That's better. So that's a better candidate.

Another three-root case: -1, -3, 11. Product is (-1)*(-3)*11=33. Sum is -1 -3 +11=7. |sum|=7. So |a_{n-1}|=14. Even better. So that's better.

Wait, so that's sum 7. Let me check again. Roots: -1, -3, 11. Product is (-1)*(-3)*11=33. Sum is (-1) + (-3) + 11 = 7. Yes. So |sum| is 7. So |a_{n-1}| is 14. That's better than 18. So that's better.

Another three-root case: -1, 1, -33. Product is (-1)*1*(-33)=33. Sum is -1 + 1 -33 = -33. |sum|=33. Not good.

How about -1, 1, 3, -11? Wait, that's four roots. Let me check that. Wait, product is (-1)*1*3*(-11)=33. Sum is -1 + 1 + 3 -11 = -8. |sum|=8. So |a_{n-1}|=16. Which is worse than 14. So not better.

Wait, but maybe there are other three-root combinations. Let me think. For example, 3, 11, 1. Wait, that's same as before. What about 3, -1, -11? Wait, that's same as -1, 3, -11. Which we already considered. Sum is -9. So |sum|=9. So not better than 7.

Another three-root case: 1, -1, 33. Product is 1*(-1)*33=-33. Wait, but we are considering product 33. So that's not allowed. So need to have product 33. So maybe other combinations?

How about 3, 1, 11. Product is 3*1*11=33. Sum is 15. Not better.

Alternatively, 3, -1, -11. Which we had before. Sum is -9. So not better than 7.

Wait, so the best three-root case so far is -1, -3, 11 with sum 7. Let me check if there are other three-root combinations. For example, 1, -1, -33. Product is 1*(-1)*(-33)=33. Sum is 1 -1 -33 = -33. Not good. How about 1, 3, -11? Product is 1*3*(-11)=-33. Not allowed. So no. How about 1, -3, 11? Product is 1*(-3)*11=-33. Not allowed. So no. So maybe that's the only three-root case with product 33 and sum 7.

Wait, but maybe with more roots? Let's try four roots. Let me think. For example, including 1, -1, 3, -11. Product is 1*(-1)*3*(-11)=33. Sum is 1 -1 + 3 -11 = -8. |sum|=8. Which is worse than 7. How about 1, -1, -3, 11. Product is 1*(-1)*(-3)*11=33. Sum is 1 -1 -3 +11=8. |sum|=8. Same as above.

Another four-root case: 1, -1, 3, -11. Same as above. How about 1, -1, 1, -33? But roots must be distinct. So can't have two 1s. So invalid.

Alternatively, 1, -1, 3, 11. Product is 1*(-1)*3*11=-33. Not allowed. So need to have product 33. So maybe 1, -1, -3, -11. Product is 1*(-1)*(-3)*(-11)= -33. Not allowed. So no.

Alternatively, 1, -1, 3, -11. Product is 33. Sum is -8. So same as before.

Alternatively, maybe other four roots? Let me think. For example, 1, -1, 1, -33. No, duplicates. How about 1, -1, 3, -11. As above. How about 1, -1, -3, 11. Sum is 8. So same.

Alternatively, maybe 1, -1, 1, -33. Not allowed. So seems like four roots can't get better than |sum|=8. So worse than 7.

What about five roots? Let me think. For example, including 1, -1, 3, -1, ... but duplicates again. Not allowed. So need distinct integers. So maybe 1, -1, 3, -11, something? Let me see. If I have five roots, their product is 33. Let me think. Let me try to include 1, -1, 3, -11, and then another number. Let me compute the product so far: 1*(-1)*3*(-11) = 33. So if I have five roots, then the fifth root must be 1, but that's duplicate. Or maybe another number. Wait, if I have 1, -1, 3, -11, and then another root, say x. Then the product is 1*(-1)*3*(-11)*x = 33*x. But we need the product to be 33, so 33*x = 33 => x=1. But 1 is already a root. So duplicate. Not allowed. Therefore, impossible. Similarly, trying to have five roots would require introducing another factor, but since the product is fixed, we can't. Therefore, five roots is not possible. Similarly, more roots would require introducing more factors, but since the product is fixed, we can't. Therefore, the maximum number of roots is four, but even then, the minimal |sum| is 8, which is worse than 7.

So for product 33, the minimal |sum| is 7, leading to |a_{n-1}|=14. Let me check if there are other possibilities with more roots. Wait, maybe with negative numbers?

Wait, for example, if I take roots as -1, -1, 3, 11. But duplicates again. Not allowed. So no. How about -1, 1, -3, 11. Wait, that's four roots. Product is (-1)*1*(-3)*11=33. Sum is -1 +1 -3 +11=8. Which is same as before.

Alternatively, maybe other combinations. Let me think. For example, -1, -3, 1, 11. Same as above. So same sum. So no improvement.

So for product 33, the minimal |sum| is 7. Now let's check the case where the product of roots is -33. Maybe that gives a better |sum|.

Case 2: Product of roots is -33.

Similarly, we need to find distinct integers whose product is -33. Let me think about possible factorizations.

First, single root: -33. Then sum is -33. |sum|=33. |a_{n-1}|=66. Not better.

Two roots: Let's think. For product -33, possible pairs:

- 1 and -33. Product is -33. Sum is 1 + (-33) = -32. |sum|=32.

- 3 and -11. Product is -33. Sum is 3 + (-11) = -8. |sum|=8.

- -1 and 33. Product is -33. Sum is -1 + 33 = 32. |sum|=32.

- -3 and 11. Product is -33. Sum is -3 + 11 = 8. |sum|=8.

So for two roots, the minimal |sum| is 8, leading to |a_{n-1}|=16. Which is worse than the 14 from the previous case.

Three roots. Let me think. For product -33. Let me try to find three distinct integers.

For example:

- 1, 3, -11. Product is 1*3*(-11)=-33. Sum is 1 + 3 -11 = -7. |sum|=7. So |a_{n-1}|=14. Same as the previous case.

Another three roots: 1, -3, 11. Product is 1*(-3)*11=-33. Sum is 1 -3 +11=9. |sum|=9. Worse.

Another three roots: -1, 3, 11. Product is (-1)*3*11=-33. Sum is -1 +3 +11=13. |sum|=13. Worse.

Another three roots: -1, -3, -11. Product is (-1)*(-3)*(-11)=-33. Sum is -1 -3 -11=-15. |sum|=15. Worse.

Another three roots: -1, 1, 33. Product is (-1)*1*33=-33. Sum is -1 +1 +33=33. |sum|=33. Worse.

Another three roots: -1, -1, 33. Not allowed, duplicates.

How about 1, -1, 33. Product is -33. Sum is 1 -1 +33=33. Same as above.

Another three roots: 1, -1, -33. Product is 1*(-1)*(-33)=33. Not allowed, since we need product -33.

Wait, so the best three roots for product -33 is 1, 3, -11 with sum -7. Which gives |sum|=7, same as the previous case. Therefore, |a_{n-1}|=14. So same as the case when product was 33. So same value.

Now, let's check if there are better possibilities with more roots.

Four roots. Let me think. For product -33. Let me try to find four distinct integers.

For example, 1, -1, 3, -11. Product is 1*(-1)*3*(-11)=33. Not -33. So need to adjust. Let me try 1, -1, 3, 11. Product is 1*(-1)*3*11=-33. Sum is 1 -1 +3 +11=14. |sum|=14. Which is worse than 7.

Another four roots: 1, -1, -3, 11. Product is 1*(-1)*(-3)*11=33. Not -33. So need to adjust. How about 1, -1, -3, -11. Product is 1*(-1)*(-3)*(-11)=-33. Sum is 1 -1 -3 -11=-14. |sum|=14. Same as above.

Another four roots: 1, -1, 3, -11. Product is 1*(-1)*3*(-11)=33. Not -33. So no. How about 1, -1, -3, 11. Product is 1*(-1)*(-3)*11=33. Not -33. So no. How about 1, -1, -1, 33. Duplicates. Not allowed.

Alternatively, 1, -1, 3, -11. Product is 33. Not -33. So no. How about 1, -1, -3, 11. Product is 33. Not -33. So no. How about 1, -1, 3, 11. Product is -33. Sum is 1 -1 +3 +11=14. So same as before.

Another four roots: -1, 1, -3, 11. Same as above. Sum is -1 +1 -3 +11=8. Wait, no. Wait, -1 +1 is 0, -3 +11 is 8. So total sum is 8. Wait, but earlier I thought it was 14. Wait, no. Wait, if the roots are -1, 1, -3, 11, then the product is (-1)*1*(-3)*11=33. Not -33. So that's not allowed. Wait, so if I want product -33, then maybe -1, 1, 3, 11. Product is (-1)*1*3*11=-33. Sum is -1 +1 +3 +11=14. So |sum|=14. So same as before.

Alternatively, maybe other four roots. Let me think. For example, -1, -1, 3, 11. Duplicates. Not allowed. How about -1, 3, 11, 1. Same as above. So no improvement.

Alternatively, maybe more roots? Let me try five roots. For example, 1, -1, 3, -1, ... but duplicates. Not allowed. How about 1, -1, 3, -11, something. Let me compute the product so far: 1*(-1)*3*(-11)=33. So to get product -33, need another root x such that 33*x = -33 => x = -1. But -1 is already a root. So duplicate. Not allowed. Similarly, other combinations would require duplicates. Therefore, five roots is not possible. So four roots is the maximum. And the minimal |sum| for four roots is 8 or 14. So worse than 7.

So for product -33, the minimal |sum| is 7, same as for product 33. Therefore, in both cases, the minimal |sum| is 7, leading to |a_{n-1}|=14. So is 14 the minimal possible value?

Wait, but let me check if there are other factorizations with more roots. For example, maybe with more roots, but with some combination that gives a smaller |sum|.

Wait, for example, if we take roots as 1, -1, 3, -1, ... but duplicates. Not allowed. How about 1, -1, 3, -11, and another root? Wait, as before, that would require another root which duplicates. So no.

Alternatively, maybe using more factors. For example, 1, -1, 3, -11, and 1. But duplicates again. Not allowed.

Alternatively, maybe using 1, -1, 3, -11, and another number. Wait, but as before, the product would be 33*x, so to get -33, x=-1. Which is duplicate. So no.

Alternatively, maybe using more numbers. For example, 1, -1, 3, -11, and 1. No, duplicates. So no.

Alternatively, maybe using fractions? But roots have to be integers. So no.

So seems like the minimal |sum| is 7, leading to |a_{n-1}|=14. But let me check if there are other three-root combinations with product ±33 and smaller |sum|.

Wait, for example, in the case of product -33, we had roots 1, 3, -11 with sum -7. What about other three roots?

How about -1, 3, 11? Product is -1*3*11=-33. Sum is -1 +3 +11=13. |sum|=13. Worse.

How about -1, -3, 11? Product is (-1)*(-3)*11=33. Not -33. So no.

How about -1, -3, -11? Product is (-1)*(-3)*(-11)=-33. Sum is -1 -3 -11=-15. |sum|=15. Worse.

How about 1, -1, -33? Product is 1*(-1)*(-33)=33. Not -33. So no.

How about 1, -3, 11? Product is 1*(-3)*11=-33. Sum is 1 -3 +11=9. |sum|=9. Worse than 7.

How about 1, -1, 3, -11? Wait, that's four roots. Product is 1*(-1)*3*(-11)=33. Not -33. So no.

So seems like the minimal |sum| is indeed 7. Therefore, |a_{n-1}|=14. But let me check if there are other factorizations with more roots. For example, maybe with more than three roots?

Wait, for example, if I take roots as 1, -1, 3, -1, ... but duplicates. Not allowed. How about 1, -1, 3, -11, and another root? As before, that would require another root which duplicates. So no.

Alternatively, maybe using more numbers. For example, 1, -1, 3, -11, and 1. No, duplicates. So no.

Alternatively, maybe using 1, -1, 3, -11, and something else. Wait, but as before, the product would be 33*x, so to get -33, x=-1. Which is duplicate. So no.

Alternatively, maybe using 1, -1, 3, -11, and 1. No. So no improvement.

Therefore, seems like the minimal |sum| is 7, leading to |a_{n-1}|=14. But let me check if there are other three-root combinations with product ±33 and smaller |sum|.

Wait, for example, if I take roots as 1, -1, and something else. Wait, for product 33, if I take 1, -1, and x. Then product is 1*(-1)*x = -x = 33 => x = -33. So roots are 1, -1, -33. Sum is 1 -1 -33 = -33. |sum|=33. Not better.

For product -33, if I take 1, -1, x. Then product is 1*(-1)*x = -x = -33 => x=33. So roots are 1, -1, 33. Sum is 1 -1 +33=33. |sum|=33. Not better.

So no improvement.

Another thought: maybe using more than three roots with some combination. For example, if I take roots as 1, -1, 3, -11, and then another root. Wait, but as before, that would require another root which duplicates. So no.

Alternatively, maybe using 1, -1, 3, -11, and 1. No, duplicates. Not allowed.

Alternatively, maybe using 1, -1, 3, -11, and -1. Again duplicates. Not allowed.

So no improvement.

Wait, but what if we take roots as 1, -1, 3, -11, and another root which is 1? No, duplicates. So no.

Alternatively, maybe using more numbers. For example, 1, -1, 3, -11, and 1. No. Not allowed.

So seems like no improvement. Therefore, the minimal |sum| is 7, leading to |a_{n-1}|=14. But let me check if there are other three-root combinations with product ±33 and smaller |sum|.

Wait, for example, if I take roots as 3, -1, -11. Product is 3*(-1)*(-11)=33. Sum is 3 -1 -11 = -9. |sum|=9. Which is worse than 7.

Or roots as -3, 1, 11. Product is -3*1*11=-33. Sum is -3 +1 +11=9. |sum|=9. Worse than 7.

How about roots as -1, -3, 11. Product is (-1)*(-3)*11=33. Sum is -1 -3 +11=7. Which is the case we had before. So that's the same as the case with product 33. So that's the minimal.

Wait, so in both cases, whether the product is 33 or -33, we can get |sum|=7. Therefore, the minimal |a_{n-1}| is 14. But let me check if there are other possibilities with more roots. For example, maybe with four roots, but with some combination that gives a smaller |sum|.

Wait, for example, if I take roots as -1, 1, -3, 11. Product is (-1)*1*(-3)*11=33. Sum is -1 +1 -3 +11=8. |sum|=8. Which is worse than 7.

Another four roots: -1, 1, 3, -11. Product is (-1)*1*3*(-11)=33. Sum is -1 +1 +3 -11=-8. |sum|=8. Same.

Another four roots: -1, -1, 3, 11. Duplicates. Not allowed.

Another four roots: -1, 3, -11, 1. Same as above.

So no improvement.

Alternatively, maybe with five roots? As discussed before, not possible.

Therefore, seems like 14 is the minimal possible value. But let me check if there are other factorizations with different numbers.

Wait, for example, if I take roots as 1, -1, 3, -11, and then another root. Wait, but as before, that would require another root which duplicates. So no.

Alternatively, maybe using more numbers. For example, 1, -1, 3, -11, and 1. No. Not allowed.

Alternatively, maybe using 1, -1, 3, -11, and something else. Wait, but product would be 33*x. To get product ±33, need x=±1. Which duplicates. So no.

Alternatively, maybe using 1, -1, 3, -11, and 1. No. Not allowed.

So no improvement.

Wait, but let me think again. What if the polynomial has more roots? For example, if we take roots as 1, -1, 3, -11, and another root which is 1. But duplicates. Not allowed. So no.

Alternatively, maybe using 1, -1, 3, -11, and another root which is -1. Again duplicates. Not allowed.

So no. Therefore, seems like 14 is the minimal possible value. However, let me check if there are other three-root combinations with product ±33 and smaller |sum|.

Wait, for example, if I take roots as 1, -1, and something else. Wait, we tried that. For product 33, it's 1, -1, -33. Sum is -33. For product -33, it's 1, -1, 33. Sum is 33. Both worse.

How about roots as 3, -1, -11. Product is 3*(-1)*(-11)=33. Sum is 3 -1 -11=-9. |sum|=9. Worse than 7.

How about roots as -3, 1, 11. Product is -3*1*11=-33. Sum is -3 +1 +11=9. |sum|=9. Worse than 7.

How about roots as -1, -3, 11. Product is (-1)*(-3)*11=33. Sum is -1 -3 +11=7. Which is the case we had. So that's the minimal.

Alternatively, roots as -1, 3, -11. Product is (-1)*3*(-11)=33. Sum is -1 +3 -11=-9. |sum|=9. Worse.

So yes, the minimal |sum| is 7. Therefore, |a_{n-1}|=14. But let me check if there are other factorizations with more roots. For example, maybe with four roots, but with some combination that gives a smaller |sum|.

Wait, for example, if I take roots as -1, 1, -3, 11. Product is (-1)*1*(-3)*11=33. Sum is -1 +1 -3 +11=8. |sum|=8. Worse than 7.

Another four roots: -1, 1, 3, -11. Product is (-1)*1*3*(-11)=33. Sum is -1 +1 +3 -11=-8. |sum|=8. Same.

Another four roots: -1, -1, 3, 11. Duplicates. Not allowed.

Another four roots: -1, -3, 1, 11. Product is (-1)*(-3)*1*11=33. Sum is -1 -3 +1 +11=8. Same as above.

So no improvement.

Therefore, seems like 14 is the minimal possible value. However, let me check if there are other factorizations with different numbers. For example, maybe using 2 as a root? Wait, but 2 is not a factor of 33. Since 33 is 3*11, so if we include 2 as a root, then the product would have a factor of 2, which is not in 33. Therefore, unless we have another root that cancels it out. Wait, but since the product is ±33, which is odd, so all roots must be odd integers. Because if there is an even root, then the product would be even, but 33 is odd. Therefore, all roots must be odd integers. Therefore, 2 cannot be a root. Similarly, 4, 6, etc., cannot be roots. Therefore, all roots must be odd integers. Therefore, the roots can only be ±1, ±3, ±11, ±33, and combinations thereof. Therefore, there are no other roots to consider. Therefore, the previous analysis holds.

Therefore, the minimal |a_{n-1}| is 14. However, let me check once again if there are other three-root combinations. For example, if I take roots as -1, -3, 11 (sum 7) or -1, 3, -11 (sum -9). Wait, but if I take roots as -1, -3, 11, then the product is (-1)*(-3)*11=33. So that's allowed. Similarly, if I take roots as -1, 3, -11, product is (-1)*3*(-11)=33. So that's allowed. So both cases are allowed. So the minimal |sum| is 7. Therefore, |a_{n-1}|=14.

But let me check if there is a case with more roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11, and another root. Wait, but as before, that would require another root which duplicates. So no. Alternatively, maybe with more roots, but with some cancellation. For example, if I take roots as -1, 1, -3, 11, and another root which is 1. But duplicates. Not allowed. So no.

Alternatively, maybe take roots as -1, 1, -3, 11, and -1. Again duplicates. Not allowed.

Therefore, no improvement.

Wait, but what if we take roots as -1, 1, -3, 11, and 1? No, duplicates. Not allowed.

Alternatively, maybe take roots as -1, 1, -3, 11, and something else. But as before, the product would be 33*x, so to get product ±33, x=±1. Which duplicates. So no.

Therefore, I think that 14 is indeed the minimal possible value. However, let me check if there are other three-root combinations with product ±33 and smaller |sum|.

Wait, for example, if I take roots as -1, -1, 33. But duplicates. Not allowed. How about -1, 1, 33. Product is -33. Sum is -1 +1 +33=33. |sum|=33. Worse.

How about -1, 1, -33. Product is 33. Sum is -1 +1 -33=-33. |sum|=33. Worse.

How about -1, 3, 11. Product is -33. Sum is -1 +3 +11=13. |sum|=13. Worse than 7.

How about -1, -3, -11. Product is -33. Sum is -1 -3 -11=-15. |sum|=15. Worse.

How about 1, -3, 11. Product is -33. Sum is 1 -3 +11=9. |sum|=9. Worse.

So no improvement. Therefore, I think that 14 is indeed the minimal possible value. However, let me check if there are other factorizations with more roots. For example, if I take roots as -1, 1, -3, 11, and another root which is 1. But duplicates. Not allowed. So no.

Alternatively, maybe take roots as -1, 1, -3, 11, and -1. Again duplicates. Not allowed.

Therefore, I think that the minimal |a_{n-1}| is 14. However, let me check once again if there is a case with four roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11. Sum is -1 +1 -3 +11=8. |sum|=8. Which is worse than 7. How about roots as -1, 1, 3, -11. Sum is -1 +1 +3 -11=-8. |sum|=8. Same. How about roots as -1, -1, 3, 11. Duplicates. Not allowed. So no.

Alternatively, roots as -1, 3, -11, 1. Same as above. So no improvement.

Therefore, I think that 14 is the minimal possible value. However, let me check if there is a case with more roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11, and another root which is 1. But duplicates. Not allowed. So no.

Alternatively, maybe take roots as -1, 1, -3, 11, and another root which is -1. Again duplicates. Not allowed.

Therefore, I think that 14 is indeed the minimal possible value. However, let me check if there is a case with four roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11. Sum is 8. If I take roots as -1, 1, -3, 11, and another root which is something else. Wait, but that would require another root which duplicates. So no.

Alternatively, maybe take roots as -1, 1, -3, 11, and 1. No. Not allowed.

Therefore, I think that the minimal |a_{n-1}| is 14. However, let me check if there is a case with three roots where the sum is smaller. For example, if I take roots as -1, -3, 11 (sum 7) or -1, 3, -11 (sum -9). Is there a way to get a smaller |sum|?

For example, if I take roots as -1, -1, 33. But duplicates. Not allowed. How about -1, -3, 11. Sum is 7. If I take roots as -1, -3, 11, and another root. Wait, but that would require another root which duplicates. So no.

Alternatively, maybe take roots as -1, -3, 11, and 1. Then product is (-1)*(-3)*11*1=33. Sum is -1 -3 +11 +1=8. |sum|=8. Worse than 7.

Therefore, no improvement.

Therefore, I think that the minimal |a_{n-1}| is 14. However, let me check if there is a case with more roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11, and another root which is something else. Wait, but as before, that would require another root which duplicates. So no.

Alternatively, maybe take roots as -1, 1, -3, 11, and 1. No. Not allowed.

Therefore, I think that 14 is indeed the minimal possible value. However, let me check once again if there are other three-root combinations. For example, if I take roots as -1, -1, 33. Duplicates. Not allowed. How about -1, 1, 33. Product is -33. Sum is -1 +1 +33=33. |sum|=33. Worse.

How about -1, 3, 11. Product is -33. Sum is -1 +3 +11=13. |sum|=13. Worse.

How about -1, -3, 11. Product is 33. Sum is -1 -3 +11=7. Which is the case we had. So that's the minimal.

Therefore, I think that the answer is 14. However, let me check if there is a case with four roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11. Sum is 8. If I take roots as -1, 1, -3, 11, and another root. But duplicates. Not allowed. So no.

Alternatively, maybe take roots as -1, 1, -3, 11, and 1. No. Not allowed.

Therefore, I think that the minimal |a_{n-1}| is 14. However, let me check if there is a case with more roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11, and another root which is 1. No. Not allowed.

Alternatively, maybe take roots as -1, 1, -3, 11, and -1. Again duplicates. Not allowed.

Therefore, I think that 14 is indeed the minimal possible value. Therefore, the answer is 14. However, let me check once again if there is a case with three roots where the sum is smaller. For example, if I take roots as -1, -3, 11 (sum 7) or -1, 3, -11 (sum -9). Is there a way to get a sum with absolute value less than 7?

For example, sum of 0? Let me see. If I can find three distinct integers whose product is ±33 and sum is 0. Let me check.

Suppose sum is 0. Then, the roots are a, b, -a - b. Their product is a*b*(-a - b) = -a*b*(a + b) = ±33. Let me see if there are integers a, b such that this holds.

Let me try small integers. For example, let me take a=1, b=-1. Then product is -1*(-1)*(1 + (-1)) = -1*0=0. Not ±33. Not good.

How about a=1, b=3. Then product is -1*3*(1 + 3) = -3*4=-12. Not 33 or -33.

a=1, b= -3. Product is -1*(-3)*(1 + (-3)) = 3*(-2)=-6. Not.

a=1, b= -11. Product is -1*(-11)*(1 + (-11))=11*(-10)=-110. Not.

a=3, b=-1. Product is -3*(-1)*(3 + (-1))=3*2=6. Not.

a=3, b=-3. Product is -3*(-3)*(3 + (-3))=9*0=0. Not.

a=3, b=-11. Product is -3*(-11)*(3 + (-11))=33*(-8)=-264. Not.

a=1, b= - something. Let me think. Suppose a=1, b= x. Then product is -1*x*(1 + x) = ±33. So -x(1 + x) = ±33. So x(1 + x) = ∓33. Let me solve for integer x.

Case 1: x(1 + x) = -33. So x^2 + x + 33 = 0. Discriminant 1 - 132 = -131. No real solutions.

Case 2: x(1 + x) = 33. So x^2 + x - 33 = 0. Discriminant 1 + 132 = 133. Not a perfect square. Therefore, no integer solutions.

Therefore, no such integers a, b. Therefore, no three roots with sum 0.

How about sum of 1? Let me check. Suppose sum is 1. Then roots are a, b, 1 - a - b. Product is a*b*(1 - a - b) = ±33. Seems complicated. Maybe not worth checking. Similarly for other small sums. It's probably not possible. Therefore, the minimal |sum| is indeed 7.

Therefore, the minimal |a_{n-1}| is 14. Therefore, the answer is \boxed{14}.

**Final Answer**
\boxed{14}

To solve the problem, we are given a polynomial of the form:

$$
f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0
$$

with the following conditions:

- The polynomial has **integer coefficients**.
- All **roots are distinct integers**.
- The **leading coefficient** is $ a_n = 2 $.
- The **constant term** is $ a_0 = 66 $.

We are to find the **least possible value of $ |a_{n-1}| $**.

---

### Step 1: Use Vieta's Formula

From Vieta's formulas, for a polynomial of degree $ n $ with leading coefficient $ a_n $, the coefficient $ a_{n-1} $ is related to the **sum of the roots** as follows:

$$
a_{n-1} = -a_n \cdot (\text{sum of roots})
$$

Given $ a_n = 2 $, we have:

$$
a_{n-1} = -2 \cdot (\text{sum of roots}) \Rightarrow |a_{n-1}| = 2 \cdot |\text{sum of roots}|
$$

So, our goal is to **minimize** $ |a_{n-1}| $, which is equivalent to **minimizing** $ |\text{sum of roots}| $.

---

### Step 2: Use the Constant Term

The constant term $ a_0 $ is given by:

$$
a_0 = a_n \cdot (-1)^n \cdot (\text{product of roots})
$$

Substituting the known values:

$$
66 = 2 \cdot (-1)^n \cdot (\text{product of roots}) \Rightarrow \text{product of roots} = \frac{66}{2 \cdot (-1)^n} = \frac{33}{(-1)^n}
$$

So the **product of the roots** is either $ 33 $ or $ -33 $, depending on whether $ n $ is even or odd.

---

### Step 3: Find Sets of Distinct Integers with Product ±33

We now look for **sets of distinct integers** whose product is either $ 33 $ or $ -33 $, and compute the **sum of the roots** for each such set. Our aim is to find the set with the **smallest absolute sum**.

#### Case 1: Product = 33

We consider all possible sets of **distinct integers** whose product is 33. The prime factorization of 33 is $ 3 \times 11 $, so the possible factorizations include:

- $ \{1, 3, 11\} $: product = 33, sum = $ 1 + 3 + 11 = 15 $
- $ \{-1, -3, 11\} $: product = 33, sum = $ -1 - 3 + 11 = 7 $
- $ \{-1, 3, -11\} $: product = 33, sum = $ -1 + 3 - 11 = -9 $
- $ \{1, -1, -33\} $: product = 33, sum = $ 1 - 1 - 33 = -33 $

The **smallest absolute sum** is $ |7| = 7 $, from the set $ \{-1, -3, 11\} $.

#### Case 2: Product = -33

Similarly, we consider sets of distinct integers whose product is -33. The factorizations include:

- $ \{1, 3, -11\} $: product = -33, sum = $ 1 + 3 - 11 = -7 $
- $ \{-1, 3, 11\} $: product = -33, sum = $ -1 + 3 + 11 = 13 $
- $ \{-1, -3, -11\} $: product = -33, sum = $ -1 - 3 - 11 = -15 $
- $ \{1, -1, 33\} $: product = -33, sum = $ 1 - 1 + 33 = 33 $

The **smallest absolute sum** is again $ |-7| = 7 $, from the set $ \{1, 3, -11\} $.

---

### Step 4: Compute the Minimum $ |a_{n-1}| $

From the above, the **smallest absolute sum of roots** is $ 7 $, and since:

$$
|a_{n-1}| = 2 \cdot |\text{sum of roots}| = 2 \cdot 7 = 14
$$

---

### Final Answer

$$
\boxed{14}
$$

[Assistant]
Explanation:
Okay, so I need to find the least possible value of |a_{n-1}| for a polynomial with integer coefficients, distinct integer roots, leading coefficient 2, and constant term 66. Let me start by recalling some concepts.

First, since the polynomial has integer coefficients and distinct integer roots, it can be factored as f(x) = 2(x - r_1)(x - r_2)...(x - r_n), where each r_i is a distinct integer. The constant term a_0 is the product of the roots multiplied by the leading coefficient. Wait, actually, the constant term is obtained by multiplying all the roots and then multiplying by the leading coefficient. Let me check that.

If f(x) = a_n(x - r_1)(x - r_2)...(x - r_n), then expanding this, the constant term is a_n * (-1)^n * r_1 * r_2 * ... * r_n. So in our case, a_0 = 2 * (-1)^n * (product of roots). Given that a_0 is 66, so:

2 * (-1)^n * (product of roots) = 66

Which implies that the product of the roots is 66 / [2 * (-1)^n] = 33 / (-1)^n. Since the product of the roots is an integer (because all roots are integers), and 33 is an integer, so this is okay. But since the roots are integers, their product must be an integer, which it is. However, the sign depends on n. Let me note that.

But maybe I can think of the product of the roots as ±33. Because 66 divided by 2 is 33, and then multiplied by (-1)^n. So depending on whether n is even or odd, the product of the roots is either 33 or -33. So the product of the roots is ±33. Let me write that down:

product of roots = ±33.

But since the roots are distinct integers, I need to find all possible sets of distinct integers whose product is ±33, and then for each such set, compute the coefficient a_{n-1}, which is related to the sum of the roots multiplied by the leading coefficient. Wait, let me recall Vieta's formula.

For a polynomial f(x) = a_n x^n + a_{n-1}x^{n-1} + ... + a_0, Vieta's formula tells us that:

a_{n-1} = -a_n * (sum of roots)

Wait, no. Wait, Vieta's formula for the coefficient of x^{n-1} is (-1)^{1} * a_n * (sum of roots). Wait, let me check again.

Actually, Vieta's formula for a monic polynomial (leading coefficient 1) is that the coefficient of x^{n-1} is - (sum of roots). But in our case, the leading coefficient is 2, so the polynomial is 2*(x - r_1)...(x - r_n). So expanding this, the coefficient of x^{n-1} would be 2 * (-1) * (sum of roots). Wait, let me think.

Let me take a simple case. Suppose n=2. Then f(x) = 2(x - r1)(x - r2) = 2x^2 - 2(r1 + r2)x + 2r1r2. So here, a_{n-1} = -2(r1 + r2). So in general, for degree n, the coefficient a_{n-1} is -a_n * (sum of roots). Wait, because if you have f(x) = a_n(x - r1)...(x - rn), then expanding, the coefficient of x^{n-1} is -a_n*(r1 + r2 + ... + rn). Therefore, in our case, since a_n = 2, then a_{n-1} = -2*(sum of roots). Therefore, |a_{n-1}| = 2*|sum of roots|. Therefore, our goal is to find the minimal possible value of 2*|sum of roots|, given that the product of the roots is ±33, and all roots are distinct integers.

Therefore, the problem reduces to: find a set of distinct integers whose product is ±33, and such that the absolute value of their sum is as small as possible. Then multiply that minimal |sum| by 2 to get the minimal |a_{n-1}|.

So first, I need to find all possible sets of distinct integers (with possible positive and negative numbers) whose product is ±33, and then find the one with the minimal |sum|.

But first, let me note that 33 factors into primes as 3 * 11. So the positive divisors of 33 are 1, 3, 11, 33. Similarly, the negative divisors are -1, -3, -11, -33. So the product of the roots is either 33 or -33. Let me consider both cases.

But since the roots are distinct integers, we need to consider all possible factorizations of 33 and -33 into distinct integers. Also, note that the number of roots n can vary. For example, if we take the roots as 1, 3, 11, 33, then n=4. But maybe we can have more roots by including 1 and -1, but then the product would be adjusted accordingly.

Wait, but the product of the roots is fixed as ±33. So if we have more roots, we need to include 1s and -1s, but since the roots must be distinct, we can't have repeated roots. So for example, if we want to have more roots, we need to include factors that multiply to 1 or -1, but since they have to be distinct integers, we can include 1 and -1, but then we need to adjust the other factors accordingly.

But let me think step by step.

First, let me consider the case where the product of the roots is 33. Then, the roots are distinct integers whose product is 33. Similarly, for product -33.

So first, let me list all possible sets of distinct integers whose product is 33. Let me start with positive product first.

Case 1: Product of roots is 33.

Possible factorizations into distinct integers:

First, note that 33 can be written as:

- 33 = 33 (single root, but then n=1, but the polynomial would be degree 1, but the leading coefficient is 2, so f(x) = 2x - 2*33? Wait, but the problem says the roots are distinct integers. If n=1, there's only one root, which is 33. But the problem says "roots are distinct integers", but for a degree 1 polynomial, there's only one root, so it's trivially distinct. However, the problem says "distinct integers", but maybe they mean more than one? Wait, the problem says "its roots are distinct integers". So if the polynomial is degree 1, then it has one root, which is distinct. But maybe the problem allows for n=1? However, the question is about a_{n-1}, which for n=1 would be a_0, but given that a_0 is 66. Wait, but in our case, a_0 is 66, which is given. Let me check.

Wait, if the polynomial is degree n, then a_0 is the constant term. For example, if n=1, then f(x) = 2x + a_0. But in that case, the constant term is a_0 = 2*(-r_1). Wait, earlier we had that a_0 = 2*(-1)^n * product of roots. For n=1, that would be 2*(-1)^1 * r_1 = -2r_1. But given that a_0 is 66, so -2r_1 = 66 => r_1 = -33. So the root is -33, and the polynomial is 2x - 66. Then, a_{n-1} is a_0, which is 66? Wait, no. Wait, for n=1, the polynomial is 2x + a_0. Wait, but Vieta's formula says that for f(x) = a_n x^n + ... + a_0, the constant term is a_n * (-1)^n * product of roots. So for n=1, a_0 = 2 * (-1)^1 * r_1 = -2r_1. Therefore, if a_0 is 66, then -2r_1 = 66 => r_1 = -33. Then, the coefficient a_{n-1} is a_0? Wait, no. For n=1, the polynomial is 2x + a_0. So the coefficients are a_1 = 2, a_0 = 66. Therefore, a_{n-1} is a_0? Wait, n=1, so a_{n-1} is a_0. But the problem says "the least possible value of |a_{n-1}|". So in this case, |a_{n-1}| = |a_0| = 66. But maybe there are other polynomials with higher degree where |a_{n-1}| is smaller. So maybe we need to check all possibilities.

But let's not get ahead of ourselves. Let me first consider all possible factorizations of 33 and -33 into distinct integers, and then compute the sum of roots for each case, then find the minimal |sum|, then multiply by 2 to get |a_{n-1}|.

So starting with product 33.

First, possible factorizations into distinct integers:

Option 1: Single root: 33. Then sum is 33. |sum| = 33. Then |a_{n-1}| = 2*33 = 66.

Option 2: Two roots: 3 and 11. Product is 3*11=33. Sum is 3 + 11 = 14. |sum| = 14. Then |a_{n-1}| = 2*14 = 28.

Option 3: Two roots: 1 and 33. Product is 1*33=33. Sum is 1 + 33 = 34. |sum| = 34. |a_{n-1}| = 68. Worse than 28.

Option 4: Two roots: -1 and -33. Product is (-1)*(-33)=33. Sum is -1 + (-33) = -34. |sum| = 34. Same as above.

Option 5: Two roots: -3 and -11. Product is (-3)*(-11)=33. Sum is -3 + (-11) = -14. |sum| = 14. Same as option 2.

So for two roots, the minimal |sum| is 14, leading to |a_{n-1}|=28.

Option 6: Three roots. Let's think. Since 33 factors into 3*11, but to have three distinct integers, we need to include 1 or -1. Let me try:

For example, 1, 3, 11. Product is 1*3*11=33. Sum is 1 + 3 + 11 = 15. |sum| = 15. |a_{n-1}| = 30. Which is worse than 28.

Alternatively, 1, -1, -33. Product is 1*(-1)*(-33)=33. Sum is 1 + (-1) + (-33) = -33. |sum| = 33. Worse.

Or 1, -3, -11. Product is 1*(-3)*(-11)=33. Sum is 1 - 3 - 11 = -13. |sum| = 13. That's better than 15. So |a_{n-1}| = 2*13=26. That's better than 28. So this is a better candidate.

Wait, so that's a three-root case with sum -13. Let me check that again. 1, -3, -11. Product is 1 * (-3) * (-11) = 33. Yes. Sum is 1 - 3 - 11 = -13. So |sum| is 13. So |a_{n-1}| is 26. That's better.

Another three-root case: -1, 3, -11. Product is (-1)*3*(-11)=33. Sum is -1 + 3 -11 = -9. |sum| = 9. That's even better. So |a_{n-1}| = 18. Wait, that's better. Let me check that.

Wait, roots are -1, 3, -11. Product is (-1)*3*(-11) = 33. Sum is (-1) + 3 + (-11) = -9. So |sum| is 9. So |a_{n-1}| is 2*9=18. That's better. So that's a better candidate.

Another three-root case: -1, -3, 11. Product is (-1)*(-3)*11=33. Sum is -1 -3 +11=7. |sum|=7. So |a_{n-1}|=14. Even better. So that's better.

Wait, so that's sum 7. Let me check again. Roots: -1, -3, 11. Product is (-1)*(-3)*11=33. Sum is (-1) + (-3) + 11 = 7. Yes. So |sum| is 7. So |a_{n-1}| is 14. That's better than 18. So that's better.

Another three-root case: -1, 1, -33. Product is (-1)*1*(-33)=33. Sum is -1 + 1 -33 = -33. |sum|=33. Not good.

How about -1, 1, 3, -11? Wait, that's four roots. Let me check that. Wait, product is (-1)*1*3*(-11)=33. Sum is -1 + 1 + 3 -11 = -8. |sum|=8. So |a_{n-1}|=16. Which is worse than 14. So not better.

Wait, but maybe there are other three-root combinations. Let me think. For example, 3, 11, 1. Wait, that's same as before. What about 3, -1, -11? Wait, that's same as -1, 3, -11. Which we already considered. Sum is -9. So |sum|=9. So not better than 7.

Another three-root case: 1, -1, 33. Product is 1*(-1)*33=-33. Wait, but we are considering product 33. So that's not allowed. So need to have product 33. So maybe other combinations?

How about 3, 1, 11. Product is 3*1*11=33. Sum is 15. Not better.

Alternatively, 3, -1, -11. Which we had before. Sum is -9. So not better than 7.

Wait, so the best three-root case so far is -1, -3, 11 with sum 7. Let me check if there are other three-root combinations. For example, 1, -1, -33. Product is 1*(-1)*(-33)=33. Sum is 1 -1 -33 = -33. Not good. How about 1, 3, -11? Product is 1*3*(-11)=-33. Not allowed. So no. How about 1, -3, 11? Product is 1*(-3)*11=-33. Not allowed. So no. So maybe that's the only three-root case with product 33 and sum 7.

Wait, but maybe with more roots? Let's try four roots. Let me think. For example, including 1, -1, 3, -11. Product is 1*(-1)*3*(-11)=33. Sum is 1 -1 + 3 -11 = -8. |sum|=8. Which is worse than 7. How about 1, -1, -3, 11. Product is 1*(-1)*(-3)*11=33. Sum is 1 -1 -3 +11=8. |sum|=8. Same as above.

Another four-root case: 1, -1, 3, -11. Same as above. How about 1, -1, 1, -33? But roots must be distinct. So can't have two 1s. So invalid.

Alternatively, 1, -1, 3, 11. Product is 1*(-1)*3*11=-33. Not allowed. So need to have product 33. So maybe 1, -1, -3, -11. Product is 1*(-1)*(-3)*(-11)= -33. Not allowed. So no.

Alternatively, 1, -1, 3, -11. Product is 33. Sum is -8. So same as before.

Alternatively, maybe other four roots? Let me think. For example, 1, -1, 1, -33. No, duplicates. How about 1, -1, 3, -11. As above. How about 1, -1, -3, 11. Sum is 8. So same.

Alternatively, maybe 1, -1, 1, -33. Not allowed. So seems like four roots can't get better than |sum|=8. So worse than 7.

What about five roots? Let me think. For example, including 1, -1, 3, -1, ... but duplicates again. Not allowed. So need distinct integers. So maybe 1, -1, 3, -11, something? Let me see. If I have five roots, their product is 33. Let me think. Let me try to include 1, -1, 3, -11, and then another number. Let me compute the product so far: 1*(-1)*3*(-11) = 33. So if I have five roots, then the fifth root must be 1, but that's duplicate. Or maybe another number. Wait, if I have 1, -1, 3, -11, and then another root, say x. Then the product is 1*(-1)*3*(-11)*x = 33*x. But we need the product to be 33, so 33*x = 33 => x=1. But 1 is already a root. So duplicate. Not allowed. Therefore, impossible. Similarly, trying to have five roots would require introducing another factor, but since the product is fixed, we can't. Therefore, five roots is not possible. Similarly, more roots would require introducing more factors, but since the product is fixed, we can't. Therefore, the maximum number of roots is four, but even then, the minimal |sum| is 8, which is worse than 7.

So for product 33, the minimal |sum| is 7, leading to |a_{n-1}|=14. Let me check if there are other possibilities with more roots. Wait, maybe with negative numbers?

Wait, for example, if I take roots as -1, -1, 3, 11. But duplicates again. Not allowed. So no. How about -1, 1, -3, 11. Wait, that's four roots. Product is (-1)*1*(-3)*11=33. Sum is -1 +1 -3 +11=8. Which is same as before.

Alternatively, maybe other combinations. Let me think. For example, -1, -3, 1, 11. Same as above. So same sum. So no improvement.

So for product 33, the minimal |sum| is 7. Now let's check the case where the product of roots is -33. Maybe that gives a better |sum|.

Case 2: Product of roots is -33.

Similarly, we need to find distinct integers whose product is -33. Let me think about possible factorizations.

First, single root: -33. Then sum is -33. |sum|=33. |a_{n-1}|=66. Not better.

Two roots: Let's think. For product -33, possible pairs:

- 1 and -33. Product is -33. Sum is 1 + (-33) = -32. |sum|=32.

- 3 and -11. Product is -33. Sum is 3 + (-11) = -8. |sum|=8.

- -1 and 33. Product is -33. Sum is -1 + 33 = 32. |sum|=32.

- -3 and 11. Product is -33. Sum is -3 + 11 = 8. |sum|=8.

So for two roots, the minimal |sum| is 8, leading to |a_{n-1}|=16. Which is worse than the 14 from the previous case.

Three roots. Let me think. For product -33. Let me try to find three distinct integers.

For example:

- 1, 3, -11. Product is 1*3*(-11)=-33. Sum is 1 + 3 -11 = -7. |sum|=7. So |a_{n-1}|=14. Same as the previous case.

Another three roots: 1, -3, 11. Product is 1*(-3)*11=-33. Sum is 1 -3 +11=9. |sum|=9. Worse.

Another three roots: -1, 3, 11. Product is (-1)*3*11=-33. Sum is -1 +3 +11=13. |sum|=13. Worse.

Another three roots: -1, -3, -11. Product is (-1)*(-3)*(-11)=-33. Sum is -1 -3 -11=-15. |sum|=15. Worse.

Another three roots: -1, 1, 33. Product is (-1)*1*33=-33. Sum is -1 +1 +33=33. |sum|=33. Worse.

Another three roots: -1, -1, 33. Not allowed, duplicates.

How about 1, -1, 33. Product is -33. Sum is 1 -1 +33=33. Same as above.

Another three roots: 1, -1, -33. Product is 1*(-1)*(-33)=33. Not allowed, since we need product -33.

Wait, so the best three roots for product -33 is 1, 3, -11 with sum -7. Which gives |sum|=7, same as the previous case. Therefore, |a_{n-1}|=14. So same as the case when product was 33. So same value.

Now, let's check if there are better possibilities with more roots.

Four roots. Let me think. For product -33. Let me try to find four distinct integers.

For example, 1, -1, 3, -11. Product is 1*(-1)*3*(-11)=33. Not -33. So need to adjust. Let me try 1, -1, 3, 11. Product is 1*(-1)*3*11=-33. Sum is 1 -1 +3 +11=14. |sum|=14. Which is worse than 7.

Another four roots: 1, -1, -3, 11. Product is 1*(-1)*(-3)*11=33. Not -33. So need to adjust. How about 1, -1, -3, -11. Product is 1*(-1)*(-3)*(-11)=-33. Sum is 1 -1 -3 -11=-14. |sum|=14. Same as above.

Another four roots: 1, -1, 3, -11. Product is 1*(-1)*3*(-11)=33. Not -33. So no. How about 1, -1, -3, 11. Product is 1*(-1)*(-3)*11=33. Not -33. So no. How about 1, -1, -1, 33. Duplicates. Not allowed.

Alternatively, 1, -1, 3, -11. Product is 33. Not -33. So no. How about 1, -1, -3, 11. Product is 33. Not -33. So no. How about 1, -1, 3, 11. Product is -33. Sum is 1 -1 +3 +11=14. So same as before.

Another four roots: -1, 1, -3, 11. Same as above. Sum is -1 +1 -3 +11=8. Wait, no. Wait, -1 +1 is 0, -3 +11 is 8. So total sum is 8. Wait, but earlier I thought it was 14. Wait, no. Wait, if the roots are -1, 1, -3, 11, then the product is (-1)*1*(-3)*11=33. Not -33. So that's not allowed. Wait, so if I want product -33, then maybe -1, 1, 3, 11. Product is (-1)*1*3*11=-33. Sum is -1 +1 +3 +11=14. So |sum|=14. So same as before.

Alternatively, maybe other four roots. Let me think. For example, -1, -1, 3, 11. Duplicates. Not allowed. How about -1, 3, 11, 1. Same as above. So no improvement.

Alternatively, maybe more roots? Let me try five roots. For example, 1, -1, 3, -1, ... but duplicates. Not allowed. How about 1, -1, 3, -11, something. Let me compute the product so far: 1*(-1)*3*(-11)=33. So to get product -33, need another root x such that 33*x = -33 => x = -1. But -1 is already a root. So duplicate. Not allowed. Similarly, other combinations would require duplicates. Therefore, five roots is not possible. So four roots is the maximum. And the minimal |sum| for four roots is 8 or 14. So worse than 7.

So for product -33, the minimal |sum| is 7, same as for product 33. Therefore, in both cases, the minimal |sum| is 7, leading to |a_{n-1}|=14. So is 14 the minimal possible value?

Wait, but let me check if there are other factorizations with more roots. For example, maybe with more roots, but with some combination that gives a smaller |sum|.

Wait, for example, if we take roots as 1, -1, 3, -1, ... but duplicates. Not allowed. How about 1, -1, 3, -11, and another root? Wait, as before, that would require another root which duplicates. So no.

Alternatively, maybe using more factors. For example, 1, -1, 3, -11, and 1. But duplicates again. Not allowed.

Alternatively, maybe using 1, -1, 3, -11, and another number. Wait, but as before, the product would be 33*x, so to get -33, x=-1. Which is duplicate. So no.

Alternatively, maybe using more numbers. For example, 1, -1, 3, -11, and 1. No, duplicates. So no.

Alternatively, maybe using fractions? But roots have to be integers. So no.

So seems like the minimal |sum| is 7, leading to |a_{n-1}|=14. But let me check if there are other three-root combinations with product ±33 and smaller |sum|.

Wait, for example, in the case of product -33, we had roots 1, 3, -11 with sum -7. What about other three roots?

How about -1, 3, 11? Product is -1*3*11=-33. Sum is -1 +3 +11=13. |sum|=13. Worse.

How about -1, -3, 11? Product is (-1)*(-3)*11=33. Not -33. So no.

How about -1, -3, -11? Product is (-1)*(-3)*(-11)=-33. Sum is -1 -3 -11=-15. |sum|=15. Worse.

How about 1, -1, -33? Product is 1*(-1)*(-33)=33. Not -33. So no.

How about 1, -3, 11? Product is 1*(-3)*11=-33. Sum is 1 -3 +11=9. |sum|=9. Worse than 7.

How about 1, -1, 3, -11? Wait, that's four roots. Product is 1*(-1)*3*(-11)=33. Not -33. So no.

So seems like the minimal |sum| is indeed 7. Therefore, |a_{n-1}|=14. But let me check if there are other factorizations with more roots. For example, maybe with more than three roots?

Wait, for example, if I take roots as 1, -1, 3, -1, ... but duplicates. Not allowed. How about 1, -1, 3, -11, and another root? As before, that would require another root which duplicates. So no.

Alternatively, maybe using more numbers. For example, 1, -1, 3, -11, and 1. No, duplicates. So no.

Alternatively, maybe using 1, -1, 3, -11, and something else. Wait, but as before, the product would be 33*x, so to get -33, x=-1. Which is duplicate. So no.

Alternatively, maybe using 1, -1, 3, -11, and 1. No. So no improvement.

Therefore, seems like the minimal |sum| is 7, leading to |a_{n-1}|=14. But let me check if there are other three-root combinations with product ±33 and smaller |sum|.

Wait, for example, if I take roots as 1, -1, and something else. Wait, for product 33, if I take 1, -1, and x. Then product is 1*(-1)*x = -x = 33 => x = -33. So roots are 1, -1, -33. Sum is 1 -1 -33 = -33. |sum|=33. Not better.

For product -33, if I take 1, -1, x. Then product is 1*(-1)*x = -x = -33 => x=33. So roots are 1, -1, 33. Sum is 1 -1 +33=33. |sum|=33. Not better.

So no improvement.

Another thought: maybe using more than three roots with some combination. For example, if I take roots as 1, -1, 3, -11, and then another root. Wait, but as before, that would require another root which duplicates. So no.

Alternatively, maybe using 1, -1, 3, -11, and 1. No, duplicates. Not allowed.

Alternatively, maybe using 1, -1, 3, -11, and -1. Again duplicates. Not allowed.

So no improvement.

Wait, but what if we take roots as 1, -1, 3, -11, and another root which is 1? No, duplicates. So no.

Alternatively, maybe using more numbers. For example, 1, -1, 3, -11, and 1. No. Not allowed.

So seems like no improvement. Therefore, the minimal |sum| is 7, leading to |a_{n-1}|=14. But let me check if there are other three-root combinations with product ±33 and smaller |sum|.

Wait, for example, if I take roots as 3, -1, -11. Product is 3*(-1)*(-11)=33. Sum is 3 -1 -11 = -9. |sum|=9. Which is worse than 7.

Or roots as -3, 1, 11. Product is -3*1*11=-33. Sum is -3 +1 +11=9. |sum|=9. Worse than 7.

How about roots as -1, -3, 11. Product is (-1)*(-3)*11=33. Sum is -1 -3 +11=7. Which is the case we had before. So that's the same as the case with product 33. So that's the minimal.

Wait, so in both cases, whether the product is 33 or -33, we can get |sum|=7. Therefore, the minimal |a_{n-1}| is 14. But let me check if there are other possibilities with more roots. For example, maybe with four roots, but with some combination that gives a smaller |sum|.

Wait, for example, if I take roots as -1, 1, -3, 11. Product is (-1)*1*(-3)*11=33. Sum is -1 +1 -3 +11=8. |sum|=8. Which is worse than 7.

Another four roots: -1, 1, 3, -11. Product is (-1)*1*3*(-11)=33. Sum is -1 +1 +3 -11=-8. |sum|=8. Same.

Another four roots: -1, -1, 3, 11. Duplicates. Not allowed.

Another four roots: -1, 3, -11, 1. Same as above.

So no improvement.

Alternatively, maybe with five roots? As discussed before, not possible.

Therefore, seems like 14 is the minimal possible value. But let me check if there are other factorizations with different numbers.

Wait, for example, if I take roots as 1, -1, 3, -11, and then another root. Wait, but as before, that would require another root which duplicates. So no.

Alternatively, maybe using more numbers. For example, 1, -1, 3, -11, and 1. No. Not allowed.

Alternatively, maybe using 1, -1, 3, -11, and something else. Wait, but product would be 33*x. To get product ±33, need x=±1. Which duplicates. So no.

Alternatively, maybe using 1, -1, 3, -11, and 1. No. Not allowed.

So no improvement.

Wait, but let me think again. What if the polynomial has more roots? For example, if we take roots as 1, -1, 3, -11, and another root which is 1. But duplicates. Not allowed. So no.

Alternatively, maybe using 1, -1, 3, -11, and another root which is -1. Again duplicates. Not allowed.

So no. Therefore, seems like 14 is the minimal possible value. However, let me check if there are other three-root combinations with product ±33 and smaller |sum|.

Wait, for example, if I take roots as 1, -1, and something else. Wait, we tried that. For product 33, it's 1, -1, -33. Sum is -33. For product -33, it's 1, -1, 33. Sum is 33. Both worse.

How about roots as 3, -1, -11. Product is 3*(-1)*(-11)=33. Sum is 3 -1 -11=-9. |sum|=9. Worse than 7.

How about roots as -3, 1, 11. Product is -3*1*11=-33. Sum is -3 +1 +11=9. |sum|=9. Worse than 7.

How about roots as -1, -3, 11. Product is (-1)*(-3)*11=33. Sum is -1 -3 +11=7. Which is the case we had. So that's the minimal.

Alternatively, roots as -1, 3, -11. Product is (-1)*3*(-11)=33. Sum is -1 +3 -11=-9. |sum|=9. Worse.

So yes, the minimal |sum| is 7. Therefore, |a_{n-1}|=14. But let me check if there are other factorizations with more roots. For example, maybe with four roots, but with some combination that gives a smaller |sum|.

Wait, for example, if I take roots as -1, 1, -3, 11. Product is (-1)*1*(-3)*11=33. Sum is -1 +1 -3 +11=8. |sum|=8. Worse than 7.

Another four roots: -1, 1, 3, -11. Product is (-1)*1*3*(-11)=33. Sum is -1 +1 +3 -11=-8. |sum|=8. Same.

Another four roots: -1, -1, 3, 11. Duplicates. Not allowed.

Another four roots: -1, -3, 1, 11. Product is (-1)*(-3)*1*11=33. Sum is -1 -3 +1 +11=8. Same as above.

So no improvement.

Therefore, seems like 14 is the minimal possible value. However, let me check if there are other factorizations with different numbers. For example, maybe using 2 as a root? Wait, but 2 is not a factor of 33. Since 33 is 3*11, so if we include 2 as a root, then the product would have a factor of 2, which is not in 33. Therefore, unless we have another root that cancels it out. Wait, but since the product is ±33, which is odd, so all roots must be odd integers. Because if there is an even root, then the product would be even, but 33 is odd. Therefore, all roots must be odd integers. Therefore, 2 cannot be a root. Similarly, 4, 6, etc., cannot be roots. Therefore, all roots must be odd integers. Therefore, the roots can only be ±1, ±3, ±11, ±33, and combinations thereof. Therefore, there are no other roots to consider. Therefore, the previous analysis holds.

Therefore, the minimal |a_{n-1}| is 14. However, let me check once again if there are other three-root combinations. For example, if I take roots as -1, -3, 11 (sum 7) or -1, 3, -11 (sum -9). Wait, but if I take roots as -1, -3, 11, then the product is (-1)*(-3)*11=33. So that's allowed. Similarly, if I take roots as -1, 3, -11, product is (-1)*3*(-11)=33. So that's allowed. So both cases are allowed. So the minimal |sum| is 7. Therefore, |a_{n-1}|=14.

But let me check if there is a case with more roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11, and another root. Wait, but as before, that would require another root which duplicates. So no. Alternatively, maybe with more roots, but with some cancellation. For example, if I take roots as -1, 1, -3, 11, and another root which is 1. But duplicates. Not allowed. So no.

Alternatively, maybe take roots as -1, 1, -3, 11, and -1. Again duplicates. Not allowed.

Therefore, no improvement.

Wait, but what if we take roots as -1, 1, -3, 11, and 1? No, duplicates. Not allowed.

Alternatively, maybe take roots as -1, 1, -3, 11, and something else. But as before, the product would be 33*x, so to get product ±33, x=±1. Which duplicates. So no.

Therefore, I think that 14 is indeed the minimal possible value. However, let me check if there are other three-root combinations with product ±33 and smaller |sum|.

Wait, for example, if I take roots as -1, -1, 33. But duplicates. Not allowed. How about -1, 1, 33. Product is -33. Sum is -1 +1 +33=33. |sum|=33. Worse.

How about -1, 1, -33. Product is 33. Sum is -1 +1 -33=-33. |sum|=33. Worse.

How about -1, 3, 11. Product is -33. Sum is -1 +3 +11=13. |sum|=13. Worse than 7.

How about -1, -3, -11. Product is -33. Sum is -1 -3 -11=-15. |sum|=15. Worse.

How about 1, -3, 11. Product is -33. Sum is 1 -3 +11=9. |sum|=9. Worse.

So no improvement. Therefore, I think that 14 is indeed the minimal possible value. However, let me check if there are other factorizations with more roots. For example, if I take roots as -1, 1, -3, 11, and another root which is 1. But duplicates. Not allowed. So no.

Alternatively, maybe take roots as -1, 1, -3, 11, and -1. Again duplicates. Not allowed.

Therefore, I think that the minimal |a_{n-1}| is 14. However, let me check once again if there is a case with four roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11. Sum is -1 +1 -3 +11=8. |sum|=8. Which is worse than 7. How about roots as -1, 1, 3, -11. Sum is -1 +1 +3 -11=-8. |sum|=8. Same. How about roots as -1, -1, 3, 11. Duplicates. Not allowed. So no.

Alternatively, roots as -1, 3, -11, 1. Same as above. So no improvement.

Therefore, I think that 14 is the minimal possible value. However, let me check if there is a case with more roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11, and another root which is 1. But duplicates. Not allowed. So no.

Alternatively, maybe take roots as -1, 1, -3, 11, and another root which is -1. Again duplicates. Not allowed.

Therefore, I think that 14 is indeed the minimal possible value. However, let me check if there is a case with four roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11. Sum is 8. If I take roots as -1, 1, -3, 11, and another root which is something else. Wait, but that would require another root which duplicates. So no.

Alternatively, maybe take roots as -1, 1, -3, 11, and 1. No. Not allowed.

Therefore, I think that the minimal |a_{n-1}| is 14. However, let me check if there is a case with three roots where the sum is smaller. For example, if I take roots as -1, -3, 11 (sum 7) or -1, 3, -11 (sum -9). Is there a way to get a smaller |sum|?

For example, if I take roots as -1, -1, 33. But duplicates. Not allowed. How about -1, -3, 11. Sum is 7. If I take roots as -1, -3, 11, and another root. Wait, but that would require another root which duplicates. So no.

Alternatively, maybe take roots as -1, -3, 11, and 1. Then product is (-1)*(-3)*11*1=33. Sum is -1 -3 +11 +1=8. |sum|=8. Worse than 7.

Therefore, no improvement.

Therefore, I think that the minimal |a_{n-1}| is 14. However, let me check if there is a case with more roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11, and another root which is something else. Wait, but as before, that would require another root which duplicates. So no.

Alternatively, maybe take roots as -1, 1, -3, 11, and 1. No. Not allowed.

Therefore, I think that 14 is indeed the minimal possible value. However, let me check once again if there are other three-root combinations. For example, if I take roots as -1, -1, 33. Duplicates. Not allowed. How about -1, 1, 33. Product is -33. Sum is -1 +1 +33=33. |sum|=33. Worse.

How about -1, 3, 11. Product is -33. Sum is -1 +3 +11=13. |sum|=13. Worse.

How about -1, -3, 11. Product is 33. Sum is -1 -3 +11=7. Which is the case we had. So that's the minimal.

Therefore, I think that the answer is 14. However, let me check if there is a case with four roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11. Sum is 8. If I take roots as -1, 1, -3, 11, and another root. But duplicates. Not allowed. So no.

Alternatively, maybe take roots as -1, 1, -3, 11, and 1. No. Not allowed.

Therefore, I think that the minimal |a_{n-1}| is 14. However, let me check if there is a case with more roots where the sum is smaller. For example, if I take roots as -1, 1, -3, 11, and another root which is 1. No. Not allowed.

Alternatively, maybe take roots as -1, 1, -3, 11, and -1. Again duplicates. Not allowed.

Therefore, I think that 14 is indeed the minimal possible value. Therefore, the answer is 14. However, let me check once again if there is a case with three roots where the sum is smaller. For example, if I take roots as -1, -3, 11 (sum 7) or -1, 3, -11 (sum -9). Is there a way to get a sum with absolute value less than 7?

For example, sum of 0? Let me see. If I can find three distinct integers whose product is ±33 and sum is 0. Let me check.

Suppose sum is 0. Then, the roots are a, b, -a - b. Their product is a*b*(-a - b) = -a*b*(a + b) = ±33. Let me see if there are integers a, b such that this holds.

Let me try small integers. For example, let me take a=1, b=-1. Then product is -1*(-1)*(1 + (-1)) = -1*0=0. Not ±33. Not good.

How about a=1, b=3. Then product is -1*3*(1 + 3) = -3*4=-12. Not 33 or -33.

a=1, b= -3. Product is -1*(-3)*(1 + (-3)) = 3*(-2)=-6. Not.

a=1, b= -11. Product is -1*(-11)*(1 + (-11))=11*(-10)=-110. Not.

a=3, b=-1. Product is -3*(-1)*(3 + (-1))=3*2=6. Not.

a=3, b=-3. Product is -3*(-3)*(3 + (-3))=9*0=0. Not.

a=3, b=-11. Product is -3*(-11)*(3 + (-11))=33*(-8)=-264. Not.

a=1, b= - something. Let me think. Suppose a=1, b= x. Then product is -1*x*(1 + x) = ±33. So -x(1 + x) = ±33. So x(1 + x) = ∓33. Let me solve for integer x.

Case 1: x(1 + x) = -33. So x^2 + x + 33 = 0. Discriminant 1 - 132 = -131. No real solutions.

Case 2: x(1 + x) = 33. So x^2 + x - 33 = 0. Discriminant 1 + 132 = 133. Not a perfect square. Therefore, no integer solutions.

Therefore, no such integers a, b. Therefore, no three roots with sum 0.

How about sum of 1? Let me check. Suppose sum is 1. Then roots are a, b, 1 - a - b. Product is a*b*(1 - a - b) = ±33. Seems complicated. Maybe not worth checking. Similarly for other small sums. It's probably not possible. Therefore, the minimal |sum| is indeed 7.

Therefore, the minimal |a_{n-1}| is 14. Therefore, the answer is \boxed{14}.

**Final Answer**
\boxed{14}

To solve the problem, we are given a polynomial of the form:

$$
f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0
$$

with the following conditions:

- The polynomial has **integer coefficients**.
- All **roots are distinct integers**.
- The **leading coefficient** is $ a_n = 2 $.
- The **constant term** is $ a_0 = 66 $.

We are to find the **least possible value of $ |a_{n-1}| $**.

---

### Step 1: Use Vieta's Formula

From Vieta's formulas, for a polynomial of degree $ n $ with leading coefficient $ a_n $, the coefficient $ a_{n-1} $ is related to the **sum of the roots** as follows:

$$
a_{n-1} = -a_n \cdot (\text{sum of roots})
$$

Given $ a_n = 2 $, we have:

$$
a_{n-1} = -2 \cdot (\text{sum of roots}) \Rightarrow |a_{n-1}| = 2 \cdot |\text{sum of roots}|
$$

So, our goal is to **minimize** $ |a_{n-1}| $, which is equivalent to **minimizing** $ |\text{sum of roots}| $.

---

### Step 2: Use the Constant Term

The constant term $ a_0 $ is given by:

$$
a_0 = a_n \cdot (-1)^n \cdot (\text{product of roots})
$$

Substituting the known values:

$$
66 = 2 \cdot (-1)^n \cdot (\text{product of roots}) \Rightarrow \text{product of roots} = \frac{66}{2 \cdot (-1)^n} = \frac{33}{(-1)^n}
$$

So the **product of the roots** is either $ 33 $ or $ -33 $, depending on whether $ n $ is even or odd.

---

### Step 3: Find Sets of Distinct Integers with Product ±33

We now look for **sets of distinct integers** whose product is either $ 33 $ or $ -33 $, and compute the **sum of the roots** for each such set. Our aim is to find the set with the **smallest absolute sum**.

#### Case 1: Product = 33

We consider all possible sets of **distinct integers** whose product is 33. The prime factorization of 33 is $ 3 \times 11 $, so the possible factorizations include:

- $ \{1, 3, 11\} $: product = 33, sum = $ 1 + 3 + 11 = 15 $
- $ \{-1, -3, 11\} $: product = 33, sum = $ -1 - 3 + 11 = 7 $
- $ \{-1, 3, -11\} $: product = 33, sum = $ -1 + 3 - 11 = -9 $
- $ \{1, -1, -33\} $: product = 33, sum = $ 1 - 1 - 33 = -33 $

The **smallest absolute sum** is $ |7| = 7 $, from the set $ \{-1, -3, 11\} $.

#### Case 2: Product = -33

Similarly, we consider sets of distinct integers whose product is -33. The factorizations include:

- $ \{1, 3, -11\} $: product = -33, sum = $ 1 + 3 - 11 = -7 $
- $ \{-1, 3, 11\} $: product = -33, sum = $ -1 + 3 + 11 = 13 $
- $ \{-1, -3, -11\} $: product = -33, sum = $ -1 - 3 - 11 = -15 $
- $ \{1, -1, 33\} $: product = -33, sum = $ 1 - 1 + 33 = 33 $

The **smallest absolute sum** is again $ |-7| = 7 $, from the set $ \{1, 3, -11\} $.

---

### Step 4: Compute the Minimum $ |a_{n-1}| $

From the above, the **smallest absolute sum of roots** is $ 7 $, and since:

$$
|a_{n-1}| = 2 \cdot |\text{sum of roots}| = 2 \cdot 7 = 14
$$

---

### Final Answer

$$
\boxed{14}
$$

Solution:
Since $f(x)$ has integer coefficients, the Integer Root Theorem tells us that all integer roots of $f(x)$ must divide the constant term $66=2\cdot 3\cdot 11$. Thus, the possible integer roots of $f(x)$ are
$$\pm 1,~\pm 2,~\pm 3,~\pm 6,~\pm 11,~\pm 22,~\pm 33,~\pm 66.$$Moreover, since we know that all roots of $f(x)$ are integers, we know that all roots of $f(x)$ appear in the list above.

Now we apply Vieta's formulas. The product of the roots of $f(x)$ is $(-1)^n\cdot\frac{a_0}{a_n}$, which is $33$ or $-33$. Also, the sum of the roots is $-\frac{a_{n-1}}{a_n}=-\frac{a_{n-1}}2$. Thus, in order to minimize $|a_{n-1}|$, we should make the absolute value of the sum of the roots as small as possible, working under the constraint that the product of the roots must be $33$ or $-33$.

We now consider two cases.

Case 1 is that one of $33,-33$ is a root, in which case the only other possible roots are $\pm 1$. In this case, the absolute value of the sum of the roots is at least $32$.

The alternative, Case 2, is that one of $11,-11$ is a root and one of $3,-3$ is a root. Again, the only other possible roots are $\pm 1$, so the absolute value of the sum of the roots is at least $11-3-1=7$, which is better than the result of Case 1. If the absolute value of the sum of the roots is $7$, then $|a_{n-1}|=7|a_n|=7\cdot 2=14$.

Therefore, we have shown that $|a_{n-1}|\ge 14$, and we can check that equality is achieved by
\begin{align*}
f(x) &= 2(x+11)(x-3)(x-1) \\
&= 2x^3+14x^2-82x+66,
\end{align*}which has integer coefficients and integer roots. So the least possible value of $|a_{n-1}|$ is $\boxed{14}$.


Explanation:
Okay, so Tom has this Mr. Potato Head, and he wants to figure out how many different wacky personalities he can make. Let me try to break this down step by step. First, let me list out all the parts he has:

- 3 hairstyles
- 2 sets of eyebrows
- 1 pair of googly eyes
- 2 sets of ears
- 2 sets of lips
- 1 pair of regular shoes
- 1 bonus pair of clown shoes

And the question says a complete personality includes eyebrows, eyes, ears, lips, shoes, and optionally hair. Also, note that you can't mix and match parts, like you can't take one eyebrow from one pair and another from another pair. So each set is considered as a whole, right? So for eyebrows, since there are 2 sets, each set is probably a left and right eyebrow? Or maybe each set is a pair? Wait, the problem says "2 sets of eyebrows". Hmm. Let me think.

Wait, the note says you can't mix and match. For example, you can't take the left eyebrow from one pair and the right from another. So that suggests that each set is a complete pair? Or maybe each set is a single eyebrow? Wait, but if you have two sets of eyebrows, and you can't mix and match, then maybe each set is a pair? Like, each set is a left and right eyebrow? But then if you have two sets, you could choose either set, but not mix them. So for eyebrows, you have two options: either use the first set or the second set. So that would be 2 choices for eyebrows.

But let me check again. The problem says "2 sets of eyebrows". If each set is a pair (left and right), then you can't mix them. So you have two options for eyebrows: choose one set or the other. So that would be 2 choices. Similarly, for the ears: 2 sets of ears. So same thing, each set is a pair? Or maybe each set is a single ear? Wait, but if you have two sets of ears, and you can't mix and match, then maybe each set is a pair? Like, each set is left and right ears? So again, you have two options for ears. Similarly for lips: 2 sets of lips. So again, two options.

But then, for the eyes, there's 1 pair of googly eyes. So that's one set. So you can only use that. So eyes are fixed? Or is there a choice? Wait, the problem says "1 pair of googly eyes". So that's one set. So you have to use that? Or can you choose not to use eyes? Wait, no. Wait, the problem says a complete personality includes eyebrows, eyes, ears, lips, shoes, and optionally hair. So eyes are required. So you have to have eyes. So since there's only one pair of googly eyes, that's the only option for eyes. So eyes are fixed. So that's 1 choice for eyes.

Similarly, shoes: there's a pair of regular shoes and a bonus pair of clown shoes. So that's two options for shoes. So shoes can be either regular or clown. So that's 2 choices.

Hair is optional. So Tom can choose to have a hairstyle or not. There are 3 hairstyles. So if he chooses to have hair, he can pick one of the 3, otherwise, he's bald. So that's 4 choices for hair (3 + 1 for bald).

So putting this all together, the total number of different personalities would be the product of the number of choices for each component. Let me check each component again:

- Eyebrows: 2 sets, so 2 choices
- Eyes: 1 pair, so 1 choice
- Ears: 2 sets, so 2 choices
- Lips: 2 sets, so 2 choices
- Shoes: 2 options (regular or clown), so 2 choices
- Hair: 3 + 1 = 4 choices

Therefore, total number of personalities is 2 * 1 * 2 * 2 * 2 * 4. Let me compute that.

First, 2 (eyebrows) * 1 (eyes) = 2

Then 2 * 2 (ears) = 4

Then 4 * 2 (lips) = 8

Then 8 * 2 (shoes) = 16

Then 16 * 4 (hair) = 64

Wait, so 64 different personalities? Let me check again to make sure I didn't miss anything.

Wait, let me verify each component again:

Eyebrows: 2 sets. Since you can't mix and match, you have to choose one set. So 2 choices. Correct.

Eyes: 1 pair. Since it's required, only 1 choice. Correct.

Ears: 2 sets. Each set is a pair? So you have to choose one set. So 2 choices. Correct.

Lips: 2 sets. Similarly, 2 choices. Correct.

Shoes: 2 options (regular or clown). Correct.

Hair: 3 options plus the option of being bald. So 4. Correct.

So multiplying all together: 2 * 1 * 2 * 2 * 2 * 4. Let me compute again:

2 (eyebrows) * 1 (eyes) = 2

2 (ears) * 2 (lips) = 4, so 2 * 4 = 8?

Wait, no. Wait, the order is:

Eyebrows (2) * Eyes (1) * Ears (2) * Lips (2) * Shoes (2) * Hair (4)

So 2 * 1 = 2

2 * 2 = 4

4 * 2 = 8

8 * 2 = 16

16 * 4 = 64. Yes, same as before.

So 64 different personalities. Hmm. But let me think again if there's something I missed.

Wait, the problem says "a complete Mr. Potato Head personality includes eyebrows, eyes, ears, lips, shoes and optionally hair". So all of those components are required except hair. So yes, we have to include eyebrows, eyes, ears, lips, shoes. So each of those components must be present. So the only optional is hair. So that's correct.

But let me check if there's any other components. The problem says:

"3 hairstyles, 2 sets of eyebrows, 1 pair of googly eyes, 2 sets of ears, and 2 sets of lips, a pair of regular shoes, and a bonus pair of clown shoes."

So that's all the parts. So the components are:

- Hair (optional)
- Eyebrows (required, 2 choices)
- Eyes (required, 1 choice)
- Ears (required, 2 choices)
- Lips (required, 2 choices)
- Shoes (required, 2 choices)

So yes, that's all. So the calculation is correct. So 2*1*2*2*2*4=64.

Wait, but let me think again about the eyes. The problem says "1 pair of googly eyes". So that's one set. So you have to use that. So eyes are fixed. So that's 1 choice. Correct.

Similarly, for the shoes, there's a pair of regular shoes and a bonus pair of clown shoes. So two options. So that's 2 choices. Correct.

So, seems like 64 is the answer. But let me check again if there's something wrong with the interpretation of the parts.

Wait, the problem says "2 sets of eyebrows". If each set is a single eyebrow, then you have two eyebrows, but since you can't mix and match, maybe you have to take both? Wait, but that would be confusing. Wait, the note says you can't mix and match. For example, you can't take the left eyebrow from one pair and the right from another. So that suggests that each set is a pair. So if you have two sets of eyebrows, each set is a pair (left and right). Therefore, you can choose either set, but not mix them. Therefore, 2 choices for eyebrows. So that's correct.

Alternatively, if each set is a single eyebrow, then you have two eyebrows, but since you can't mix and match, maybe you have to take both? But that would be confusing. Wait, but in reality, Mr. Potato Head has two eyebrows, each on the left and right. So if you have two sets of eyebrows, maybe each set is a pair? Or maybe each set is a single eyebrow? Wait, but if you have two sets, and you can't mix and match, then maybe each set is a pair. For example, if you have two different pairs of eyebrows, like one pair is bushy and another is thin. So you can choose either pair, but not mix them. So that would be 2 choices. So that seems correct.

Alternatively, if each set is a single eyebrow, then you have two eyebrows, but since you can't mix and match, you have to take both? But that would mean that you have two eyebrows, but since they are from different sets, you can't mix. Wait, but if you have two sets, each with one eyebrow, then you can't mix them. So maybe you have to choose one set? But that would be confusing. Wait, maybe the problem is that each set is a pair. For example, each set is a left and right eyebrow. So if you have two sets, you can choose either set. So that would be 2 choices. So that seems more logical.

Therefore, I think my original calculation is correct. So 2*1*2*2*2*4=64. Therefore, the answer is 64. But let me check again.

Alternatively, maybe the problem is that for the eyebrows, since there are two sets, each set is a pair, so you have two options. Similarly for ears and lips. So that's 2*2*2=8 for those three components. Then eyes is 1, shoes is 2, hair is 4. So total is 8*1*2*4=64. Yes, same result.

Alternatively, maybe I made a mistake with the hair? Let me check. The problem says "3 hairstyles, ... and optionally hair". So hair is optional. So Tom can choose to have none, or one of the three. So that's 4 choices. Correct. So 3 + 1 = 4. So that's right.

Alternatively, maybe the problem is that the hairstyles are different, so each hairstyle is a separate option. So if you choose a hairstyle, you have 3 options, and if you don't, you have 1 option (bald). So total 4. Correct.

So, I think that's right. Therefore, the total number of different wacky personalities is 64. Therefore, the answer is \boxed{64}.

Wait, but let me check again. Maybe I made a mistake with the eyes? The problem says "1 pair of googly eyes". So that's one pair. So you have to use that. So eyes are fixed. So that's 1 choice. Correct.

Alternatively, maybe there's a choice for eyes? But the problem says "a complete personality includes eyebrows, eyes, ears, lips, shoes and optionally hair". So eyes are required. So since there's only one pair, you have to use that. So 1 choice. Correct.

Another thing: the problem says "you cannot 'mix and match'". For example, you cannot take the left eyebrow from one pair and the right eyebrow from another pair. So that suggests that each set is a complete pair. So if you have two sets of eyebrows, each set is a pair (left and right). Therefore, you can choose either set. So 2 choices. So that's correct.

Similarly, for ears: 2 sets of ears. Each set is a pair (left and right). So you can choose either set. So 2 choices. Same with lips: 2 sets of lips. Each set is a pair (left and right). So 2 choices. So that's correct.

So, all components are accounted for. So 2 (eyebrows) * 1 (eyes) * 2 (ears) * 2 (lips) * 2 (shoes) * 4 (hair) = 2*1*2*2*2*4 = 64. So yes, 64. Therefore, I think that is the correct answer.

But just to be thorough, let me think of another way. Let's think of each component:

Eyebrows: 2 options.

Eyes: 1 option.

Ears: 2 options.

Lips: 2 options.

Shoes: 2 options.

Hair: 4 options.

So total combinations: 2*1*2*2*2*4 = 64. Yep.

Alternatively, maybe the problem is that the eyes are a pair, so you have to use both eyes? But since it's a pair, you can't choose not to have eyes? But the problem says that eyes are required. So you have to have eyes. So since there's only one pair, you have to use that. So that's 1 choice. So that's correct.

Another thing: the problem says "a complete Mr. Potato Head personality includes eyebrows, eyes, ears, lips, shoes and optionally hair". So all of those are required except hair. So yes, that's what I considered.

Therefore, I think my answer is correct. 64 different personalities.

**Final Answer**
The number of different wacky personalities Tom can come up with is \boxed{64}.

To determine how many different **wacky personalities** Tom can create with his Mr. Potato Head, we need to analyze the components he has and how they can be combined, keeping in mind the constraints provided.

---

### **Step 1: List the Components and Their Options**

Here are the parts that come with the Mr. Potato Head and the number of choices for each:

- **Eyebrows**: 2 sets
  - Since you **cannot mix and match**, you must choose **one full set**.
  - **Choices**: 2

- **Eyes**: 1 pair of googly eyes
  - This is a **single, fixed set**.
  - **Choices**: 1

- **Ears**: 2 sets
  - Again, you must choose **one full set**.
  - **Choices**: 2

- **Lips**: 2 sets
  - Same logic as above.
  - **Choices**: 2

- **Shoes**: 1 pair of regular shoes + 1 bonus pair of clown shoes
  - Two distinct options.
  - **Choices**: 2

- **Hair**: 3 hairstyles + option to be **bald**
  - 3 choices for hairstyles, and 1 for being bald.
  - **Choices**: 4

---

### **Step 2: Apply the Multiplication Principle**

Since each component is independent and must be included (except hair, which is optional), we multiply the number of choices for each component:

$$
\text{Total combinations} = 2 \times 1 \times 2 \times 2 \times 2 \times 4
$$

Let’s compute this step by step:

- $2 \times 1 = 2$
- $2 \times 2 = 4$
- $4 \times 2 = 8$
- $8 \times 2 = 16$
- $16 \times 4 = 64$

---

### **Final Answer**

$$
\boxed{64}
$$

Tom can create **64 different wacky personalities** with his Mr. Potato Head.

[Assistant]
Explanation:
Okay, so Tom has this Mr. Potato Head, and he wants to figure out how many different wacky personalities he can make. Let me try to break this down step by step. First, let me list out all the parts he has:

- 3 hairstyles
- 2 sets of eyebrows
- 1 pair of googly eyes
- 2 sets of ears
- 2 sets of lips
- 1 pair of regular shoes
- 1 bonus pair of clown shoes

And the question says a complete personality includes eyebrows, eyes, ears, lips, shoes, and optionally hair. Also, note that you can't mix and match parts, like you can't take one eyebrow from one pair and another from another pair. So each set is considered as a whole, right? So for eyebrows, since there are 2 sets, each set is probably a left and right eyebrow? Or maybe each set is a pair? Wait, the problem says "2 sets of eyebrows". Hmm. Let me think.

Wait, the note says you can't mix and match. For example, you can't take the left eyebrow from one pair and the right from another. So that suggests that each set is a complete pair? Or maybe each set is a single eyebrow? Wait, but if you have two sets of eyebrows, and you can't mix and match, then maybe each set is a pair? Like, each set is a left and right eyebrow? But then if you have two sets, you could choose either set, but not mix them. So for eyebrows, you have two options: either use the first set or the second set. So that would be 2 choices for eyebrows.

But let me check again. The problem says "2 sets of eyebrows". If each set is a pair (left and right), then you can't mix them. So you have two options for eyebrows: choose one set or the other. So that would be 2 choices. Similarly, for the ears: 2 sets of ears. So same thing, each set is a pair? Or maybe each set is a single ear? Wait, but if you have two sets of ears, and you can't mix and match, then maybe each set is a pair? Like, each set is left and right ears? So again, you have two options for ears. Similarly for lips: 2 sets of lips. So again, two options.

But then, for the eyes, there's 1 pair of googly eyes. So that's one set. So you can only use that. So eyes are fixed? Or is there a choice? Wait, the problem says "1 pair of googly eyes". So that's one set. So you have to use that? Or can you choose not to use eyes? Wait, no. Wait, the problem says a complete personality includes eyebrows, eyes, ears, lips, shoes, and optionally hair. So eyes are required. So you have to have eyes. So since there's only one pair of googly eyes, that's the only option for eyes. So eyes are fixed. So that's 1 choice for eyes.

Similarly, shoes: there's a pair of regular shoes and a bonus pair of clown shoes. So that's two options for shoes. So shoes can be either regular or clown. So that's 2 choices.

Hair is optional. So Tom can choose to have a hairstyle or not. There are 3 hairstyles. So if he chooses to have hair, he can pick one of the 3, otherwise, he's bald. So that's 4 choices for hair (3 + 1 for bald).

So putting this all together, the total number of different personalities would be the product of the number of choices for each component. Let me check each component again:

- Eyebrows: 2 sets, so 2 choices
- Eyes: 1 pair, so 1 choice
- Ears: 2 sets, so 2 choices
- Lips: 2 sets, so 2 choices
- Shoes: 2 options (regular or clown), so 2 choices
- Hair: 3 + 1 = 4 choices

Therefore, total number of personalities is 2 * 1 * 2 * 2 * 2 * 4. Let me compute that.

First, 2 (eyebrows) * 1 (eyes) = 2

Then 2 * 2 (ears) = 4

Then 4 * 2 (lips) = 8

Then 8 * 2 (shoes) = 16

Then 16 * 4 (hair) = 64

Wait, so 64 different personalities? Let me check again to make sure I didn't miss anything.

Wait, let me verify each component again:

Eyebrows: 2 sets. Since you can't mix and match, you have to choose one set. So 2 choices. Correct.

Eyes: 1 pair. Since it's required, only 1 choice. Correct.

Ears: 2 sets. Each set is a pair? So you have to choose one set. So 2 choices. Correct.

Lips: 2 sets. Similarly, 2 choices. Correct.

Shoes: 2 options (regular or clown). Correct.

Hair: 3 options plus the option of being bald. So 4. Correct.

So multiplying all together: 2 * 1 * 2 * 2 * 2 * 4. Let me compute again:

2 (eyebrows) * 1 (eyes) = 2

2 (ears) * 2 (lips) = 4, so 2 * 4 = 8?

Wait, no. Wait, the order is:

Eyebrows (2) * Eyes (1) * Ears (2) * Lips (2) * Shoes (2) * Hair (4)

So 2 * 1 = 2

2 * 2 = 4

4 * 2 = 8

8 * 2 = 16

16 * 4 = 64. Yes, same as before.

So 64 different personalities. Hmm. But let me think again if there's something I missed.

Wait, the problem says "a complete Mr. Potato Head personality includes eyebrows, eyes, ears, lips, shoes and optionally hair". So all of those components are required except hair. So yes, we have to include eyebrows, eyes, ears, lips, shoes. So each of those components must be present. So the only optional is hair. So that's correct.

But let me check if there's any other components. The problem says:

"3 hairstyles, 2 sets of eyebrows, 1 pair of googly eyes, 2 sets of ears, and 2 sets of lips, a pair of regular shoes, and a bonus pair of clown shoes."

So that's all the parts. So the components are:

- Hair (optional)
- Eyebrows (required, 2 choices)
- Eyes (required, 1 choice)
- Ears (required, 2 choices)
- Lips (required, 2 choices)
- Shoes (required, 2 choices)

So yes, that's all. So the calculation is correct. So 2*1*2*2*2*4=64.

Wait, but let me think again about the eyes. The problem says "1 pair of googly eyes". So that's one set. So you have to use that. So eyes are fixed. So that's 1 choice. Correct.

Similarly, for the shoes, there's a pair of regular shoes and a bonus pair of clown shoes. So two options. So that's 2 choices. Correct.

So, seems like 64 is the answer. But let me check again if there's something wrong with the interpretation of the parts.

Wait, the problem says "2 sets of eyebrows". If each set is a single eyebrow, then you have two eyebrows, but since you can't mix and match, maybe you have to take both? Wait, but that would be confusing. Wait, the note says you can't mix and match. For example, you can't take the left eyebrow from one pair and the right from another. So that suggests that each set is a pair. So if you have two sets of eyebrows, each set is a pair (left and right). Therefore, you can choose either set, but not mix them. Therefore, 2 choices for eyebrows. So that's correct.

Alternatively, if each set is a single eyebrow, then you have two eyebrows, but since you can't mix and match, maybe you have to take both? But that would be confusing. Wait, but in reality, Mr. Potato Head has two eyebrows, each on the left and right. So if you have two sets of eyebrows, maybe each set is a pair? Or maybe each set is a single eyebrow? Wait, but if you have two sets, and you can't mix and match, then maybe each set is a pair. For example, if you have two different pairs of eyebrows, like one pair is bushy and another is thin. So you can choose either pair, but not mix them. So that would be 2 choices. So that seems correct.

Alternatively, if each set is a single eyebrow, then you have two eyebrows, but since you can't mix and match, you have to take both? But that would mean that you have two eyebrows, but since they are from different sets, you can't mix. Wait, but if you have two sets, each with one eyebrow, then you can't mix them. So maybe you have to choose one set? But that would be confusing. Wait, maybe the problem is that each set is a pair. For example, each set is a left and right eyebrow. So if you have two sets, you can choose either set. So that would be 2 choices. So that seems more logical.

Therefore, I think my original calculation is correct. So 2*1*2*2*2*4=64. Therefore, the answer is 64. But let me check again.

Alternatively, maybe the problem is that for the eyebrows, since there are two sets, each set is a pair, so you have two options. Similarly for ears and lips. So that's 2*2*2=8 for those three components. Then eyes is 1, shoes is 2, hair is 4. So total is 8*1*2*4=64. Yes, same result.

Alternatively, maybe I made a mistake with the hair? Let me check. The problem says "3 hairstyles, ... and optionally hair". So hair is optional. So Tom can choose to have none, or one of the three. So that's 4 choices. Correct. So 3 + 1 = 4. So that's right.

Alternatively, maybe the problem is that the hairstyles are different, so each hairstyle is a separate option. So if you choose a hairstyle, you have 3 options, and if you don't, you have 1 option (bald). So total 4. Correct.

So, I think that's right. Therefore, the total number of different wacky personalities is 64. Therefore, the answer is \boxed{64}.

Wait, but let me check again. Maybe I made a mistake with the eyes? The problem says "1 pair of googly eyes". So that's one pair. So you have to use that. So eyes are fixed. So that's 1 choice. Correct.

Alternatively, maybe there's a choice for eyes? But the problem says "a complete personality includes eyebrows, eyes, ears, lips, shoes and optionally hair". So eyes are required. So since there's only one pair, you have to use that. So 1 choice. Correct.

Another thing: the problem says "you cannot 'mix and match'". For example, you cannot take the left eyebrow from one pair and the right eyebrow from another pair. So that suggests that each set is a complete pair. So if you have two sets of eyebrows, each set is a pair (left and right). Therefore, you can choose either set. So 2 choices. So that's correct.

Similarly, for ears: 2 sets of ears. Each set is a pair (left and right). So you can choose either set. So 2 choices. Same with lips: 2 sets of lips. Each set is a pair (left and right). So 2 choices. So that's correct.

So, all components are accounted for. So 2 (eyebrows) * 1 (eyes) * 2 (ears) * 2 (lips) * 2 (shoes) * 4 (hair) = 2*1*2*2*2*4 = 64. So yes, 64. Therefore, I think that is the correct answer.

But just to be thorough, let me think of another way. Let's think of each component:

Eyebrows: 2 options.

Eyes: 1 option.

Ears: 2 options.

Lips: 2 options.

Shoes: 2 options.

Hair: 4 options.

So total combinations: 2*1*2*2*2*4 = 64. Yep.

Alternatively, maybe the problem is that the eyes are a pair, so you have to use both eyes? But since it's a pair, you can't choose not to have eyes? But the problem says that eyes are required. So you have to have eyes. So since there's only one pair, you have to use that. So that's 1 choice. So that's correct.

Another thing: the problem says "a complete Mr. Potato Head personality includes eyebrows, eyes, ears, lips, shoes and optionally hair". So all of those are required except hair. So yes, that's what I considered.

Therefore, I think my answer is correct. 64 different personalities.

**Final Answer**
The number of different wacky personalities Tom can come up with is \boxed{64}.

To determine how many different **wacky personalities** Tom can create with his Mr. Potato Head, we need to analyze the components he has and how they can be combined, keeping in mind the constraints provided.

---

### **Step 1: List the Components and Their Options**

Here are the parts that come with the Mr. Potato Head and the number of choices for each:

- **Eyebrows**: 2 sets
  - Since you **cannot mix and match**, you must choose **one full set**.
  - **Choices**: 2

- **Eyes**: 1 pair of googly eyes
  - This is a **single, fixed set**.
  - **Choices**: 1

- **Ears**: 2 sets
  - Again, you must choose **one full set**.
  - **Choices**: 2

- **Lips**: 2 sets
  - Same logic as above.
  - **Choices**: 2

- **Shoes**: 1 pair of regular shoes + 1 bonus pair of clown shoes
  - Two distinct options.
  - **Choices**: 2

- **Hair**: 3 hairstyles + option to be **bald**
  - 3 choices for hairstyles, and 1 for being bald.
  - **Choices**: 4

---

### **Step 2: Apply the Multiplication Principle**

Since each component is independent and must be included (except hair, which is optional), we multiply the number of choices for each component:

$$
\text{Total combinations} = 2 \times 1 \times 2 \times 2 \times 2 \times 4
$$

Let’s compute this step by step:

- $2 \times 1 = 2$
- $2 \times 2 = 4$
- $4 \times 2 = 8$
- $8 \times 2 = 16$
- $16 \times 4 = 64$

---

### **Final Answer**

$$
\boxed{64}
$$

Tom can create **64 different wacky personalities** with his Mr. Potato Head.

Solution: There are 2 choices of eyebrows, 1 choice for eyes, 2 choices for ears, 2 choices for lips, 2 choices for shoes, and 4 choices for hair (3 choices of hairstyles, or bald), and each can be picked independent of the others, so the number of combinations is the product of those, $2\cdot 1\cdot 2\cdot 2\cdot 2\cdot 4 = \boxed{64}$.