Explanation:
Okay, so I need to simplify this expression: the square root of the cube root of the square root of 1/729. And then write the result with a rational denominator. Hmm, let me break this down step by step.

First, let me write down the expression again to make sure I have it right: √(∛(√(1/729))). So, starting from the inside, there's a square root of 1/729, then a cube root of that result, and then a square root of that.

Let me recall that roots can be expressed as exponents. Remember that the nth root of something is the same as raising it to the power of 1/n. So maybe I can rewrite all these roots as exponents to make it easier to handle. Let me try that.

Starting with the innermost part: √(1/729). The square root is the same as raising to the 1/2 power. So that would be (1/729)^(1/2). Then, the next layer is the cube root of that, so that would be [(1/729)^(1/2)]^(1/3). And then the outermost square root is the square root of that result, so that would be [ [(1/729)^(1/2)]^(1/3) ]^(1/2).

Wait, exponents multiply when you have something raised to a power and then another power. So maybe I can combine all those exponents. Let me do that step by step.

Starting with the innermost part: (1/729)^(1/2). Then, taking the cube root of that is the same as raising it to the 1/3 power. So combining those exponents: (1/729)^(1/2 * 1/3) = (1/729)^(1/6). Then, taking the square root of that result would be raising it to the 1/2 power again. So overall exponent is 1/6 * 1/2 = 1/12.

Therefore, the entire expression simplifies to (1/729)^(1/12). Wait, is that right? Let me check again.

Original expression: √(∛(√(1/729))).

Breaking it down:

Start with √(1/729) = (1/729)^(1/2)

Then ∛(that) = [(1/729)^(1/2)]^(1/3) = (1/729)^(1/2 * 1/3) = (1/729)^(1/6)

Then √(that) = [(1/729)^(1/6)]^(1/2) = (1/729)^(1/6 * 1/2) = (1/729)^(1/12)

Yes, that seems correct. So the entire expression simplifies to (1/729)^(1/12).

But the question says to write the result with a rational denominator. Hmm. Wait, but (1/729)^(1/12) is already a single term. Maybe I need to express it as a radical with a rational denominator? Or perhaps rationalize the denominator if there are any denominators? Wait, but in this case, the expression is just a single term. Let me think.

Alternatively, maybe I need to express it as a fraction with a rational denominator. Let me recall that when you have roots, sometimes you need to rationalize denominators. But in this case, the expression is (1/729)^(1/12). Let me write 729 as a power of some number. 729 is 9^3, right? Because 9*9=81, 81*9=729. So 9^3. Alternatively, 729 is 3^6, since 3^6 = 729. Let me check: 3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243, 3^6=729. Yes, that's correct. So 729 is 3^6. Therefore, 1/729 is 3^(-6).

So (1/729)^(1/12) is equal to (3^(-6))^(1/12) = 3^(-6*(1/12)) = 3^(-1/2). Which is 1/(3^(1/2)) or 1/√3.

Wait, so that simplifies to 1 over square root of 3. But the question says to write the result with a rational denominator. So if I have 1/√3, I need to rationalize the denominator.

So rationalizing 1/√3 would involve multiplying numerator and denominator by √3, resulting in √3 / 3.

But wait, let me check if my previous steps are correct. Let me go back.

Original expression: √(∛(√(1/729))).

First, √(1/729) is (1/729)^(1/2). Then, ∛ of that is (1/729)^(1/2 * 1/3) = (1/729)^(1/6). Then, √ of that is (1/729)^(1/6 * 1/2) = (1/729)^(1/12).

Then, since 729 is 3^6, so 1/729 is 3^(-6). Therefore, (3^(-6))^(1/12) = 3^(-6/12) = 3^(-1/2) = 1 / 3^(1/2) = 1 / √3.

Yes, that's correct. So the simplified form is 1 over square root of 3, but we need to rationalize the denominator.

So rationalizing 1/√3: multiply numerator and denominator by √3:

(1 * √3) / (√3 * √3) = √3 / 3.

So the result with a rational denominator is √3 / 3.

But let me check if there's another way to approach this problem, maybe by simplifying the radicals step by step instead of converting to exponents. Let me try that as a verification.

Starting with the innermost radical: √(1/729).

√(1/729) is √1 / √729 = 1 / 27, since √729 is 27.

Then, the next layer is the cube root of that: ∛(1/27).

Cube root of 1 is 1, and cube root of 27 is 3, so ∛(1/27) = 1/3.

Then, the outermost square root is √(1/3).

Which is √1 / √3 = 1 / √3.

Again, same result. So then rationalizing gives √3 / 3.

Therefore, the answer is √3 / 3.

But let me check if the question wants the answer in terms of a single radical or something else. Wait, the original expression was a nested radical, but after simplifying, we have √3 / 3.

But maybe I need to write it as a single radical with a rational denominator? Wait, but √3 / 3 is already with a rational denominator. The denominator is 3, which is rational. So maybe that's the answer.

But let me check again. The problem says "simplify and write the result with a rational denominator". So if the original expression simplifies to 1/√3, then rationalizing gives √3 / 3, which has a rational denominator.

Alternatively, maybe there's a different way to express the original expression as a single radical? Let me think.

Original expression: √(∛(√(1/729))).

We can think of this as a combination of exponents. Let me write all the exponents:

Starting from the inside: √(1/729) is (1/729)^{1/2}

Then, ∛ of that is [(1/729)^{1/2}]^{1/3} = (1/729)^{1/6}

Then, √ of that is [(1/729)^{1/6}]^{1/2} = (1/729)^{1/12}

So, as before, (1/729)^{1/12} = 1 / 729^{1/12}

But 729 is 3^6, so 729^{1/12} = (3^6)^{1/12} = 3^{6/12} = 3^{1/2} = √3.

Therefore, 1 / 729^{1/12} = 1 / √3, which again is the same as before.

So, again, rationalizing gives √3 / 3.

Therefore, the answer is √3 over 3.

But let me check if there's a different interpretation of the original expression. The original expression is written as sqrt(cbrt(sqrt(1/729))). Is there a possibility that the order of operations is different? For example, is it sqrt(cbrt(sqrt(1/729))) or something else? But according to standard order of operations, the innermost radical is evaluated first, then moving outwards. So the way I did it is correct.

Alternatively, maybe the problem expects the answer expressed as a single radical with a rational denominator? Let me think.

If I have √3 / 3, that's already simplified. But if I wanted to write it as a single radical, maybe? Let me see.

But √3 / 3 is equal to 1 / √3, which is the same as before. But since the question says to write with a rational denominator, √3 / 3 is the correct form.

Alternatively, maybe I need to express the entire original expression as a single radical? Let me think.

Original expression: sqrt(cbrt(sqrt(1/729)))

Which we simplified to 1 / √3, which is √3 / 3.

Alternatively, if I wanted to write this as a single radical, maybe? Let me see.

But √3 / 3 is already a simplified expression.

Alternatively, maybe the problem expects the answer in terms of exponents? But the question says "write the result with a rational denominator", which suggests that the answer should be a fraction with a rational denominator, not necessarily a radical.

But √3 / 3 is a fraction with a rational denominator. So that's probably the answer.

Wait, but let me check if there's a mistake in my calculation. Let me verify with decimal approximations.

Original expression: sqrt(cbrt(sqrt(1/729))).

First, compute 1/729. 1 divided by 729 is approximately 0.001371742112.

Then, sqrt(0.001371742112) is sqrt(1/729) = 1/27 ≈ 0.037037037.

Then, cube root of that: cube root of 1/27 is 1/3 ≈ 0.333333333.

Then, square root of that: sqrt(1/3) ≈ 0.57735.

Now, √3 / 3 is approximately 1.732 / 3 ≈ 0.5773, which matches.

Alternatively, 1 / √3 ≈ 0.5773, same thing.

So that seems correct.

But let me check if there's another way to write √3 / 3 with a different radical expression. For example, maybe as a 12th root or something?

But the problem says "simplify and write the result with a rational denominator". Since √3 / 3 is already simplified and has a rational denominator, that's probably the answer.

Alternatively, maybe the problem expects the answer expressed as a single radical? Let me think.

If I have √3 / 3, that can be written as (3^{1/2}) / 3^1 = 3^{-1/2}, which is 1 / 3^{1/2} = 1 / √3. But that's going back again.

Alternatively, if I wanted to write it as a single radical with a rational denominator, maybe? For example, if I have √3 / 3, that's equivalent to √3 / √9, since 3 is √9. Then, combining the radicals: √3 / √9 = √(3/9) = √(1/3). Wait, but that's going back to 1 / √3 again.

So that doesn't help.

Alternatively, maybe rationalizing the denominator is the key here. Since the original expression simplifies to 1 / √3, which has an irrational denominator, so we need to rationalize it.

Therefore, the answer is √3 / 3.

But let me check if there's a different approach where I can combine all the roots into a single radical.

Original expression: sqrt(cbrt(sqrt(1/729)))

Let me think of each root as exponents.

Starting from the inside: sqrt(1/729) is (1/729)^{1/2}

Then, cbrt of that is [(1/729)^{1/2}]^{1/3} = (1/729)^{1/6}

Then, sqrt of that is [(1/729)^{1/6}]^{1/2} = (1/729)^{1/12}

So, as before, (1/729)^{1/12} = 1 / 729^{1/12}

But 729 is 3^6, so 729^{1/12} = (3^6)^{1/12} = 3^{6/12} = 3^{1/2} = √3.

Therefore, 1 / √3.

So again, same result.

Therefore, the answer is √3 / 3.

Alternatively, if I wanted to write this as a single radical with a rational denominator, maybe? Let me think.

Suppose I have √3 / 3. If I want to write this as a single radical, perhaps?

But √3 / 3 is already a simplified expression.

Alternatively, if I wanted to write it as a 6th root or something? Let me see.

But √3 is 3^{1/2}, so √3 / 3 is 3^{1/2} / 3^1 = 3^{-1/2} = 1 / 3^{1/2} = 1 / √3.

So, again, same thing.

Therefore, I think the answer is √3 / 3.

But let me check if there's a different way to interpret the original expression. For example, maybe the expression is sqrt( cbrt( sqrt(1/729) ) ), which is what I did.

Alternatively, could it be interpreted as sqrt( cbrt( sqrt(1) / 729 ) )? But that's the same as sqrt( cbrt( 1 / 729 ) ), which is the same as before.

Alternatively, maybe the original expression is written with different grouping? But the way it's written is sqrt( cbrt( sqrt(1/729) ) ).

Therefore, I think my approach is correct.

Therefore, the simplified expression with a rational denominator is √3 / 3.

But let me check once more.

Original expression: sqrt( cbrt( sqrt(1/729) ) )

Step 1: sqrt(1/729) = 1/27

Step 2: cbrt(1/27) = 1/3

Step 3: sqrt(1/3) = 1/√3 = √3 / 3

Yes, that's correct.

Therefore, the answer is √3 over 3.

But let me check if the problem expects the answer in some other form. For example, maybe as a 12th root?

But 1 / √3 is equal to 3^{-1/2}, which is also equal to 3^{-6/12} = 1 / 3^{6/12} = 1 / 3^{1/2} = same thing.

Alternatively, if I wanted to write it as a 12th root, it would be 1 / 729^{1/12}, but that's not helpful.

Alternatively, maybe the problem expects the answer expressed as a radical with a rational denominator? For example, if I have √3 / 3, that's already with a rational denominator.

Alternatively, if I wanted to write it as a single fraction with a radical in the numerator, which is already the case.

Therefore, I think √3 / 3 is the correct answer.

But let me check if there's a mistake in my calculation. For example, maybe I miscalculated the exponents?

Original expression: sqrt(cbrt(sqrt(1/729)))

Let me write all exponents:

Starting with 1/729. Let me write 1/729 as 729^{-1}.

Then sqrt(729^{-1}) = (729^{-1})^{1/2} = 729^{-1/2}

Then cbrt(729^{-1/2}) = (729^{-1/2})^{1/3} = 729^{-1/6}

Then sqrt(729^{-1/6}) = (729^{-1/6})^{1/2} = 729^{-1/12}

So 729^{-1/12} = 1 / 729^{1/12}

As before.

Since 729 = 3^6, then 729^{1/12} = (3^6)^{1/12} = 3^{6/12} = 3^{1/2} = √3.

Therefore, 1 / √3.

So again, same result.

Therefore, I think my answer is correct.

Therefore, the simplified expression with a rational denominator is √3 / 3.

But let me check if the problem wants the answer in terms of a radical with a rational denominator, maybe expressed as a single radical? For example, if I have √3 / 3, can I write that as a single radical?

Well, √3 / 3 is equal to √3 / √9 = √(3/9) = √(1/3). But that's going back to 1 / √3, which is not helpful.

Alternatively, if I wanted to write √3 / 3 as a single radical with a rational denominator, maybe?

But √3 / 3 is already simplified.

Alternatively, if I wanted to write it as a 6th root or something?

But I don't think that's necessary.

Therefore, I think the answer is √3 / 3.

Therefore, I think that's the final answer.

**Final Answer**
The simplified expression with a rational denominator is \boxed{\dfrac{\sqrt{3}}{3}}.

To simplify the expression
$$
\sqrt{\sqrt[3]{\sqrt{\frac{1}{729}}}}
$$
and write the result with a **rational denominator**, we proceed step by step, using properties of exponents and radicals.

---

### Step 1: Convert all radicals to exponents

We begin by expressing each radical as a power:

- The innermost expression is $\sqrt{\frac{1}{729}} = \left(\frac{1}{729}\right)^{1/2}$
- Then, the cube root of that is $\left(\left(\frac{1}{729}\right)^{1/2}\right)^{1/3} = \left(\frac{1}{729}\right)^{1/6}$
- Finally, the square root of that is $\left(\left(\frac{1}{729}\right)^{1/6}\right)^{1/2} = \left(\frac{1}{729}\right)^{1/12}$

So the entire expression simplifies to:
$$
\left(\frac{1}{729}\right)^{1/12}
$$

---

### Step 2: Simplify the base

We know that:
$$
729 = 3^6
$$
So:
$$
\left(\frac{1}{729}\right)^{1/12} = \left(3^{-6}\right)^{1/12} = 3^{-6/12} = 3^{-1/2} = \frac{1}{\sqrt{3}}
$$

---

### Step 3: Rationalize the denominator

The expression $\frac{1}{\sqrt{3}}$ has an irrational denominator. To rationalize it, we multiply numerator and denominator by $\sqrt{3}$:

$$
\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}
$$

---

### Final Answer

The simplified expression with a rational denominator is:

$$
\boxed{\dfrac{\sqrt{3}}{3}}
$$

[Assistant]
Explanation:
Okay, so I need to simplify this expression: the square root of the cube root of the square root of 1/729. And then write the result with a rational denominator. Hmm, let me break this down step by step.

First, let me write down the expression again to make sure I have it right: √(∛(√(1/729))). So, starting from the inside, there's a square root of 1/729, then a cube root of that result, and then a square root of that.

Let me recall that roots can be expressed as exponents. Remember that the nth root of something is the same as raising it to the power of 1/n. So maybe I can rewrite all these roots as exponents to make it easier to handle. Let me try that.

Starting with the innermost part: √(1/729). The square root is the same as raising to the 1/2 power. So that would be (1/729)^(1/2). Then, the next layer is the cube root of that, so that would be [(1/729)^(1/2)]^(1/3). And then the outermost square root is the square root of that result, so that would be [ [(1/729)^(1/2)]^(1/3) ]^(1/2).

Wait, exponents multiply when you have something raised to a power and then another power. So maybe I can combine all those exponents. Let me do that step by step.

Starting with the innermost part: (1/729)^(1/2). Then, taking the cube root of that is the same as raising it to the 1/3 power. So combining those exponents: (1/729)^(1/2 * 1/3) = (1/729)^(1/6). Then, taking the square root of that result would be raising it to the 1/2 power again. So overall exponent is 1/6 * 1/2 = 1/12.

Therefore, the entire expression simplifies to (1/729)^(1/12). Wait, is that right? Let me check again.

Original expression: √(∛(√(1/729))).

Breaking it down:

Start with √(1/729) = (1/729)^(1/2)

Then ∛(that) = [(1/729)^(1/2)]^(1/3) = (1/729)^(1/2 * 1/3) = (1/729)^(1/6)

Then √(that) = [(1/729)^(1/6)]^(1/2) = (1/729)^(1/6 * 1/2) = (1/729)^(1/12)

Yes, that seems correct. So the entire expression simplifies to (1/729)^(1/12).

But the question says to write the result with a rational denominator. Hmm. Wait, but (1/729)^(1/12) is already a single term. Maybe I need to express it as a radical with a rational denominator? Or perhaps rationalize the denominator if there are any denominators? Wait, but in this case, the expression is just a single term. Let me think.

Alternatively, maybe I need to express it as a fraction with a rational denominator. Let me recall that when you have roots, sometimes you need to rationalize denominators. But in this case, the expression is (1/729)^(1/12). Let me write 729 as a power of some number. 729 is 9^3, right? Because 9*9=81, 81*9=729. So 9^3. Alternatively, 729 is 3^6, since 3^6 = 729. Let me check: 3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243, 3^6=729. Yes, that's correct. So 729 is 3^6. Therefore, 1/729 is 3^(-6).

So (1/729)^(1/12) is equal to (3^(-6))^(1/12) = 3^(-6*(1/12)) = 3^(-1/2). Which is 1/(3^(1/2)) or 1/√3.

Wait, so that simplifies to 1 over square root of 3. But the question says to write the result with a rational denominator. So if I have 1/√3, I need to rationalize the denominator.

So rationalizing 1/√3 would involve multiplying numerator and denominator by √3, resulting in √3 / 3.

But wait, let me check if my previous steps are correct. Let me go back.

Original expression: √(∛(√(1/729))).

First, √(1/729) is (1/729)^(1/2). Then, ∛ of that is (1/729)^(1/2 * 1/3) = (1/729)^(1/6). Then, √ of that is (1/729)^(1/6 * 1/2) = (1/729)^(1/12).

Then, since 729 is 3^6, so 1/729 is 3^(-6). Therefore, (3^(-6))^(1/12) = 3^(-6/12) = 3^(-1/2) = 1 / 3^(1/2) = 1 / √3.

Yes, that's correct. So the simplified form is 1 over square root of 3, but we need to rationalize the denominator.

So rationalizing 1/√3: multiply numerator and denominator by √3:

(1 * √3) / (√3 * √3) = √3 / 3.

So the result with a rational denominator is √3 / 3.

But let me check if there's another way to approach this problem, maybe by simplifying the radicals step by step instead of converting to exponents. Let me try that as a verification.

Starting with the innermost radical: √(1/729).

√(1/729) is √1 / √729 = 1 / 27, since √729 is 27.

Then, the next layer is the cube root of that: ∛(1/27).

Cube root of 1 is 1, and cube root of 27 is 3, so ∛(1/27) = 1/3.

Then, the outermost square root is √(1/3).

Which is √1 / √3 = 1 / √3.

Again, same result. So then rationalizing gives √3 / 3.

Therefore, the answer is √3 / 3.

But let me check if the question wants the answer in terms of a single radical or something else. Wait, the original expression was a nested radical, but after simplifying, we have √3 / 3.

But maybe I need to write it as a single radical with a rational denominator? Wait, but √3 / 3 is already with a rational denominator. The denominator is 3, which is rational. So maybe that's the answer.

But let me check again. The problem says "simplify and write the result with a rational denominator". So if the original expression simplifies to 1/√3, then rationalizing gives √3 / 3, which has a rational denominator.

Alternatively, maybe there's a different way to express the original expression as a single radical? Let me think.

Original expression: √(∛(√(1/729))).

We can think of this as a combination of exponents. Let me write all the exponents:

Starting from the inside: √(1/729) is (1/729)^{1/2}

Then, ∛ of that is [(1/729)^{1/2}]^{1/3} = (1/729)^{1/6}

Then, √ of that is [(1/729)^{1/6}]^{1/2} = (1/729)^{1/12}

So, as before, (1/729)^{1/12} = 1 / 729^{1/12}

But 729 is 3^6, so 729^{1/12} = (3^6)^{1/12} = 3^{6/12} = 3^{1/2} = √3.

Therefore, 1 / 729^{1/12} = 1 / √3, which again is the same as before.

So, again, rationalizing gives √3 / 3.

Therefore, the answer is √3 over 3.

But let me check if there's a different interpretation of the original expression. The original expression is written as sqrt(cbrt(sqrt(1/729))). Is there a possibility that the order of operations is different? For example, is it sqrt(cbrt(sqrt(1/729))) or something else? But according to standard order of operations, the innermost radical is evaluated first, then moving outwards. So the way I did it is correct.

Alternatively, maybe the problem expects the answer expressed as a single radical with a rational denominator? Let me think.

If I have √3 / 3, that's already simplified. But if I wanted to write it as a single radical, maybe? Let me see.

But √3 / 3 is equal to 1 / √3, which is the same as before. But since the question says to write with a rational denominator, √3 / 3 is the correct form.

Alternatively, maybe I need to express the entire original expression as a single radical? Let me think.

Original expression: sqrt(cbrt(sqrt(1/729)))

Which we simplified to 1 / √3, which is √3 / 3.

Alternatively, if I wanted to write this as a single radical, maybe? Let me see.

But √3 / 3 is already a simplified expression.

Alternatively, maybe the problem expects the answer in terms of exponents? But the question says "write the result with a rational denominator", which suggests that the answer should be a fraction with a rational denominator, not necessarily a radical.

But √3 / 3 is a fraction with a rational denominator. So that's probably the answer.

Wait, but let me check if there's a mistake in my calculation. Let me verify with decimal approximations.

Original expression: sqrt(cbrt(sqrt(1/729))).

First, compute 1/729. 1 divided by 729 is approximately 0.001371742112.

Then, sqrt(0.001371742112) is sqrt(1/729) = 1/27 ≈ 0.037037037.

Then, cube root of that: cube root of 1/27 is 1/3 ≈ 0.333333333.

Then, square root of that: sqrt(1/3) ≈ 0.57735.

Now, √3 / 3 is approximately 1.732 / 3 ≈ 0.5773, which matches.

Alternatively, 1 / √3 ≈ 0.5773, same thing.

So that seems correct.

But let me check if there's another way to write √3 / 3 with a different radical expression. For example, maybe as a 12th root or something?

But the problem says "simplify and write the result with a rational denominator". Since √3 / 3 is already simplified and has a rational denominator, that's probably the answer.

Alternatively, maybe the problem expects the answer expressed as a single radical? Let me think.

If I have √3 / 3, that can be written as (3^{1/2}) / 3^1 = 3^{-1/2}, which is 1 / 3^{1/2} = 1 / √3. But that's going back again.

Alternatively, if I wanted to write it as a single radical with a rational denominator, maybe? For example, if I have √3 / 3, that's equivalent to √3 / √9, since 3 is √9. Then, combining the radicals: √3 / √9 = √(3/9) = √(1/3). Wait, but that's going back to 1 / √3 again.

So that doesn't help.

Alternatively, maybe rationalizing the denominator is the key here. Since the original expression simplifies to 1 / √3, which has an irrational denominator, so we need to rationalize it.

Therefore, the answer is √3 / 3.

But let me check if there's a different approach where I can combine all the roots into a single radical.

Original expression: sqrt(cbrt(sqrt(1/729)))

Let me think of each root as exponents.

Starting from the inside: sqrt(1/729) is (1/729)^{1/2}

Then, cbrt of that is [(1/729)^{1/2}]^{1/3} = (1/729)^{1/6}

Then, sqrt of that is [(1/729)^{1/6}]^{1/2} = (1/729)^{1/12}

So, as before, (1/729)^{1/12} = 1 / 729^{1/12}

But 729 is 3^6, so 729^{1/12} = (3^6)^{1/12} = 3^{6/12} = 3^{1/2} = √3.

Therefore, 1 / √3.

So again, same result.

Therefore, the answer is √3 / 3.

Alternatively, if I wanted to write this as a single radical with a rational denominator, maybe? Let me think.

Suppose I have √3 / 3. If I want to write this as a single radical, perhaps?

But √3 / 3 is already a simplified expression.

Alternatively, if I wanted to write it as a 6th root or something? Let me see.

But √3 is 3^{1/2}, so √3 / 3 is 3^{1/2} / 3^1 = 3^{-1/2} = 1 / 3^{1/2} = 1 / √3.

So, again, same thing.

Therefore, I think the answer is √3 / 3.

But let me check if there's a different way to interpret the original expression. For example, maybe the expression is sqrt( cbrt( sqrt(1/729) ) ), which is what I did.

Alternatively, could it be interpreted as sqrt( cbrt( sqrt(1) / 729 ) )? But that's the same as sqrt( cbrt( 1 / 729 ) ), which is the same as before.

Alternatively, maybe the original expression is written with different grouping? But the way it's written is sqrt( cbrt( sqrt(1/729) ) ).

Therefore, I think my approach is correct.

Therefore, the simplified expression with a rational denominator is √3 / 3.

But let me check once more.

Original expression: sqrt( cbrt( sqrt(1/729) ) )

Step 1: sqrt(1/729) = 1/27

Step 2: cbrt(1/27) = 1/3

Step 3: sqrt(1/3) = 1/√3 = √3 / 3

Yes, that's correct.

Therefore, the answer is √3 over 3.

But let me check if the problem expects the answer in some other form. For example, maybe as a 12th root?

But 1 / √3 is equal to 3^{-1/2}, which is also equal to 3^{-6/12} = 1 / 3^{6/12} = 1 / 3^{1/2} = same thing.

Alternatively, if I wanted to write it as a 12th root, it would be 1 / 729^{1/12}, but that's not helpful.

Alternatively, maybe the problem expects the answer expressed as a radical with a rational denominator? For example, if I have √3 / 3, that's already with a rational denominator.

Alternatively, if I wanted to write it as a single fraction with a radical in the numerator, which is already the case.

Therefore, I think √3 / 3 is the correct answer.

But let me check if there's a mistake in my calculation. For example, maybe I miscalculated the exponents?

Original expression: sqrt(cbrt(sqrt(1/729)))

Let me write all exponents:

Starting with 1/729. Let me write 1/729 as 729^{-1}.

Then sqrt(729^{-1}) = (729^{-1})^{1/2} = 729^{-1/2}

Then cbrt(729^{-1/2}) = (729^{-1/2})^{1/3} = 729^{-1/6}

Then sqrt(729^{-1/6}) = (729^{-1/6})^{1/2} = 729^{-1/12}

So 729^{-1/12} = 1 / 729^{1/12}

As before.

Since 729 = 3^6, then 729^{1/12} = (3^6)^{1/12} = 3^{6/12} = 3^{1/2} = √3.

Therefore, 1 / √3.

So again, same result.

Therefore, I think my answer is correct.

Therefore, the simplified expression with a rational denominator is √3 / 3.

But let me check if the problem wants the answer in terms of a radical with a rational denominator, maybe expressed as a single radical? For example, if I have √3 / 3, can I write that as a single radical?

Well, √3 / 3 is equal to √3 / √9 = √(3/9) = √(1/3). But that's going back to 1 / √3, which is not helpful.

Alternatively, if I wanted to write √3 / 3 as a single radical with a rational denominator, maybe?

But √3 / 3 is already simplified.

Alternatively, if I wanted to write it as a 6th root or something?

But I don't think that's necessary.

Therefore, I think the answer is √3 / 3.

Therefore, I think that's the final answer.

**Final Answer**
The simplified expression with a rational denominator is \boxed{\dfrac{\sqrt{3}}{3}}.

To simplify the expression
$$
\sqrt{\sqrt[3]{\sqrt{\frac{1}{729}}}}
$$
and write the result with a **rational denominator**, we proceed step by step, using properties of exponents and radicals.

---

### Step 1: Convert all radicals to exponents

We begin by expressing each radical as a power:

- The innermost expression is $\sqrt{\frac{1}{729}} = \left(\frac{1}{729}\right)^{1/2}$
- Then, the cube root of that is $\left(\left(\frac{1}{729}\right)^{1/2}\right)^{1/3} = \left(\frac{1}{729}\right)^{1/6}$
- Finally, the square root of that is $\left(\left(\frac{1}{729}\right)^{1/6}\right)^{1/2} = \left(\frac{1}{729}\right)^{1/12}$

So the entire expression simplifies to:
$$
\left(\frac{1}{729}\right)^{1/12}
$$

---

### Step 2: Simplify the base

We know that:
$$
729 = 3^6
$$
So:
$$
\left(\frac{1}{729}\right)^{1/12} = \left(3^{-6}\right)^{1/12} = 3^{-6/12} = 3^{-1/2} = \frac{1}{\sqrt{3}}
$$

---

### Step 3: Rationalize the denominator

The expression $\frac{1}{\sqrt{3}}$ has an irrational denominator. To rationalize it, we multiply numerator and denominator by $\sqrt{3}$:

$$
\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}
$$

---

### Final Answer

The simplified expression with a rational denominator is:

$$
\boxed{\dfrac{\sqrt{3}}{3}}
$$

Solution: First, notice that $729=3^6$.  We can begin simplifying from the innermost square root: $$\sqrt{\sqrt[3]{\frac{1}{\sqrt{729}}}}=\sqrt{\sqrt[3]{\frac{1}{27}}}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}=\boxed{\frac{\sqrt{3}}{3}}$$


Explanation:
Okay, so I need to find the length of the median QM in triangle PQR. The triangle has sides PQ = QR = 34 and PR = 32. Let me first visualize the triangle. Since PQ and QR are both 34, that means triangle PQR is isosceles with PQ and QR as the two equal sides. Wait, but hold on, if PQ and QR are equal, then the base would be PR, which is 32. So the triangle has two sides of 34 and a base of 32. The median QM... Hmm, a median connects a vertex to the midpoint of the opposite side. So QM is the median from Q to the midpoint of PR. Let me confirm: since PR is the base, then M is the midpoint of PR. Therefore, PM = MR = 16, since PR is 32.

Now, how do I find the length of QM? There are a few ways to approach this. One method is using the formula for the length of a median in a triangle. I remember there's a formula that relates the lengths of the sides of the triangle to the length of a median. Let me recall... I think it's called Apollonius's theorem. Yes, Apollonius's theorem states that the sum of the squares of any two sides of a triangle equals twice the square of the median to the third side plus twice the square of the half third side.

So, in formula terms, for a triangle with sides a, b, c, and median m_a to side a, the formula is:

b² + c² = 2m_a² + 2*(a/2)²

Alternatively, maybe I should write it more clearly. Let me check again. Apollonius's theorem says that the sum of the squares of two sides is equal to twice the square of the median to the third side plus twice the square of half the third side. So, if we have triangle PQR, and we want the median from Q to PR, which is side PR. Let me denote the sides:

Let me label the triangle properly. Let me consider triangle PQR with PQ = QR = 34. Wait, hold on, if PQ = QR, then the triangle is isosceles with base PR. Therefore, vertex Q is the apex, and the base is PR. Therefore, the median from Q to PR would be the line from Q to the midpoint M of PR.

So, applying Apollonius's theorem here. Let me recall the formula again. For any triangle, the length of the median can be calculated by:

m_a² = (2b² + 2c² - a²)/4

Where a, b, c are the lengths of the sides of the triangle, and m_a is the median to side a.

So, in this case, if we want the median from Q to PR, which is side PR. Let me denote the sides:

Let me assign labels. Let me call side PR as side a, so a = 32. Then sides PQ and QR are both 34. Wait, but in Apollonius's theorem, the median is to side a, so the formula would be:

m_a² = (2b² + 2c² - a²)/4

But here, sides b and c would be the other two sides. Wait, in triangle PQR, if PR is side a, then sides PQ and QR are sides b and c? Wait, maybe I need to be careful with notation. Let me think.

Let me denote the triangle with vertices P, Q, R. Let me consider side PR as the side to which the median is drawn. So, the median is from Q to PR. Therefore, in Apollonius's theorem, the formula for the median from Q to PR would be:

QM² = (2PQ² + 2QR² - PR²)/4

Wait, is that correct? Let me check. If the median is to side PR, then the formula would be:

m² = (2b² + 2c² - a²)/4, where a is the side to which the median is drawn, and b and c are the other two sides.

Yes, so in this case, a = PR = 32, and the other two sides are PQ = 34 and QR = 34. Therefore, substituting into the formula:

QM² = (2*(34)² + 2*(34)² - (32)²)/4

Let me compute that step by step. First, compute 2*(34)² + 2*(34)². That is 2*34² + 2*34² = 4*34². Then subtract (32)², so:

QM² = (4*34² - 32²)/4

Then divide by 4. Alternatively, we can factor this expression. Let me compute the numerator first:

4*34² - 32². Let me compute 34 squared. 34 squared is 1156. So 4*1156 is 4624. Then 32 squared is 1024. Therefore, 4624 - 1024 = 3600. Therefore, QM² = 3600 / 4 = 900. Therefore, QM is sqrt(900) = 30.

Wait, so the length of the median QM is 30? That seems straightforward. Let me check if I did that correctly.

Alternatively, maybe I can verify using coordinates. Let me try coordinate geometry to confirm. Let me place the triangle in coordinate plane. Let me set point P at (-16, 0) and point R at (16, 0), so that PR is 32 units long, lying along the x-axis. Then the midpoint M of PR is at (0, 0). Wait, no. Wait, if P is at (-16, 0) and R is at (16, 0), then midpoint M is at ((-16 + 16)/2, (0 + 0)/2) = (0, 0). Wait, but then point Q is somewhere above the x-axis. Since PQ = QR = 34, so Q is equidistant from P and R, so it lies along the perpendicular bisector of PR, which is the y-axis. Therefore, coordinates of Q are (0, h), where h is the height of the triangle.

So, let me compute h. The distance from Q to P is 34. The coordinates of P are (-16, 0), coordinates of Q are (0, h). Therefore, distance PQ is sqrt[ (0 - (-16))² + (h - 0)² ] = sqrt[16² + h²] = sqrt(256 + h²). This is equal to 34. Therefore:

sqrt(256 + h²) = 34

Squaring both sides:

256 + h² = 1156

Therefore, h² = 1156 - 256 = 900

Therefore, h = 30. Therefore, coordinates of Q are (0, 30). Therefore, the median QM is from Q(0, 30) to M(0, 0). Wait, so the distance between (0, 30) and (0, 0) is 30 units. Therefore, QM is 30.

So that confirms the earlier result. Therefore, the length of the median QM is 30.

Alternatively, maybe I can use the Pythagorean theorem. Since the triangle is isosceles, the median from Q to PR is also the altitude and the angle bisector. Therefore, in this case, since we have the base PR = 32, the median QM splits it into two equal parts of 16 each. Then triangle PQM is a right triangle with legs QM and PM, and hypotenuse PQ. Therefore, by Pythagoras:

QM² + PM² = PQ²

Therefore, QM² + 16² = 34²

Therefore, QM² = 34² - 16² = (34 - 16)(34 + 16) = 18 * 50 = 900

Therefore, QM = sqrt(900) = 30.

Same result. So all three methods: Apollonius's theorem, coordinate geometry, and Pythagorean theorem, give me the same answer. Therefore, I can be confident that the length of the median QM is 30.

But let me just make sure I didn't mix up any sides. The problem says PQ = QR = 34. So, in triangle PQR, sides PQ and QR are both 34. Therefore, the triangle is isosceles with equal sides PQ and QR, so the base is PR. Therefore, the median from Q to PR is indeed the altitude, which splits the base into two equal parts. Therefore, the right triangle approach is valid here.

Alternatively, if I had messed up the sides, maybe I would have a different answer. For example, if I thought that the median was to another side, but no, the problem says "median QM", so Q is the vertex, and M is the midpoint of the opposite side. Since the triangle is PQR, the opposite side of Q is PR. Therefore, M is the midpoint of PR.

Therefore, all steps are correct. Therefore, the answer is 30.

**Final Answer**
The length of median $\overline{QM}$ is $\boxed{30}$.

To find the length of the median $\overline{QM}$ in triangle $PQR$, we are given the following side lengths:

- $PQ = QR = 34$
- $PR = 32$

---

### Step 1: Understand the Triangle Structure

Since $PQ = QR = 34$, triangle $PQR$ is **isosceles** with the equal sides $PQ$ and $QR$, and the base $PR = 32$. The median $\overline{QM}$ connects vertex $Q$ to the **midpoint $M$** of the base $PR$.

This makes $\overline{QM}$ not only a median but also the **altitude** and **angle bisector** of the isosceles triangle, due to the symmetry of the triangle.

---

### Step 2: Use the Pythagorean Theorem

Since $\overline{QM}$ is the altitude, it splits the base $PR$ into two equal segments:

$$
PM = MR = \frac{PR}{2} = \frac{32}{2} = 16
$$

Now consider the right triangle $\triangle PQM$, where:

- $PQ = 34$ (hypotenuse)
- $PM = 16$ (one leg)
- $QM$ (the other leg, which we want to find)

Apply the Pythagorean Theorem:

$$
QM^2 + PM^2 = PQ^2
$$
$$
QM^2 + 16^2 = 34^2
$$
$$
QM^2 + 256 = 1156
$$
$$
QM^2 = 1156 - 256 = 900
$$
$$
QM = \sqrt{900} = 30
$$

---

### Step 3: Confirm with Apollonius's Theorem (Optional)

Apollonius's Theorem states that for any triangle, the square of the length of a median to a side is equal to:

$$
m^2 = \frac{2b^2 + 2c^2 - a^2}{4}
$$

Here, the median is from $Q$ to side $PR = a = 32$, and the other two sides are $PQ = QR = 34$. So:

$$
QM^2 = \frac{2(34)^2 + 2(34)^2 - (32)^2}{4}
$$
$$
QM^2 = \frac{4(34)^2 - (32)^2}{4}
$$
$$
QM^2 = \frac{4(1156) - 1024}{4} = \frac{4624 - 1024}{4} = \frac{3600}{4} = 900
$$
$$
QM = \sqrt{900} = 30
$$

---

### Final Answer

$$
\boxed{30}
$$

[Assistant]
Explanation:
Okay, so I need to find the length of the median QM in triangle PQR. The triangle has sides PQ = QR = 34 and PR = 32. Let me first visualize the triangle. Since PQ and QR are both 34, that means triangle PQR is isosceles with PQ and QR as the two equal sides. Wait, but hold on, if PQ and QR are equal, then the base would be PR, which is 32. So the triangle has two sides of 34 and a base of 32. The median QM... Hmm, a median connects a vertex to the midpoint of the opposite side. So QM is the median from Q to the midpoint of PR. Let me confirm: since PR is the base, then M is the midpoint of PR. Therefore, PM = MR = 16, since PR is 32.

Now, how do I find the length of QM? There are a few ways to approach this. One method is using the formula for the length of a median in a triangle. I remember there's a formula that relates the lengths of the sides of the triangle to the length of a median. Let me recall... I think it's called Apollonius's theorem. Yes, Apollonius's theorem states that the sum of the squares of any two sides of a triangle equals twice the square of the median to the third side plus twice the square of the half third side.

So, in formula terms, for a triangle with sides a, b, c, and median m_a to side a, the formula is:

b² + c² = 2m_a² + 2*(a/2)²

Alternatively, maybe I should write it more clearly. Let me check again. Apollonius's theorem says that the sum of the squares of two sides is equal to twice the square of the median to the third side plus twice the square of half the third side. So, if we have triangle PQR, and we want the median from Q to PR, which is side PR. Let me denote the sides:

Let me label the triangle properly. Let me consider triangle PQR with PQ = QR = 34. Wait, hold on, if PQ = QR, then the triangle is isosceles with base PR. Therefore, vertex Q is the apex, and the base is PR. Therefore, the median from Q to PR would be the line from Q to the midpoint M of PR.

So, applying Apollonius's theorem here. Let me recall the formula again. For any triangle, the length of the median can be calculated by:

m_a² = (2b² + 2c² - a²)/4

Where a, b, c are the lengths of the sides of the triangle, and m_a is the median to side a.

So, in this case, if we want the median from Q to PR, which is side PR. Let me denote the sides:

Let me assign labels. Let me call side PR as side a, so a = 32. Then sides PQ and QR are both 34. Wait, but in Apollonius's theorem, the median is to side a, so the formula would be:

m_a² = (2b² + 2c² - a²)/4

But here, sides b and c would be the other two sides. Wait, in triangle PQR, if PR is side a, then sides PQ and QR are sides b and c? Wait, maybe I need to be careful with notation. Let me think.

Let me denote the triangle with vertices P, Q, R. Let me consider side PR as the side to which the median is drawn. So, the median is from Q to PR. Therefore, in Apollonius's theorem, the formula for the median from Q to PR would be:

QM² = (2PQ² + 2QR² - PR²)/4

Wait, is that correct? Let me check. If the median is to side PR, then the formula would be:

m² = (2b² + 2c² - a²)/4, where a is the side to which the median is drawn, and b and c are the other two sides.

Yes, so in this case, a = PR = 32, and the other two sides are PQ = 34 and QR = 34. Therefore, substituting into the formula:

QM² = (2*(34)² + 2*(34)² - (32)²)/4

Let me compute that step by step. First, compute 2*(34)² + 2*(34)². That is 2*34² + 2*34² = 4*34². Then subtract (32)², so:

QM² = (4*34² - 32²)/4

Then divide by 4. Alternatively, we can factor this expression. Let me compute the numerator first:

4*34² - 32². Let me compute 34 squared. 34 squared is 1156. So 4*1156 is 4624. Then 32 squared is 1024. Therefore, 4624 - 1024 = 3600. Therefore, QM² = 3600 / 4 = 900. Therefore, QM is sqrt(900) = 30.

Wait, so the length of the median QM is 30? That seems straightforward. Let me check if I did that correctly.

Alternatively, maybe I can verify using coordinates. Let me try coordinate geometry to confirm. Let me place the triangle in coordinate plane. Let me set point P at (-16, 0) and point R at (16, 0), so that PR is 32 units long, lying along the x-axis. Then the midpoint M of PR is at (0, 0). Wait, no. Wait, if P is at (-16, 0) and R is at (16, 0), then midpoint M is at ((-16 + 16)/2, (0 + 0)/2) = (0, 0). Wait, but then point Q is somewhere above the x-axis. Since PQ = QR = 34, so Q is equidistant from P and R, so it lies along the perpendicular bisector of PR, which is the y-axis. Therefore, coordinates of Q are (0, h), where h is the height of the triangle.

So, let me compute h. The distance from Q to P is 34. The coordinates of P are (-16, 0), coordinates of Q are (0, h). Therefore, distance PQ is sqrt[ (0 - (-16))² + (h - 0)² ] = sqrt[16² + h²] = sqrt(256 + h²). This is equal to 34. Therefore:

sqrt(256 + h²) = 34

Squaring both sides:

256 + h² = 1156

Therefore, h² = 1156 - 256 = 900

Therefore, h = 30. Therefore, coordinates of Q are (0, 30). Therefore, the median QM is from Q(0, 30) to M(0, 0). Wait, so the distance between (0, 30) and (0, 0) is 30 units. Therefore, QM is 30.

So that confirms the earlier result. Therefore, the length of the median QM is 30.

Alternatively, maybe I can use the Pythagorean theorem. Since the triangle is isosceles, the median from Q to PR is also the altitude and the angle bisector. Therefore, in this case, since we have the base PR = 32, the median QM splits it into two equal parts of 16 each. Then triangle PQM is a right triangle with legs QM and PM, and hypotenuse PQ. Therefore, by Pythagoras:

QM² + PM² = PQ²

Therefore, QM² + 16² = 34²

Therefore, QM² = 34² - 16² = (34 - 16)(34 + 16) = 18 * 50 = 900

Therefore, QM = sqrt(900) = 30.

Same result. So all three methods: Apollonius's theorem, coordinate geometry, and Pythagorean theorem, give me the same answer. Therefore, I can be confident that the length of the median QM is 30.

But let me just make sure I didn't mix up any sides. The problem says PQ = QR = 34. So, in triangle PQR, sides PQ and QR are both 34. Therefore, the triangle is isosceles with equal sides PQ and QR, so the base is PR. Therefore, the median from Q to PR is indeed the altitude, which splits the base into two equal parts. Therefore, the right triangle approach is valid here.

Alternatively, if I had messed up the sides, maybe I would have a different answer. For example, if I thought that the median was to another side, but no, the problem says "median QM", so Q is the vertex, and M is the midpoint of the opposite side. Since the triangle is PQR, the opposite side of Q is PR. Therefore, M is the midpoint of PR.

Therefore, all steps are correct. Therefore, the answer is 30.

**Final Answer**
The length of median $\overline{QM}$ is $\boxed{30}$.

To find the length of the median $\overline{QM}$ in triangle $PQR$, we are given the following side lengths:

- $PQ = QR = 34$
- $PR = 32$

---

### Step 1: Understand the Triangle Structure

Since $PQ = QR = 34$, triangle $PQR$ is **isosceles** with the equal sides $PQ$ and $QR$, and the base $PR = 32$. The median $\overline{QM}$ connects vertex $Q$ to the **midpoint $M$** of the base $PR$.

This makes $\overline{QM}$ not only a median but also the **altitude** and **angle bisector** of the isosceles triangle, due to the symmetry of the triangle.

---

### Step 2: Use the Pythagorean Theorem

Since $\overline{QM}$ is the altitude, it splits the base $PR$ into two equal segments:

$$
PM = MR = \frac{PR}{2} = \frac{32}{2} = 16
$$

Now consider the right triangle $\triangle PQM$, where:

- $PQ = 34$ (hypotenuse)
- $PM = 16$ (one leg)
- $QM$ (the other leg, which we want to find)

Apply the Pythagorean Theorem:

$$
QM^2 + PM^2 = PQ^2
$$
$$
QM^2 + 16^2 = 34^2
$$
$$
QM^2 + 256 = 1156
$$
$$
QM^2 = 1156 - 256 = 900
$$
$$
QM = \sqrt{900} = 30
$$

---

### Step 3: Confirm with Apollonius's Theorem (Optional)

Apollonius's Theorem states that for any triangle, the square of the length of a median to a side is equal to:

$$
m^2 = \frac{2b^2 + 2c^2 - a^2}{4}
$$

Here, the median is from $Q$ to side $PR = a = 32$, and the other two sides are $PQ = QR = 34$. So:

$$
QM^2 = \frac{2(34)^2 + 2(34)^2 - (32)^2}{4}
$$
$$
QM^2 = \frac{4(34)^2 - (32)^2}{4}
$$
$$
QM^2 = \frac{4(1156) - 1024}{4} = \frac{4624 - 1024}{4} = \frac{3600}{4} = 900
$$
$$
QM = \sqrt{900} = 30
$$

---

### Final Answer

$$
\boxed{30}
$$

Solution:
Because $\triangle PQR$ is isosceles with $PQ=QR$, median $\overline{QM}$ is also an altitude:

[asy]
size(100);
pair P,Q,R,M,NN;
P = (0,0);
Q = (0.5,0.9);
R = (1,0);
NN = (0.5,0);
M = (Q+R)/2;
draw(rightanglemark(Q,NN,P,2.5));
draw(P--Q--R--P);
draw(Q--NN);
label("$P$",P,SW);
label("$R$",R,SE);
label("$Q$",Q,N);
label("$M$",NN,S);
[/asy]

We have $MP = PR/2 = 16$, so right triangle $PQM$ gives us  \begin{align*}
QM &= \sqrt{PQ^2 - PM^2}\\
&= \sqrt{34^2 - 16^2}\\
&= \sqrt{(34-16)(34+16)}\\
& = \boxed{30}.
\end{align*}  (We might also have recognized that $PM/PQ = 8/17$, so $QM/PQ = 15/17$.)


Explanation:
Okay, so I need to find the number of different possible rational roots for this polynomial: $2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1 = 0$, where the coefficients $a_3$, $a_2$, and $a_1$ are integers. Hmm, let me recall the Rational Root Theorem.

From what I remember, the Rational Root Theorem states that any possible rational root of a polynomial equation with integer coefficients, say $p(x) = 0$, is of the form $\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. So, in this case, the constant term is 1, and the leading coefficient is 2.

Let me write that down. The possible rational roots are $\frac{p}{q}$, where $p$ divides 1 and $q$ divides 2. Since 1 has only two factors, 1 and -1, and 2 has factors 1, -1, 2, -2. Therefore, the possible rational roots would be $\pm1$, $\pm\frac{1}{2}$.

Wait, so that would be four possible rational roots? But the question says "the number of different possible rational roots of this polynomial." But hold on, does that mean that all of these are possible? Or are there some restrictions because the coefficients $a_3$, $a_2$, $a_1$ are integers?

Wait, the Rational Root Theorem gives all possible candidates for rational roots, regardless of the specific coefficients. However, the coefficients here are not fixed; they are variables (integers). So, does that mean that all these possible rational roots are possible? Or are there some that can't be achieved because of the coefficients?

Wait, the question is asking for the number of different possible rational roots that this polynomial can have. So, depending on the values of $a_3$, $a_2$, and $a_1$, different rational roots might be possible. But the question is not asking for the number of rational roots for a specific polynomial, but rather the number of different rational roots that could exist for any such polynomial with integer coefficients.

So, in other words, we need to find all rational numbers that can be roots of some polynomial of the given form. Therefore, the answer would be the set of all possible $\frac{p}{q}$ where $p$ divides 1 and $q$ divides 2, which are $\pm1$, $\pm\frac{1}{2}$. Therefore, there are four possible rational roots.

But wait, let me check if all of these are actually possible. For example, can 1 be a root? Let's substitute x=1 into the polynomial:

$2(1)^4 + a_3(1)^3 + a_2(1)^2 + a_1(1) + 1 = 2 + a_3 + a_2 + a_1 + 1 = a_3 + a_2 + a_1 + 3$. For this to be zero, we need $a_3 + a_2 + a_1 = -3$. Since $a_3$, $a_2$, and $a_1$ are integers, this is possible. For example, set $a_3 = -3$, $a_2 = 0$, $a_1 = 0$. Then the polynomial becomes $2x^4 - 3x^3 + 0x^2 + 0x + 1 = 0$, which has 1 as a root.

Similarly, check x = -1:

$2(-1)^4 + a_3(-1)^3 + a_2(-1)^2 + a_1(-1) + 1 = 2(1) + a_3(-1) + a_2(1) + a_1(-1) + 1 = 2 - a_3 + a_2 - a_1 + 1 = 3 - a_3 + a_2 - a_1$. Setting this equal to zero: $ -a_3 + a_2 - a_1 = -3 $, so $a_3 - a_2 + a_1 = 3$. Again, possible with integer coefficients. For example, set $a_3 = 3$, $a_2 = 0$, $a_1 = 0$. Then the polynomial is $2x^4 + 3x^3 + 0x^2 + 0x + 1 = 0$, which has -1 as a root.

Now check x = 1/2. Let's substitute x = 1/2:

$2(1/2)^4 + a_3(1/2)^3 + a_2(1/2)^2 + a_1(1/2) + 1 = 2*(1/16) + a_3*(1/8) + a_2*(1/4) + a_1*(1/2) + 1 = 1/8 + a_3/8 + a_2/4 + a_1/2 + 1$. Let me combine terms:

Convert all to eighths:

1/8 + a_3/8 + 2a_2/8 + 4a_1/8 + 8/8 = [1 + a_3 + 2a_2 + 4a_1 + 8]/8 = [9 + a_3 + 2a_2 + 4a_1]/8.

Set this equal to zero: 9 + a_3 + 2a_2 + 4a_1 = 0. So, a_3 + 2a_2 + 4a_1 = -9. Since a_3, a_2, a_1 are integers, this is possible. For example, set a_1 = -2, then 4a_1 = -8. Then a_3 + 2a_2 = -1. Let me choose a_2 = 0, then a_3 = -1. So coefficients would be a_3 = -1, a_2 = 0, a_1 = -2. Then the polynomial is 2x^4 - x^3 + 0x^2 - 2x + 1. Let me check if x=1/2 is a root:

2*(1/2)^4 - (1/2)^3 - 2*(1/2) + 1 = 2*(1/16) - 1/8 - 1 + 1 = 1/8 - 1/8 -1 +1 = 0. Yes, that works. So x=1/2 is possible.

Similarly, check x = -1/2. Substitute x = -1/2:

2*(-1/2)^4 + a_3*(-1/2)^3 + a_2*(-1/2)^2 + a_1*(-1/2) + 1 = 2*(1/16) + a_3*(-1/8) + a_2*(1/4) + a_1*(-1/2) + 1 = 1/8 - a_3/8 + a_2/4 - a_1/2 + 1. Convert to eighths:

1/8 - a_3/8 + 2a_2/8 - 4a_1/8 + 8/8 = [1 - a_3 + 2a_2 - 4a_1 + 8]/8 = [9 - a_3 + 2a_2 - 4a_1]/8. Set equal to zero: 9 - a_3 + 2a_2 - 4a_1 = 0 => -a_3 + 2a_2 - 4a_1 = -9 => a_3 - 2a_2 + 4a_1 = 9. Again, possible with integers. For example, set a_1 = 2, then 4a_1 = 8. Then a_3 - 2a_2 = 1. Let me choose a_2 = 0, then a_3 = 1. So coefficients are a_3 = 1, a_2 = 0, a_1 = 2. Then the polynomial is 2x^4 + x^3 + 0x^2 + 2x + 1. Let me check x=-1/2:

2*(1/16) + (-1/8) + 0 + 2*(-1/2) + 1 = 1/8 - 1/8 -1 + 1 = 0. Yes, that works. So x=-1/2 is possible.

Therefore, all four possible rational roots are achievable with appropriate integer coefficients. Therefore, the number of different possible rational roots is 4.

Wait, but hold on. Let me check if there are any other possible rational roots. The Rational Root Theorem says that any rational root must be of the form ±1, ±1/2. So there are only four possibilities. Therefore, the answer is 4.

But let me think again. Suppose someone says, maybe there are other roots? For example, could there be a root like 2? But according to the Rational Root Theorem, since the leading coefficient is 2 and the constant term is 1, the possible roots are ±1, ±1/2. So 2 is not in that list. Therefore, 2 cannot be a rational root. Similarly, 1/3 is not in the list. Therefore, only those four.

But let me check if maybe there's a case where a root is repeated or something? But the question says "different possible rational roots", so even if a root could be achieved in multiple ways, we just count it once. So since all four are possible, the answer is 4.

Wait, but let me check if maybe some of these roots are not possible. For example, could there be a case where even though the Rational Root Theorem says it's possible, but due to the polynomial being degree 4, maybe some roots are excluded? For example, maybe if you have a root that is rational, but the polynomial factors in some way? But no, the Rational Root Theorem is about possible candidates, not about whether they actually are roots. So as long as we can find coefficients such that the root is satisfied, then it's possible.

So since we have shown that each of the four candidates can be roots with appropriate coefficients, then they are all possible. Therefore, the answer is 4.

But wait, let me check again. The problem says "the number of different possible rational roots of this polynomial." So maybe there's a trick here? For example, maybe some of these roots cannot be roots because of the degree? But no, the polynomial is degree 4, so it can have up to 4 roots. But the question is not about how many roots it can have, but how many different rational roots are possible across all such polynomials.

Alternatively, maybe there's a case where even though the Rational Root Theorem suggests a root, but due to the coefficients being integers, the root cannot be achieved? But we already checked for each of the four candidates, and found that they can be achieved.

Alternatively, maybe there's a mistake in my calculation? Let me check again for x=1/2.

Earlier, I had the equation 9 + a_3 + 2a_2 + 4a_1 = 0. So if I set a_1 = -2, then 4a_1 = -8, so 9 + a_3 + 2a_2 -8 = 0 => 1 + a_3 + 2a_2 = 0 => a_3 + 2a_2 = -1. Then choosing a_2 = 0, a_3 = -1. So the polynomial is 2x^4 -x^3 + 0x^2 -2x +1. Plugging in x=1/2:

2*(1/2)^4 - (1/2)^3 -2*(1/2) +1 = 2*(1/16) - 1/8 -1 +1 = 1/8 - 1/8 -1 +1 = 0. Correct.

Similarly for x=-1/2, the polynomial 2x^4 +x^3 + 0x^2 +2x +1. Plugging in x=-1/2:

2*(1/16) + (-1/8) + 0 + 2*(-1/2) +1 = 1/8 -1/8 -1 +1 = 0. Correct.

So all four roots are possible. Therefore, the answer is 4.

But wait, let me check if maybe there are other rational roots. For example, suppose someone says, what about 1/1? That's 1, which is already considered. Similarly, -1/1 is -1. So no others.

Alternatively, maybe fractions with denominator 2? Like 1/2 and -1/2. So that's covered.

Therefore, the possible rational roots are ±1, ±1/2. So four different ones.

Therefore, the answer is 4.

But wait, hold on. Let me check if maybe there's a case where the polynomial could have a root that is not in that list? For example, suppose someone says, what if the polynomial is 2x^4 + 0x^3 + 0x^2 + 0x +1 = 0. Then the equation is 2x^4 +1 =0. The roots are ±( -1/2)^{1/4}, which are not rational. So in this case, there are no rational roots. But the question is not about the number of rational roots for a specific polynomial, but the number of different possible rational roots that could exist for any such polynomial.

So even though some polynomials of this form don't have rational roots, the question is about the possible ones. So as long as there exists at least one polynomial of this form with that rational root, then it counts.

Therefore, since we have shown that each of ±1, ±1/2 can be roots of some polynomial of this form, then those are the only possible ones.

Therefore, the answer is 4.

But wait, let me check again. Suppose someone says, what about 2? If I try to plug in x=2 into the polynomial:

2*(2)^4 + a_3*(2)^3 + a_2*(2)^2 + a_1*(2) +1 = 2*16 + 8a_3 + 4a_2 + 2a_1 +1 = 32 + 8a_3 + 4a_2 + 2a_1 +1 = 33 + 8a_3 + 4a_2 + 2a_1. For this to be zero, 8a_3 + 4a_2 + 2a_1 = -33. But the left side is even (since all terms are multiples of 2), and the right side is odd. Therefore, impossible. Therefore, x=2 cannot be a root. Similarly, any other rational number not in the list ±1, ±1/2 would not satisfy the Rational Root Theorem conditions, so they can't be roots.

Therefore, the only possible rational roots are ±1, ±1/2. Therefore, the number of different possible rational roots is 4.

Therefore, the answer is 4.

But wait, let me check if maybe there's a case where a root is repeated or something? But the question says "different possible rational roots", so even if a root can be achieved in multiple ways, it's still counted once.

Therefore, I think the answer is 4.

But let me check once more. The Rational Root Theorem says that any rational root p/q must have p dividing 1 and q dividing 2. Therefore, p is ±1, and q is ±1, ±2. Therefore, the possible roots are ±1, ±1/2. So four roots.

Therefore, the answer is 4.

But wait, the problem says "the number of different possible rational roots of this polynomial." So maybe there's a trick here? For example, maybe some of these roots cannot be achieved because of the polynomial's degree? For example, if the polynomial is of degree 4, can it have, say, 1 as a root? Well, yes, as we saw.

Alternatively, maybe the polynomial could have multiple roots? But the question is not about the number of roots, but the number of different possible rational roots.

Alternatively, maybe there's a case where even though the Rational Root Theorem says it's possible, but due to the coefficients being integers, the root cannot be achieved? But we already checked for each of the four roots, and found that they can be achieved.

Therefore, I think the answer is 4.

But let me check once again. Suppose someone says, what about 1/1? That's 1, which is already considered. Similarly, -1/1 is -1. So no other fractions.

Therefore, the answer is 4.

**Final Answer**
\boxed{4}

To determine the number of **different possible rational roots** of the polynomial

$$
2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1 = 0,
$$

where $ a_3, a_2, a_1 \in \mathbb{Z} $, we apply the **Rational Root Theorem**.

---

### Step 1: Apply the Rational Root Theorem

The Rational Root Theorem states that any rational root $ \frac{p}{q} $ of a polynomial with integer coefficients must satisfy:

- $ p $ divides the **constant term** (here, 1),
- $ q $ divides the **leading coefficient** (here, 2).

So, the possible values of $ p $ are the divisors of 1: $ \pm1 $, and the possible values of $ q $ are the divisors of 2: $ \pm1, \pm2 $.

Thus, the possible rational roots are:

$$
\frac{p}{q} = \pm1, \pm\frac{1}{2}.
$$

So, the **set of possible rational roots** is:

$$
\left\{1, -1, \frac{1}{2}, -\frac{1}{2} \right\}.
$$

---

### Step 2: Determine if These Roots Are Actually Possible

We now verify that **each of these four values** can be a root of some polynomial of the given form with integer coefficients.

#### Case 1: $ x = 1 $

Substitute into the polynomial:

$$
2(1)^4 + a_3(1)^3 + a_2(1)^2 + a_1(1) + 1 = 2 + a_3 + a_2 + a_1 + 1 = a_3 + a_2 + a_1 + 3.
$$

Set this equal to 0:

$$
a_3 + a_2 + a_1 = -3.
$$

This is possible with integer values (e.g., $ a_3 = -3, a_2 = 0, a_1 = 0 $).

#### Case 2: $ x = -1 $

Substitute:

$$
2(-1)^4 + a_3(-1)^3 + a_2(-1)^2 + a_1(-1) + 1 = 2 - a_3 + a_2 - a_1 + 1 = 3 - a_3 + a_2 - a_1.
$$

Set equal to 0:

$$
a_3 - a_2 + a_1 = 3.
$$

Again, this is possible with integer values (e.g., $ a_3 = 3, a_2 = 0, a_1 = 0 $).

#### Case 3: $ x = \frac{1}{2} $

Substitute:

$$
2\left(\frac{1}{2}\right)^4 + a_3\left(\frac{1}{2}\right)^3 + a_2\left(\frac{1}{2}\right)^2 + a_1\left(\frac{1}{2}\right) + 1 = \frac{1}{8} + \frac{a_3}{8} + \frac{a_2}{4} + \frac{a_1}{2} + 1.
$$

Convert to a common denominator (8):

$$
\frac{1 + a_3 + 2a_2 + 4a_1 + 8}{8} = \frac{9 + a_3 + 2a_2 + 4a_1}{8}.
$$

Set equal to 0:

$$
9 + a_3 + 2a_2 + 4a_1 = 0 \Rightarrow a_3 + 2a_2 + 4a_1 = -9.
$$

This is possible with integer values (e.g., $ a_1 = -2, a_2 = 0, a_3 = -1 $).

#### Case 4: $ x = -\frac{1}{2} $

Substitute:

$$
2\left(-\frac{1}{2}\right)^4 + a_3\left(-\frac{1}{2}\right)^3 + a_2\left(-\frac{1}{2}\right)^2 + a_1\left(-\frac{1}{2}\right) + 1 = \frac{1}{8} - \frac{a_3}{8} + \frac{a_2}{4} - \frac{a_1}{2} + 1.
$$

Convert to a common denominator (8):

$$
\frac{1 - a_3 + 2a_2 - 4a_1 + 8}{8} = \frac{9 - a_3 + 2a_2 - 4a_1}{8}.
$$

Set equal to 0:

$$
9 - a_3 + 2a_2 - 4a_1 = 0 \Rightarrow a_3 - 2a_2 + 4a_1 = 9.
$$

This is possible with integer values (e.g., $ a_1 = 2, a_2 = 0, a_3 = 1 $).

---

### Step 3: Conclusion

All four values $ \pm1, \pm\frac{1}{2} $ are **possible rational roots** of the polynomial for some choice of integer coefficients $ a_3, a_2, a_1 $. No other rational numbers can be roots, as they would not satisfy the Rational Root Theorem conditions.

Thus, the number of **different possible rational roots** is:

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to find the number of different possible rational roots for this polynomial: $2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1 = 0$, where the coefficients $a_3$, $a_2$, and $a_1$ are integers. Hmm, let me recall the Rational Root Theorem.

From what I remember, the Rational Root Theorem states that any possible rational root of a polynomial equation with integer coefficients, say $p(x) = 0$, is of the form $\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. So, in this case, the constant term is 1, and the leading coefficient is 2.

Let me write that down. The possible rational roots are $\frac{p}{q}$, where $p$ divides 1 and $q$ divides 2. Since 1 has only two factors, 1 and -1, and 2 has factors 1, -1, 2, -2. Therefore, the possible rational roots would be $\pm1$, $\pm\frac{1}{2}$.

Wait, so that would be four possible rational roots? But the question says "the number of different possible rational roots of this polynomial." But hold on, does that mean that all of these are possible? Or are there some restrictions because the coefficients $a_3$, $a_2$, $a_1$ are integers?

Wait, the Rational Root Theorem gives all possible candidates for rational roots, regardless of the specific coefficients. However, the coefficients here are not fixed; they are variables (integers). So, does that mean that all these possible rational roots are possible? Or are there some that can't be achieved because of the coefficients?

Wait, the question is asking for the number of different possible rational roots that this polynomial can have. So, depending on the values of $a_3$, $a_2$, and $a_1$, different rational roots might be possible. But the question is not asking for the number of rational roots for a specific polynomial, but rather the number of different rational roots that could exist for any such polynomial with integer coefficients.

So, in other words, we need to find all rational numbers that can be roots of some polynomial of the given form. Therefore, the answer would be the set of all possible $\frac{p}{q}$ where $p$ divides 1 and $q$ divides 2, which are $\pm1$, $\pm\frac{1}{2}$. Therefore, there are four possible rational roots.

But wait, let me check if all of these are actually possible. For example, can 1 be a root? Let's substitute x=1 into the polynomial:

$2(1)^4 + a_3(1)^3 + a_2(1)^2 + a_1(1) + 1 = 2 + a_3 + a_2 + a_1 + 1 = a_3 + a_2 + a_1 + 3$. For this to be zero, we need $a_3 + a_2 + a_1 = -3$. Since $a_3$, $a_2$, and $a_1$ are integers, this is possible. For example, set $a_3 = -3$, $a_2 = 0$, $a_1 = 0$. Then the polynomial becomes $2x^4 - 3x^3 + 0x^2 + 0x + 1 = 0$, which has 1 as a root.

Similarly, check x = -1:

$2(-1)^4 + a_3(-1)^3 + a_2(-1)^2 + a_1(-1) + 1 = 2(1) + a_3(-1) + a_2(1) + a_1(-1) + 1 = 2 - a_3 + a_2 - a_1 + 1 = 3 - a_3 + a_2 - a_1$. Setting this equal to zero: $ -a_3 + a_2 - a_1 = -3 $, so $a_3 - a_2 + a_1 = 3$. Again, possible with integer coefficients. For example, set $a_3 = 3$, $a_2 = 0$, $a_1 = 0$. Then the polynomial is $2x^4 + 3x^3 + 0x^2 + 0x + 1 = 0$, which has -1 as a root.

Now check x = 1/2. Let's substitute x = 1/2:

$2(1/2)^4 + a_3(1/2)^3 + a_2(1/2)^2 + a_1(1/2) + 1 = 2*(1/16) + a_3*(1/8) + a_2*(1/4) + a_1*(1/2) + 1 = 1/8 + a_3/8 + a_2/4 + a_1/2 + 1$. Let me combine terms:

Convert all to eighths:

1/8 + a_3/8 + 2a_2/8 + 4a_1/8 + 8/8 = [1 + a_3 + 2a_2 + 4a_1 + 8]/8 = [9 + a_3 + 2a_2 + 4a_1]/8.

Set this equal to zero: 9 + a_3 + 2a_2 + 4a_1 = 0. So, a_3 + 2a_2 + 4a_1 = -9. Since a_3, a_2, a_1 are integers, this is possible. For example, set a_1 = -2, then 4a_1 = -8. Then a_3 + 2a_2 = -1. Let me choose a_2 = 0, then a_3 = -1. So coefficients would be a_3 = -1, a_2 = 0, a_1 = -2. Then the polynomial is 2x^4 - x^3 + 0x^2 - 2x + 1. Let me check if x=1/2 is a root:

2*(1/2)^4 - (1/2)^3 - 2*(1/2) + 1 = 2*(1/16) - 1/8 - 1 + 1 = 1/8 - 1/8 -1 +1 = 0. Yes, that works. So x=1/2 is possible.

Similarly, check x = -1/2. Substitute x = -1/2:

2*(-1/2)^4 + a_3*(-1/2)^3 + a_2*(-1/2)^2 + a_1*(-1/2) + 1 = 2*(1/16) + a_3*(-1/8) + a_2*(1/4) + a_1*(-1/2) + 1 = 1/8 - a_3/8 + a_2/4 - a_1/2 + 1. Convert to eighths:

1/8 - a_3/8 + 2a_2/8 - 4a_1/8 + 8/8 = [1 - a_3 + 2a_2 - 4a_1 + 8]/8 = [9 - a_3 + 2a_2 - 4a_1]/8. Set equal to zero: 9 - a_3 + 2a_2 - 4a_1 = 0 => -a_3 + 2a_2 - 4a_1 = -9 => a_3 - 2a_2 + 4a_1 = 9. Again, possible with integers. For example, set a_1 = 2, then 4a_1 = 8. Then a_3 - 2a_2 = 1. Let me choose a_2 = 0, then a_3 = 1. So coefficients are a_3 = 1, a_2 = 0, a_1 = 2. Then the polynomial is 2x^4 + x^3 + 0x^2 + 2x + 1. Let me check x=-1/2:

2*(1/16) + (-1/8) + 0 + 2*(-1/2) + 1 = 1/8 - 1/8 -1 + 1 = 0. Yes, that works. So x=-1/2 is possible.

Therefore, all four possible rational roots are achievable with appropriate integer coefficients. Therefore, the number of different possible rational roots is 4.

Wait, but hold on. Let me check if there are any other possible rational roots. The Rational Root Theorem says that any rational root must be of the form ±1, ±1/2. So there are only four possibilities. Therefore, the answer is 4.

But let me think again. Suppose someone says, maybe there are other roots? For example, could there be a root like 2? But according to the Rational Root Theorem, since the leading coefficient is 2 and the constant term is 1, the possible roots are ±1, ±1/2. So 2 is not in that list. Therefore, 2 cannot be a rational root. Similarly, 1/3 is not in the list. Therefore, only those four.

But let me check if maybe there's a case where a root is repeated or something? But the question says "different possible rational roots", so even if a root could be achieved in multiple ways, we just count it once. So since all four are possible, the answer is 4.

Wait, but let me check if maybe some of these roots are not possible. For example, could there be a case where even though the Rational Root Theorem says it's possible, but due to the polynomial being degree 4, maybe some roots are excluded? For example, maybe if you have a root that is rational, but the polynomial factors in some way? But no, the Rational Root Theorem is about possible candidates, not about whether they actually are roots. So as long as we can find coefficients such that the root is satisfied, then it's possible.

So since we have shown that each of the four candidates can be roots with appropriate coefficients, then they are all possible. Therefore, the answer is 4.

But wait, let me check again. The problem says "the number of different possible rational roots of this polynomial." So maybe there's a trick here? For example, maybe some of these roots cannot be roots because of the degree? But no, the polynomial is degree 4, so it can have up to 4 roots. But the question is not about how many roots it can have, but how many different rational roots are possible across all such polynomials.

Alternatively, maybe there's a case where even though the Rational Root Theorem suggests a root, but due to the coefficients being integers, the root cannot be achieved? But we already checked for each of the four candidates, and found that they can be achieved.

Alternatively, maybe there's a mistake in my calculation? Let me check again for x=1/2.

Earlier, I had the equation 9 + a_3 + 2a_2 + 4a_1 = 0. So if I set a_1 = -2, then 4a_1 = -8, so 9 + a_3 + 2a_2 -8 = 0 => 1 + a_3 + 2a_2 = 0 => a_3 + 2a_2 = -1. Then choosing a_2 = 0, a_3 = -1. So the polynomial is 2x^4 -x^3 + 0x^2 -2x +1. Plugging in x=1/2:

2*(1/2)^4 - (1/2)^3 -2*(1/2) +1 = 2*(1/16) - 1/8 -1 +1 = 1/8 - 1/8 -1 +1 = 0. Correct.

Similarly for x=-1/2, the polynomial 2x^4 +x^3 + 0x^2 +2x +1. Plugging in x=-1/2:

2*(1/16) + (-1/8) + 0 + 2*(-1/2) +1 = 1/8 -1/8 -1 +1 = 0. Correct.

So all four roots are possible. Therefore, the answer is 4.

But wait, let me check if maybe there are other rational roots. For example, suppose someone says, what about 1/1? That's 1, which is already considered. Similarly, -1/1 is -1. So no others.

Alternatively, maybe fractions with denominator 2? Like 1/2 and -1/2. So that's covered.

Therefore, the possible rational roots are ±1, ±1/2. So four different ones.

Therefore, the answer is 4.

But wait, hold on. Let me check if maybe there's a case where the polynomial could have a root that is not in that list? For example, suppose someone says, what if the polynomial is 2x^4 + 0x^3 + 0x^2 + 0x +1 = 0. Then the equation is 2x^4 +1 =0. The roots are ±( -1/2)^{1/4}, which are not rational. So in this case, there are no rational roots. But the question is not about the number of rational roots for a specific polynomial, but the number of different possible rational roots that could exist for any such polynomial.

So even though some polynomials of this form don't have rational roots, the question is about the possible ones. So as long as there exists at least one polynomial of this form with that rational root, then it counts.

Therefore, since we have shown that each of ±1, ±1/2 can be roots of some polynomial of this form, then those are the only possible ones.

Therefore, the answer is 4.

But wait, let me check again. Suppose someone says, what about 2? If I try to plug in x=2 into the polynomial:

2*(2)^4 + a_3*(2)^3 + a_2*(2)^2 + a_1*(2) +1 = 2*16 + 8a_3 + 4a_2 + 2a_1 +1 = 32 + 8a_3 + 4a_2 + 2a_1 +1 = 33 + 8a_3 + 4a_2 + 2a_1. For this to be zero, 8a_3 + 4a_2 + 2a_1 = -33. But the left side is even (since all terms are multiples of 2), and the right side is odd. Therefore, impossible. Therefore, x=2 cannot be a root. Similarly, any other rational number not in the list ±1, ±1/2 would not satisfy the Rational Root Theorem conditions, so they can't be roots.

Therefore, the only possible rational roots are ±1, ±1/2. Therefore, the number of different possible rational roots is 4.

Therefore, the answer is 4.

But wait, let me check if maybe there's a case where a root is repeated or something? But the question says "different possible rational roots", so even if a root can be achieved in multiple ways, it's still counted once.

Therefore, I think the answer is 4.

But let me check once more. The Rational Root Theorem says that any rational root p/q must have p dividing 1 and q dividing 2. Therefore, p is ±1, and q is ±1, ±2. Therefore, the possible roots are ±1, ±1/2. So four roots.

Therefore, the answer is 4.

But wait, the problem says "the number of different possible rational roots of this polynomial." So maybe there's a trick here? For example, maybe some of these roots cannot be achieved because of the polynomial's degree? For example, if the polynomial is of degree 4, can it have, say, 1 as a root? Well, yes, as we saw.

Alternatively, maybe the polynomial could have multiple roots? But the question is not about the number of roots, but the number of different possible rational roots.

Alternatively, maybe there's a case where even though the Rational Root Theorem says it's possible, but due to the coefficients being integers, the root cannot be achieved? But we already checked for each of the four roots, and found that they can be achieved.

Therefore, I think the answer is 4.

But let me check once again. Suppose someone says, what about 1/1? That's 1, which is already considered. Similarly, -1/1 is -1. So no other fractions.

Therefore, the answer is 4.

**Final Answer**
\boxed{4}

To determine the number of **different possible rational roots** of the polynomial

$$
2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1 = 0,
$$

where $ a_3, a_2, a_1 \in \mathbb{Z} $, we apply the **Rational Root Theorem**.

---

### Step 1: Apply the Rational Root Theorem

The Rational Root Theorem states that any rational root $ \frac{p}{q} $ of a polynomial with integer coefficients must satisfy:

- $ p $ divides the **constant term** (here, 1),
- $ q $ divides the **leading coefficient** (here, 2).

So, the possible values of $ p $ are the divisors of 1: $ \pm1 $, and the possible values of $ q $ are the divisors of 2: $ \pm1, \pm2 $.

Thus, the possible rational roots are:

$$
\frac{p}{q} = \pm1, \pm\frac{1}{2}.
$$

So, the **set of possible rational roots** is:

$$
\left\{1, -1, \frac{1}{2}, -\frac{1}{2} \right\}.
$$

---

### Step 2: Determine if These Roots Are Actually Possible

We now verify that **each of these four values** can be a root of some polynomial of the given form with integer coefficients.

#### Case 1: $ x = 1 $

Substitute into the polynomial:

$$
2(1)^4 + a_3(1)^3 + a_2(1)^2 + a_1(1) + 1 = 2 + a_3 + a_2 + a_1 + 1 = a_3 + a_2 + a_1 + 3.
$$

Set this equal to 0:

$$
a_3 + a_2 + a_1 = -3.
$$

This is possible with integer values (e.g., $ a_3 = -3, a_2 = 0, a_1 = 0 $).

#### Case 2: $ x = -1 $

Substitute:

$$
2(-1)^4 + a_3(-1)^3 + a_2(-1)^2 + a_1(-1) + 1 = 2 - a_3 + a_2 - a_1 + 1 = 3 - a_3 + a_2 - a_1.
$$

Set equal to 0:

$$
a_3 - a_2 + a_1 = 3.
$$

Again, this is possible with integer values (e.g., $ a_3 = 3, a_2 = 0, a_1 = 0 $).

#### Case 3: $ x = \frac{1}{2} $

Substitute:

$$
2\left(\frac{1}{2}\right)^4 + a_3\left(\frac{1}{2}\right)^3 + a_2\left(\frac{1}{2}\right)^2 + a_1\left(\frac{1}{2}\right) + 1 = \frac{1}{8} + \frac{a_3}{8} + \frac{a_2}{4} + \frac{a_1}{2} + 1.
$$

Convert to a common denominator (8):

$$
\frac{1 + a_3 + 2a_2 + 4a_1 + 8}{8} = \frac{9 + a_3 + 2a_2 + 4a_1}{8}.
$$

Set equal to 0:

$$
9 + a_3 + 2a_2 + 4a_1 = 0 \Rightarrow a_3 + 2a_2 + 4a_1 = -9.
$$

This is possible with integer values (e.g., $ a_1 = -2, a_2 = 0, a_3 = -1 $).

#### Case 4: $ x = -\frac{1}{2} $

Substitute:

$$
2\left(-\frac{1}{2}\right)^4 + a_3\left(-\frac{1}{2}\right)^3 + a_2\left(-\frac{1}{2}\right)^2 + a_1\left(-\frac{1}{2}\right) + 1 = \frac{1}{8} - \frac{a_3}{8} + \frac{a_2}{4} - \frac{a_1}{2} + 1.
$$

Convert to a common denominator (8):

$$
\frac{1 - a_3 + 2a_2 - 4a_1 + 8}{8} = \frac{9 - a_3 + 2a_2 - 4a_1}{8}.
$$

Set equal to 0:

$$
9 - a_3 + 2a_2 - 4a_1 = 0 \Rightarrow a_3 - 2a_2 + 4a_1 = 9.
$$

This is possible with integer values (e.g., $ a_1 = 2, a_2 = 0, a_3 = 1 $).

---

### Step 3: Conclusion

All four values $ \pm1, \pm\frac{1}{2} $ are **possible rational roots** of the polynomial for some choice of integer coefficients $ a_3, a_2, a_1 $. No other rational numbers can be roots, as they would not satisfy the Rational Root Theorem conditions.

Thus, the number of **different possible rational roots** is:

$$
\boxed{4}
$$

Solution:
By the Rational Root Theorem, the only possible rational roots are of the form $\pm \frac{a}{b},$ where $a$ divides 1 and $b$ divides 2.  Thus, the possible rational roots are
\[\pm 1, \ \pm \frac{1}{2}.\]Thus, there are $\boxed{4}$ possible rational roots.


Explanation:
Okay, so I need to figure out how many ways 6 girls and 2 boys can sit in a row, but the two boys want to sit next to each other. Hmm, let me think.

First, I remember that when dealing with permutations where certain people need to be together, we can treat them as a single unit or "block." So, if the two boys have to sit together, maybe I can think of them as one combined entity. That way, instead of having 8 individuals, I have 7 entities to arrange: the 6 girls and the 1 boy block.

Wait, let me check that. If there are 6 girls and 2 boys, normally there are 8 people. If the two boys are sitting together, they form a single block, so the total number of entities becomes 6 girls + 1 block = 7 entities. Yeah, that makes sense.

Now, how many ways can these 7 entities be arranged? Well, the number of permutations of 7 distinct entities is 7 factorial, which is 7! So that would be 7! ways. But wait, the two boys within their block can also switch places, right? Like, Boy A can be on the left and Boy B on the right, or vice versa. So, for each arrangement of the 7 entities, there are 2 ways to arrange the boys inside their block.

Therefore, the total number of arrangements should be 7! multiplied by 2. Let me write that down: Total ways = 2 * 7!.

But let me make sure I didn't miss anything. Let me think again.

Alternatively, another approach: First, consider the two boys as a single unit. So, we have 7 units total. The number of ways to arrange these 7 units is 7!. Then, the two boys can be arranged among themselves in 2! ways. So, total ways = 7! * 2! = 2 * 7!.

Yes, that seems consistent.

But let me check with another method to be safe. Suppose I don't use the block method. How else could I approach this?

Well, the total number of ways without any restrictions is 8! (since there are 8 people). But we have a restriction here. The two boys must sit together.

Alternatively, I can calculate the number of positions where the two boys can sit together and then multiply by the number of ways to arrange the girls and boys.

So, how many ways can the two boys sit together in a row of 8 seats?

First, think of the two boys as needing to occupy two adjacent seats. In a row of 8 seats, how many adjacent pairs are there? Let's see: positions 1-2, 2-3, 3-4, 4-5, 5-6, 6-7, 7-8. So that's 7 possible adjacent pairs.

For each of these pairs, the two boys can be arranged in 2! ways (since they can switch places). Then, the remaining 6 girls can be arranged in the remaining 6 seats in 6! ways.

Therefore, total number of arrangements would be 7 (positions for the boys) * 2! (arrangements of the boys) * 6! (arrangements of the girls).

Calculating that: 7 * 2 * 6! = 14 * 6!.

But wait, let me check if this is equal to 2 * 7!.

Because 7! is 7 * 6! So 2 * 7! = 2 * 7 * 6! = 14 * 6! Which matches the previous result.

So both methods give the same answer. That's reassuring.

Therefore, the answer should be 2 * 7! or 14 * 6!.

Let me compute the numerical value just to have an idea.

7! is 5040, so 2 * 5040 = 10080.

Alternatively, 6! is 720, so 14 * 720 = 10080. Yep, same result.

So, the answer is 10080.

But let me think again if there's another way someone might approach this problem and maybe get confused.

For example, someone might think that since the boys have to sit together, we can first arrange the girls and then place the boys in the gaps. Wait, but in this case, since the boys are sitting together, maybe that's not necessary.

Wait, actually, arranging the girls first and then inserting the boys. Let me try that approach.

If we first arrange the 6 girls. The number of ways to arrange the girls is 6!. Now, once the girls are seated, there are 7 possible gaps where the boys can be placed: one before the first girl, one between each pair of girls, and one after the last girl.

But since the boys have to sit together, we need to choose one of these gaps to place the two boys. There are 7 gaps. Then, within that gap, the two boys can be arranged in 2! ways.

Therefore, total number of arrangements would be 6! * 7 * 2! = same as before, 6! * 14 = 10080.

So, same answer again.

Therefore, regardless of the method, the answer is 10080.

But let me check if there's a possibility of overcounting or undercounting.

In the first method, treating the boys as a single block: 7! * 2!

In the second method, considering positions for the boys: 7 * 2! * 6!

In the third method, arranging girls first and inserting the boys: 6! * 7 * 2!

All three methods are equivalent.

So, seems like the answer is consistent.

Alternatively, if someone forgets to multiply by 2! for the boys' arrangement, they would get 7! which is 5040, but that's wrong because the boys can switch places.

Similarly, if someone thought that the number of positions is 8, they might do 8 * 6! which is 8 * 720 = 5760, which is different. But that's wrong because the number of adjacent pairs is 7, not 8.

So, that's a common mistake.

Another possible mistake is not considering the boys as a block and just subtracting something from the total, but that's more complicated.

But since we have multiple methods leading to the same answer, I think 10080 is correct.

Therefore, the number of ways is 2 * 7! = 10080.

**Final Answer**
The number of ways is \boxed{10080}.

To determine the number of ways 6 girls and 2 boys can sit in a row **with the condition that the two boys must sit next to each other**, we can approach the problem using a method that simplifies the constraint by treating the two boys as a single unit or "block."

---

### Step 1: Treat the Two Boys as a Single Unit

Since the two boys must sit together, we can consider them as a single entity or "block." This reduces the total number of entities to arrange from 8 (6 girls + 2 boys) to 7: the 6 girls and 1 block of boys.

---

### Step 2: Arrange the 7 Entities

Now, we have 7 entities to arrange: 6 girls and 1 block. The number of ways to arrange these 7 entities is:

$$
7! = 5040
$$

---

### Step 3: Account for Internal Arrangement of the Boys

Within the block of two boys, the two boys can be arranged in 2! ways (since they can switch places). So, we multiply the number of arrangements of the 7 entities by 2:

$$
2! = 2
$$

---

### Step 4: Compute the Total Number of Arrangements

$$
\text{Total arrangements} = 7! \times 2! = 5040 \times 2 = 10080
$$

---

### Final Answer

$$
\boxed{10080}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many ways 6 girls and 2 boys can sit in a row, but the two boys want to sit next to each other. Hmm, let me think.

First, I remember that when dealing with permutations where certain people need to be together, we can treat them as a single unit or "block." So, if the two boys have to sit together, maybe I can think of them as one combined entity. That way, instead of having 8 individuals, I have 7 entities to arrange: the 6 girls and the 1 boy block.

Wait, let me check that. If there are 6 girls and 2 boys, normally there are 8 people. If the two boys are sitting together, they form a single block, so the total number of entities becomes 6 girls + 1 block = 7 entities. Yeah, that makes sense.

Now, how many ways can these 7 entities be arranged? Well, the number of permutations of 7 distinct entities is 7 factorial, which is 7! So that would be 7! ways. But wait, the two boys within their block can also switch places, right? Like, Boy A can be on the left and Boy B on the right, or vice versa. So, for each arrangement of the 7 entities, there are 2 ways to arrange the boys inside their block.

Therefore, the total number of arrangements should be 7! multiplied by 2. Let me write that down: Total ways = 2 * 7!.

But let me make sure I didn't miss anything. Let me think again.

Alternatively, another approach: First, consider the two boys as a single unit. So, we have 7 units total. The number of ways to arrange these 7 units is 7!. Then, the two boys can be arranged among themselves in 2! ways. So, total ways = 7! * 2! = 2 * 7!.

Yes, that seems consistent.

But let me check with another method to be safe. Suppose I don't use the block method. How else could I approach this?

Well, the total number of ways without any restrictions is 8! (since there are 8 people). But we have a restriction here. The two boys must sit together.

Alternatively, I can calculate the number of positions where the two boys can sit together and then multiply by the number of ways to arrange the girls and boys.

So, how many ways can the two boys sit together in a row of 8 seats?

First, think of the two boys as needing to occupy two adjacent seats. In a row of 8 seats, how many adjacent pairs are there? Let's see: positions 1-2, 2-3, 3-4, 4-5, 5-6, 6-7, 7-8. So that's 7 possible adjacent pairs.

For each of these pairs, the two boys can be arranged in 2! ways (since they can switch places). Then, the remaining 6 girls can be arranged in the remaining 6 seats in 6! ways.

Therefore, total number of arrangements would be 7 (positions for the boys) * 2! (arrangements of the boys) * 6! (arrangements of the girls).

Calculating that: 7 * 2 * 6! = 14 * 6!.

But wait, let me check if this is equal to 2 * 7!.

Because 7! is 7 * 6! So 2 * 7! = 2 * 7 * 6! = 14 * 6! Which matches the previous result.

So both methods give the same answer. That's reassuring.

Therefore, the answer should be 2 * 7! or 14 * 6!.

Let me compute the numerical value just to have an idea.

7! is 5040, so 2 * 5040 = 10080.

Alternatively, 6! is 720, so 14 * 720 = 10080. Yep, same result.

So, the answer is 10080.

But let me think again if there's another way someone might approach this problem and maybe get confused.

For example, someone might think that since the boys have to sit together, we can first arrange the girls and then place the boys in the gaps. Wait, but in this case, since the boys are sitting together, maybe that's not necessary.

Wait, actually, arranging the girls first and then inserting the boys. Let me try that approach.

If we first arrange the 6 girls. The number of ways to arrange the girls is 6!. Now, once the girls are seated, there are 7 possible gaps where the boys can be placed: one before the first girl, one between each pair of girls, and one after the last girl.

But since the boys have to sit together, we need to choose one of these gaps to place the two boys. There are 7 gaps. Then, within that gap, the two boys can be arranged in 2! ways.

Therefore, total number of arrangements would be 6! * 7 * 2! = same as before, 6! * 14 = 10080.

So, same answer again.

Therefore, regardless of the method, the answer is 10080.

But let me check if there's a possibility of overcounting or undercounting.

In the first method, treating the boys as a single block: 7! * 2!

In the second method, considering positions for the boys: 7 * 2! * 6!

In the third method, arranging girls first and inserting the boys: 6! * 7 * 2!

All three methods are equivalent.

So, seems like the answer is consistent.

Alternatively, if someone forgets to multiply by 2! for the boys' arrangement, they would get 7! which is 5040, but that's wrong because the boys can switch places.

Similarly, if someone thought that the number of positions is 8, they might do 8 * 6! which is 8 * 720 = 5760, which is different. But that's wrong because the number of adjacent pairs is 7, not 8.

So, that's a common mistake.

Another possible mistake is not considering the boys as a block and just subtracting something from the total, but that's more complicated.

But since we have multiple methods leading to the same answer, I think 10080 is correct.

Therefore, the number of ways is 2 * 7! = 10080.

**Final Answer**
The number of ways is \boxed{10080}.

To determine the number of ways 6 girls and 2 boys can sit in a row **with the condition that the two boys must sit next to each other**, we can approach the problem using a method that simplifies the constraint by treating the two boys as a single unit or "block."

---

### Step 1: Treat the Two Boys as a Single Unit

Since the two boys must sit together, we can consider them as a single entity or "block." This reduces the total number of entities to arrange from 8 (6 girls + 2 boys) to 7: the 6 girls and 1 block of boys.

---

### Step 2: Arrange the 7 Entities

Now, we have 7 entities to arrange: 6 girls and 1 block. The number of ways to arrange these 7 entities is:

$$
7! = 5040
$$

---

### Step 3: Account for Internal Arrangement of the Boys

Within the block of two boys, the two boys can be arranged in 2! ways (since they can switch places). So, we multiply the number of arrangements of the 7 entities by 2:

$$
2! = 2
$$

---

### Step 4: Compute the Total Number of Arrangements

$$
\text{Total arrangements} = 7! \times 2! = 5040 \times 2 = 10080
$$

---

### Final Answer

$$
\boxed{10080}
$$

Solution: We can consider the two boys as one person, arrange the ``seven'' people first, then arrange the 2 boys.  So the number of seating arrangements in which the boys sit together is $7!\times 2!=\boxed{10,\!080}$.