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, that sounds a bit complicated, but let me break it 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 the cube root.

I remember that roots can be expressed as exponents. Maybe converting all these roots into exponents will make it easier to handle. Let me recall the rules: the nth root of something is the same as raising that something to the power of 1/n. Also, when you have exponents raised to exponents, you multiply them. So, for example, (a^m)^n = a^(m*n).

Let me start from 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 result. So, taking the cube root is raising to the 1/3 power. So that would be [(1/729)^(1/2)]^(1/3). Then, the outermost layer is the square root of that, which is raising to the 1/2 power again. So overall, the entire expression is [( ( (1/729)^(1/2) )^(1/3) )]^(1/2).

Let me write that as exponents step by step. Let me denote the innermost part first:

First, √(1/729) = (1/729)^(1/2)

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

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

So, the entire expression simplifies to (1/729)^(1/12).

Now, I need to write this with a rational denominator. Wait, but (1/729)^(1/12) is already a single term. Maybe I need to express it as a radical with a rational denominator? Let me think.

Alternatively, maybe I need to rationalize the denominator if there was a denominator in the expression. But in this case, the expression is (1/729)^(1/12). Let me see.

Alternatively, perhaps I need to express this as a fraction with a rational denominator. Let me recall that 729 is a power of 3. Let me check: 3^6 is 729 because 3^6 = 729. Let me verify:

3^1 = 3

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = 243

3^6 = 729. Yes, correct. So 729 is 3^6. Therefore, 1/729 is 1/(3^6).

So, substituting back into the expression: (1/(3^6))^(1/12) = 1/(3^(6*(1/12))) = 1/(3^(1/2)) = 1/√3.

Wait, but that seems too simple. Let me check my steps again.

Original expression: √(∛(√(1/729)))

Convert to exponents:

Start with √(1/729) = (1/729)^(1/2)

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

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

Yes, that's correct. Then, since 729 is 3^6, so (1/729)^(1/12) = (3^(-6))^(1/12) = 3^(-6*(1/12)) = 3^(-1/2) = 1/(3^(1/2)) = 1/√3.

But the problem says to write the result with a rational denominator. So, 1/√3 has an irrational denominator. To rationalize the denominator, we multiply numerator and denominator by √3:

(1/√3) * (√3/√3) = √3 / 3.

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

Wait, but let me check again. Let me verify each step once more to make sure I didn't make a mistake.

Starting with √(∛(√(1/729))).

First, √(1/729) is 1 over √729. √729 is 27, since 27^2 is 729. Wait, 27*27 is 729? Let me check: 25^2 is 625, 26^2 is 676, 27^2 is 729. Yes, correct. So √(1/729) is 1/27.

Then, the cube root of 1/27. The cube root of 1/27 is 1 over cube root of 27. Cube root of 27 is 3, since 3^3 is 27. Therefore, cube root of 1/27 is 1/3.

Then, the square root of 1/3. Which is √(1/3) = 1/√3. Which is the same as √3 / 3 after rationalizing.

So, that's the same result as before. Therefore, the answer is √3 / 3.

But let me check if there's another way to approach this. Maybe by converting all the roots into exponents from the beginning.

Original expression: √(∛(√(1/729)))

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)^(1/2 * 1/3) = (1/729)^(1/6)

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

So, same as before.

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

Yes, same result.

Therefore, the simplified expression with a rational denominator is √3 divided by 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, yes, the steps I did are correct.

Alternatively, maybe the problem expects a different form? Let me think.

Alternatively, maybe I need to write the entire expression as a single radical? Let me see.

The original expression is a square root of a cube root of a square root. So, combining all the exponents, as I did before, gives 1/729 raised to the 1/12 power. Which is the same as 729^(-1/12).

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

Same result.

Alternatively, if I wanted to write this as a radical with a rational denominator, maybe I need to express it as a fraction with a rational denominator. But since √3 / 3 is already with a rational denominator, that's probably the answer.

Wait, but let me check if there's a different way to express this. For example, maybe the problem expects a different form? Let me think.

Alternatively, maybe I need to write the entire expression as a single radical? Let me see.

The original expression is sqrt(cbrt(sqrt(1/729))). Let me think about combining all the roots.

The square root is exponent 1/2, the cube root is exponent 1/3, and the square root again is exponent 1/2. So, multiplying the exponents: 1/2 * 1/3 * 1/2? Wait, no. Wait, the exponents are applied in sequence.

Wait, starting with 1/729, then taking square root (1/2), then cube root (1/3), then square root (1/2). So, the total exponent is 1/2 * 1/3 * 1/2? Wait, no. Wait, when you have exponents, you multiply them. So, if you have (a^(1/2))^(1/3) = a^(1/2 * 1/3) = a^(1/6). Then, (a^(1/6))^(1/2) = a^(1/6 * 1/2) = a^(1/12). So, yes, total exponent is 1/12.

Therefore, the entire expression is (1/729)^(1/12) = 1/(729^(1/12)).

But 729^(1/12) is the 12th root of 729. Since 729 is 3^6, then 729^(1/12) = (3^6)^(1/12) = 3^(6/12) = 3^(1/2) = √3. Therefore, 1/(√3) = √3/3.

Same result.

Therefore, regardless of the approach, I end up with √3 / 3.

But let me check if there's a possibility that I made a mistake in the exponent multiplication. Let me verify:

Starting with 1/729, then square root: exponent 1/2. Then cube root: exponent 1/3. Then square root: exponent 1/2. So total exponent is 1/2 * 1/3 * 1/2? Wait, no. Wait, when you have exponents, each operation is multiplying the exponent.

Wait, suppose you have a number x. Then, taking square root is x^(1/2). Then taking cube root of that is (x^(1/2))^(1/3) = x^(1/2 * 1/3) = x^(1/6). Then taking square root of that is (x^(1/6))^(1/2) = x^(1/6 * 1/2) = x^(1/12). So yes, total exponent is 1/12.

Therefore, the entire expression is x^(1/12) where x is 1/729.

So, that's correct.

Alternatively, if I think of the entire expression as a single radical, it would be the 12th root of 1/729. But 1/729 is 3^(-6), so the 12th root of 3^(-6) is 3^(-6/12) = 3^(-1/2) = 1/√3.

Same result.

Therefore, I think my answer is correct.

But let me check with actual numbers. Let me compute the original expression numerically and see if it equals √3 / 3.

First, compute 1/729. Let me calculate that. 1 divided by 729. Let me approximate. 729 is 700 + 29. 1/700 is approximately 0.00142857, and 1/729 is slightly less. Let me compute 1/729:

Well, 729 * 0.00137 = 729 * 0.001 + 729 * 0.00037 = 0.729 + 0.26973 = approx 0.99873. So, 0.00137 * 729 ≈ 0.99873, which is close to 1. Therefore, 1/729 ≈ 0.001371742.

Then, take the square root of that: sqrt(0.001371742). Let me compute that. sqrt(0.001371742). Let me note that sqrt(0.001371742) is approx sqrt(1.371742 * 10^-3) = sqrt(1.371742) * 10^(-3/2). sqrt(1.371742) is approx 1.171, since 1.17^2 = 1.3689, which is close. So approx 1.171 * 10^(-1.5) = 1.171 / 10^(1.5) = 1.171 / (10 * sqrt(10)) ≈ 1.171 / (10 * 3.1623) ≈ 1.171 / 31.623 ≈ 0.037.

So sqrt(1/729) ≈ 0.037.

Then, take the cube root of that. So cube root of 0.037. Let me approximate. 0.037^(1/3). Let me note that 0.037 is approx 3.7 * 10^-2. Let me think of cube roots. 0.064 is 0.4^3, 0.027 is 0.3^3. So 0.037 is between 0.027 and 0.064. Let me try 0.33^3 = 0.035937, which is very close to 0.037. So cube root of 0.037 is approx 0.33.

Then, take the square root of that. sqrt(0.33). sqrt(0.33) is approx 0.574.

Now, let me compute √3 / 3. √3 is approx 1.732, divided by 3 is approx 0.577. Which is very close to the approximate value I just got (0.574). The slight difference is due to my approximations in the steps. Therefore, this numerical check seems to confirm that the answer is indeed √3 / 3.

Therefore, I think my answer is correct.

But just to make sure, let me check if there's another way to write √3 / 3 with a rational denominator. Wait, √3 / 3 already has a rational denominator because the denominator is 3, which is rational. The numerator is irrational, but the denominator is rational. So, that's acceptable.

Alternatively, if the problem had a more complicated expression with multiple radicals in the denominator, we might need to rationalize further, but in this case, it's already done.

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

So, I think that's the 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 exponent rules and properties of 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: Express 729 as a power of 3

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

This simplifies to:
$$
3^{-1/2} = \frac{1}{\sqrt{3}}
$$

---

### Step 3: Rationalize the denominator

To write the expression with a **rational denominator**, we rationalize $\frac{1}{\sqrt{3}}$:

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

---

### Final Answer

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

[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, that sounds a bit complicated, but let me break it 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 the cube root.

I remember that roots can be expressed as exponents. Maybe converting all these roots into exponents will make it easier to handle. Let me recall the rules: the nth root of something is the same as raising that something to the power of 1/n. Also, when you have exponents raised to exponents, you multiply them. So, for example, (a^m)^n = a^(m*n).

Let me start from 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 result. So, taking the cube root is raising to the 1/3 power. So that would be [(1/729)^(1/2)]^(1/3). Then, the outermost layer is the square root of that, which is raising to the 1/2 power again. So overall, the entire expression is [( ( (1/729)^(1/2) )^(1/3) )]^(1/2).

Let me write that as exponents step by step. Let me denote the innermost part first:

First, √(1/729) = (1/729)^(1/2)

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

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

So, the entire expression simplifies to (1/729)^(1/12).

Now, I need to write this with a rational denominator. Wait, but (1/729)^(1/12) is already a single term. Maybe I need to express it as a radical with a rational denominator? Let me think.

Alternatively, maybe I need to rationalize the denominator if there was a denominator in the expression. But in this case, the expression is (1/729)^(1/12). Let me see.

Alternatively, perhaps I need to express this as a fraction with a rational denominator. Let me recall that 729 is a power of 3. Let me check: 3^6 is 729 because 3^6 = 729. Let me verify:

3^1 = 3

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = 243

3^6 = 729. Yes, correct. So 729 is 3^6. Therefore, 1/729 is 1/(3^6).

So, substituting back into the expression: (1/(3^6))^(1/12) = 1/(3^(6*(1/12))) = 1/(3^(1/2)) = 1/√3.

Wait, but that seems too simple. Let me check my steps again.

Original expression: √(∛(√(1/729)))

Convert to exponents:

Start with √(1/729) = (1/729)^(1/2)

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

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

Yes, that's correct. Then, since 729 is 3^6, so (1/729)^(1/12) = (3^(-6))^(1/12) = 3^(-6*(1/12)) = 3^(-1/2) = 1/(3^(1/2)) = 1/√3.

But the problem says to write the result with a rational denominator. So, 1/√3 has an irrational denominator. To rationalize the denominator, we multiply numerator and denominator by √3:

(1/√3) * (√3/√3) = √3 / 3.

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

Wait, but let me check again. Let me verify each step once more to make sure I didn't make a mistake.

Starting with √(∛(√(1/729))).

First, √(1/729) is 1 over √729. √729 is 27, since 27^2 is 729. Wait, 27*27 is 729? Let me check: 25^2 is 625, 26^2 is 676, 27^2 is 729. Yes, correct. So √(1/729) is 1/27.

Then, the cube root of 1/27. The cube root of 1/27 is 1 over cube root of 27. Cube root of 27 is 3, since 3^3 is 27. Therefore, cube root of 1/27 is 1/3.

Then, the square root of 1/3. Which is √(1/3) = 1/√3. Which is the same as √3 / 3 after rationalizing.

So, that's the same result as before. Therefore, the answer is √3 / 3.

But let me check if there's another way to approach this. Maybe by converting all the roots into exponents from the beginning.

Original expression: √(∛(√(1/729)))

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)^(1/2 * 1/3) = (1/729)^(1/6)

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

So, same as before.

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

Yes, same result.

Therefore, the simplified expression with a rational denominator is √3 divided by 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, yes, the steps I did are correct.

Alternatively, maybe the problem expects a different form? Let me think.

Alternatively, maybe I need to write the entire expression as a single radical? Let me see.

The original expression is a square root of a cube root of a square root. So, combining all the exponents, as I did before, gives 1/729 raised to the 1/12 power. Which is the same as 729^(-1/12).

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

Same result.

Alternatively, if I wanted to write this as a radical with a rational denominator, maybe I need to express it as a fraction with a rational denominator. But since √3 / 3 is already with a rational denominator, that's probably the answer.

Wait, but let me check if there's a different way to express this. For example, maybe the problem expects a different form? Let me think.

Alternatively, maybe I need to write the entire expression as a single radical? Let me see.

The original expression is sqrt(cbrt(sqrt(1/729))). Let me think about combining all the roots.

The square root is exponent 1/2, the cube root is exponent 1/3, and the square root again is exponent 1/2. So, multiplying the exponents: 1/2 * 1/3 * 1/2? Wait, no. Wait, the exponents are applied in sequence.

Wait, starting with 1/729, then taking square root (1/2), then cube root (1/3), then square root (1/2). So, the total exponent is 1/2 * 1/3 * 1/2? Wait, no. Wait, when you have exponents, you multiply them. So, if you have (a^(1/2))^(1/3) = a^(1/2 * 1/3) = a^(1/6). Then, (a^(1/6))^(1/2) = a^(1/6 * 1/2) = a^(1/12). So, yes, total exponent is 1/12.

Therefore, the entire expression is (1/729)^(1/12) = 1/(729^(1/12)).

But 729^(1/12) is the 12th root of 729. Since 729 is 3^6, then 729^(1/12) = (3^6)^(1/12) = 3^(6/12) = 3^(1/2) = √3. Therefore, 1/(√3) = √3/3.

Same result.

Therefore, regardless of the approach, I end up with √3 / 3.

But let me check if there's a possibility that I made a mistake in the exponent multiplication. Let me verify:

Starting with 1/729, then square root: exponent 1/2. Then cube root: exponent 1/3. Then square root: exponent 1/2. So total exponent is 1/2 * 1/3 * 1/2? Wait, no. Wait, when you have exponents, each operation is multiplying the exponent.

Wait, suppose you have a number x. Then, taking square root is x^(1/2). Then taking cube root of that is (x^(1/2))^(1/3) = x^(1/2 * 1/3) = x^(1/6). Then taking square root of that is (x^(1/6))^(1/2) = x^(1/6 * 1/2) = x^(1/12). So yes, total exponent is 1/12.

Therefore, the entire expression is x^(1/12) where x is 1/729.

So, that's correct.

Alternatively, if I think of the entire expression as a single radical, it would be the 12th root of 1/729. But 1/729 is 3^(-6), so the 12th root of 3^(-6) is 3^(-6/12) = 3^(-1/2) = 1/√3.

Same result.

Therefore, I think my answer is correct.

But let me check with actual numbers. Let me compute the original expression numerically and see if it equals √3 / 3.

First, compute 1/729. Let me calculate that. 1 divided by 729. Let me approximate. 729 is 700 + 29. 1/700 is approximately 0.00142857, and 1/729 is slightly less. Let me compute 1/729:

Well, 729 * 0.00137 = 729 * 0.001 + 729 * 0.00037 = 0.729 + 0.26973 = approx 0.99873. So, 0.00137 * 729 ≈ 0.99873, which is close to 1. Therefore, 1/729 ≈ 0.001371742.

Then, take the square root of that: sqrt(0.001371742). Let me compute that. sqrt(0.001371742). Let me note that sqrt(0.001371742) is approx sqrt(1.371742 * 10^-3) = sqrt(1.371742) * 10^(-3/2). sqrt(1.371742) is approx 1.171, since 1.17^2 = 1.3689, which is close. So approx 1.171 * 10^(-1.5) = 1.171 / 10^(1.5) = 1.171 / (10 * sqrt(10)) ≈ 1.171 / (10 * 3.1623) ≈ 1.171 / 31.623 ≈ 0.037.

So sqrt(1/729) ≈ 0.037.

Then, take the cube root of that. So cube root of 0.037. Let me approximate. 0.037^(1/3). Let me note that 0.037 is approx 3.7 * 10^-2. Let me think of cube roots. 0.064 is 0.4^3, 0.027 is 0.3^3. So 0.037 is between 0.027 and 0.064. Let me try 0.33^3 = 0.035937, which is very close to 0.037. So cube root of 0.037 is approx 0.33.

Then, take the square root of that. sqrt(0.33). sqrt(0.33) is approx 0.574.

Now, let me compute √3 / 3. √3 is approx 1.732, divided by 3 is approx 0.577. Which is very close to the approximate value I just got (0.574). The slight difference is due to my approximations in the steps. Therefore, this numerical check seems to confirm that the answer is indeed √3 / 3.

Therefore, I think my answer is correct.

But just to make sure, let me check if there's another way to write √3 / 3 with a rational denominator. Wait, √3 / 3 already has a rational denominator because the denominator is 3, which is rational. The numerator is irrational, but the denominator is rational. So, that's acceptable.

Alternatively, if the problem had a more complicated expression with multiple radicals in the denominator, we might need to rationalize further, but in this case, it's already done.

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

So, I think that's the 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 exponent rules and properties of 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: Express 729 as a power of 3

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

This simplifies to:
$$
3^{-1/2} = \frac{1}{\sqrt{3}}
$$

---

### Step 3: Rationalize the denominator

To write the expression with a **rational denominator**, we rationalize $\frac{1}{\sqrt{3}}$:

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

---

### Final Answer

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

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 the 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? Let me check. If $p$ can be $\pm1$ and $q$ can be $\pm1$, $\pm2$, then the possible fractions are:

- $1/1 = 1$
- $-1/1 = -1$
- $1/2$
- $-1/2$

So that's four different 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 polynomial is given with integer coefficients, but the coefficients $a_3$, $a_2$, $a_1$ are variables here. So, the question is asking for the number of different possible rational roots that could exist for such a polynomial. In other words, considering all possible integer values of $a_3$, $a_2$, $a_1$, how many distinct rational roots can this polynomial have?

But actually, the Rational Root Theorem gives all possible candidates for rational roots, regardless of the coefficients. However, not all of these candidates may actually be roots for some specific polynomial. But the question is asking for the number of different possible rational roots, meaning how many different rational numbers could be roots for some polynomial of this form.

Wait, but maybe I need to think differently. Let me re-read the question: "Find the number of different possible rational roots of this polynomial." So, given that the polynomial has integer coefficients, and the form given, how many different rational roots can it have? But since the coefficients are not fixed, but can be any integers, the question is about the possible rational roots that could exist for some such polynomial.

But then, according to the Rational Root Theorem, the possible rational roots are limited to $\pm1$, $\pm\frac{1}{2}$. So, regardless of the coefficients, any rational root must be one of these four. Therefore, the number of different possible rational roots is 4? But maybe some of these are not possible? For example, can 1 be a root? Let me check.

Suppose we want 1 to be a root. Then substituting x=1 into the polynomial:

$2(1)^4 + a_3(1)^3 + a_2(1)^2 + a_1(1) + 1 = 0$

Which simplifies to:

$2 + a_3 + a_2 + a_1 + 1 = 0$

So, $a_3 + a_2 + a_1 = -3$. Since $a_3$, $a_2$, $a_1$ 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. So 1 is possible.

Similarly, check for -1:

Substitute x = -1:

$2(-1)^4 + a_3(-1)^3 + a_2(-1)^2 + a_1(-1) + 1 = 0$

Calculates to:

$2(1) + a_3(-1) + a_2(1) + a_1(-1) + 1 = 0$

Which is:

$2 - a_3 + a_2 - a_1 + 1 = 0$

Simplify:

$3 - a_3 + a_2 - a_1 = 0$ => $ -a_3 + a_2 - a_1 = -3 $ => $a_3 - a_2 + a_1 = 3$. Again, since coefficients are integers, this is possible. 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. So -1 is possible.

Now check 1/2. Let me substitute x = 1/2 into the polynomial:

$2(1/2)^4 + a_3(1/2)^3 + a_2(1/2)^2 + a_1(1/2) + 1 = 0$

Calculating each term:

$2*(1/16) = 1/8$

$a_3*(1/8)$

$a_2*(1/4)$

$a_1*(1/2)$

Plus 1. So total:

$1/8 + a_3/8 + a_2/4 + a_1/2 + 1 = 0$

Multiply all terms by 8 to eliminate denominators:

$1 + a_3 + 2a_2 + 4a_1 + 8 = 0$

Which simplifies to:

$a_3 + 2a_2 + 4a_1 + 9 = 0$

So $a_3 + 2a_2 + 4a_1 = -9$. Since $a_3$, $a_2$, $a_1$ are integers, this is possible. For example, set $a_3 = -9$, $a_2 = 0$, $a_1 = 0$. Then the polynomial becomes $2x^4 -9x^3 + 0x^2 + 0x + 1 = 0$, which has 1/2 as a root. So 1/2 is possible.

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

$2(-1/2)^4 + a_3(-1/2)^3 + a_2(-1/2)^2 + a_1(-1/2) + 1 = 0$

Calculating each term:

$2*(1/16) = 1/8$

$a_3*(-1/8)$

$a_2*(1/4)$

$a_1*(-1/2)$

Plus 1. So total:

$1/8 - a_3/8 + a_2/4 - a_1/2 + 1 = 0$

Multiply all terms by 8:

$1 - a_3 + 2a_2 - 4a_1 + 8 = 0$

Simplify:

$9 - a_3 + 2a_2 - 4a_1 = 0$ => $ -a_3 + 2a_2 - 4a_1 = -9 $ => $a_3 - 2a_2 + 4a_1 = 9$. Again, possible with integer coefficients. For example, set $a_3 = 9$, $a_2 = 0$, $a_1 = 0$. Then the polynomial is $2x^4 + 9x^3 + 0x^2 + 0x + 1 = 0$, which has -1/2 as a root. So -1/2 is possible.

Therefore, all four candidates are possible. Therefore, the number of different possible rational roots is 4. But wait, the question says "different possible rational roots". But maybe some of these are duplicates? For example, is there any overlap? The four roots are 1, -1, 1/2, -1/2. These are all distinct. So that's four different roots.

But wait, hold on. Let me think again. The question says "the number of different possible rational roots of this polynomial". But maybe there's a case where some of these roots cannot be achieved? But we just saw that each of them can be achieved by choosing appropriate coefficients. Therefore, the answer is 4?

But wait, let me check again. Wait, the polynomial is of 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. So, for example, maybe some of these roots can't be roots for some reason? But we already saw that each of them can be a root for some polynomial with integer coefficients. Therefore, the answer is 4?

But wait, I remember sometimes there are restrictions. For example, if a polynomial has a rational root, then it can be factored as (x - r) times another polynomial. But since the coefficients are integers, maybe there are some constraints? But since we can choose coefficients to make any of these roots, then they are possible.

Alternatively, maybe the question is asking for the number of possible rational roots that the polynomial can have, considering that the coefficients are integers. But since the coefficients are arbitrary integers, then the possible rational roots are exactly those given by the Rational Root Theorem, which are four. Therefore, the answer is 4.

But wait, let me check again. Let me think if there's a case where even though the Rational Root Theorem says that possible roots are those four, maybe some of them cannot be roots? For example, suppose someone says that maybe 1/2 cannot be a root? But we already constructed a polynomial where 1/2 is a root. Similarly for others. Therefore, all four are possible.

Alternatively, maybe the question is trickier? Let me think again. The polynomial is given as $2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1$. So, the constant term is 1, leading coefficient is 2. So, the possible rational roots are ±1, ±1/2. So, four possibilities. Therefore, the answer is 4.

But wait, maybe there's a mistake here. Let me check with another example. Suppose we want to have both 1 and -1 as roots. Is that possible? Let me try. If x=1 and x=-1 are both roots, then the polynomial can be factored as (x - 1)(x + 1)(quadratic). Let me compute that.

Let me suppose that the polynomial factors as (x^2 - 1)(2x^2 + bx + c). Then expanding this:

(x^2 - 1)(2x^2 + bx + c) = 2x^4 + bx^3 + cx^2 - 2x^2 - bx - c = 2x^4 + bx^3 + (c - 2)x^2 - bx - c.

Comparing with the original polynomial $2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1$, we have:

- The constant term is -c = 1 => c = -1
- The coefficient of x is -b = a_1
- The coefficient of x^2 is (c - 2) = a_2 => (-1 - 2) = -3 = a_2
- The coefficient of x^3 is b = a_3

Therefore, if we set c = -1, then the constant term is 1. Then, for example, we can choose b to be any integer, so a_3 = b, a_1 = -b, a_2 = -3. Therefore, such a polynomial exists. For example, take b = 0: then the polynomial is (x^2 - 1)(2x^2 - 1) = 2x^4 - x^2 - 2x^2 + 1 = 2x^4 - 3x^2 + 1. Which is 2x^4 + 0x^3 - 3x^2 + 0x + 1. So, this polynomial has roots at 1, -1, 1/√2, -1/√2. Wait, but 1/√2 is irrational. However, the rational roots are 1 and -1. So, even though we have a polynomial with both 1 and -1 as roots, but the other roots are irrational. So, in this case, the polynomial has two rational roots.

But the question is not about how many roots the polynomial can have, but how many different possible rational roots exist for such polynomials. So, even if some polynomials have multiple rational roots, the question is about the total number of distinct rational numbers that can be roots of some polynomial of this form.

Therefore, since we have shown that 1, -1, 1/2, -1/2 can each be roots of some polynomial with integer coefficients, and there are no other possible rational roots (as per the Rational Root Theorem), then the answer is 4.

But let me check if there's a possibility of other rational roots. Suppose someone says, what about 2? Wait, according to the Rational Root Theorem, the possible rational roots are ±1, ±1/2. Because the constant term is 1, so p must divide 1, and leading coefficient is 2, so q divides 2. Therefore, p/q can only be ±1, ±1/2. So, 2 is not a possible rational root. Similarly, 1/3 is not possible. Therefore, there are only four possible rational roots.

Therefore, the answer is 4. But wait, let me check again. Wait, the problem says "the number of different possible rational roots of this polynomial". So, maybe the question is not about all possible rational roots that can exist for any such polynomial, but rather, for a specific polynomial with integer coefficients of that form, how many different rational roots can it have? But the problem says "Find the number of different possible rational roots of this polynomial." The wording is a bit ambiguous.

But given that the coefficients are not fixed, but are variables (since they are denoted as $a_3$, $a_2$, $a_1$), the question is likely asking for the number of distinct rational numbers that could be roots of such a polynomial. Therefore, the answer is 4.

But let me check again. Suppose someone says that maybe some of these roots cannot be achieved? For example, can 1/2 be a root? As we saw earlier, yes. Similarly for others. Therefore, all four are possible.

Alternatively, maybe there's a mistake in my reasoning. Let me think again. Suppose we have a polynomial with integer coefficients, and we want to know the possible rational roots. The Rational Root Theorem says that any rational root p/q must have p dividing the constant term and q dividing the leading coefficient. Therefore, in this case, the possible rational roots are ±1, ±1/2. Therefore, there are four possible rational roots. Therefore, the answer is 4.

But maybe the question is trickier? For example, maybe some of these roots cannot be roots because of the degree? For example, if the polynomial is of degree 4, can it have, say, 1 and 1/2 as roots? Let me check. Suppose we want both 1 and 1/2 to be roots. Then, the polynomial would have factors (x - 1)(x - 1/2). Let me multiply that with another quadratic to get a quartic.

Let me compute (x - 1)(x - 1/2) = x^2 - (3/2)x + 1/2. Multiply by 2 to eliminate fractions: 2x^2 - 3x + 1. Then, multiply by another quadratic, say, (ax^2 + bx + c). Then the polynomial would be (2x^2 - 3x + 1)(ax^2 + bx + c). Let me expand this:

= 2a x^4 + (2b - 3a)x^3 + (2c - 3b + a)x^2 + (-3c + b)x + c

Comparing with the original polynomial $2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1$, we have:

- Leading coefficient: 2a = 2 => a = 1
- Constant term: c = 1
- Coefficient of x^3: 2b - 3a = a_3 => 2b - 3(1) = a_3 => 2b - 3 = a_3
- Coefficient of x^2: 2c - 3b + a = a_2 => 2(1) - 3b + 1 = a_2 => 3 - 3b = a_2
- Coefficient of x: -3c + b = a_1 => -3(1) + b = a_1 => b - 3 = a_1

So, if we set a = 1, c = 1, then we can choose b to be any integer. For example, let me choose b = 0. Then:

- a_3 = 2(0) - 3 = -3
- a_2 = 3 - 3(0) = 3
- a_1 = 0 - 3 = -3

Therefore, the polynomial is (2x^2 - 3x + 1)(x^2 + 0x + 1) = 2x^4 - 3x^3 + 3x^2 - 3x + 1. Let me check if x=1 is a root:

2(1)^4 - 3(1)^3 + 3(1)^2 - 3(1) + 1 = 2 - 3 + 3 - 3 + 1 = 0. Yes, 1 is a root. And x=1/2:

2*(1/2)^4 - 3*(1/2)^3 + 3*(1/2)^2 - 3*(1/2) + 1

= 2*(1/16) - 3*(1/8) + 3*(1/4) - 3*(1/2) + 1

= 1/8 - 3/8 + 3/4 - 3/2 + 1

Convert to eighths:

1/8 - 3/8 + 6/8 - 12/8 + 8/8 = (1 - 3 + 6 - 12 + 8)/8 = 0/8 = 0. So yes, 1/2 is a root. Therefore, this polynomial has both 1 and 1/2 as roots. Therefore, even though the polynomial has two rational roots, they are both among the four possible ones.

Therefore, this shows that multiple rational roots can exist, but the question is about the number of different possible rational roots, not how many roots a single polynomial can have. Therefore, since all four candidates are possible, the answer is 4.

But let me check once more if there's a possibility that some of these roots cannot be achieved. For example, can 1/2 be a root? As shown earlier, yes. Similarly for others. Therefore, the answer is 4.

But wait, hold on. Wait, the original polynomial is of degree 4. Suppose someone says that maybe some of these roots are not possible because of the leading coefficient or something else? For example, if we have a root 1/2, then the polynomial must satisfy certain conditions. But we already constructed such a polynomial.

Alternatively, maybe the question is in the style of a problem where you have to consider that even though the Rational Root Theorem gives possible candidates, some of them might not be possible because of the specific form of the polynomial. But in this case, since the polynomial is of the form with leading coefficient 2 and constant term 1, and the coefficients in between are arbitrary integers, there's no restriction that would prevent any of the four candidates from being roots.

Therefore, I think the answer is 4. However, I recall that sometimes problems like this might have a trick. Let me think again.

Wait, the polynomial is monic? No, the leading coefficient is 2. So, the Rational Root Theorem applies. The possible rational roots are ±1, ±1/2. Therefore, four possibilities.

Alternatively, maybe the question is asking for the number of possible rational roots that the polynomial can have, considering multiplicity? But the question says "different possible rational roots", so multiplicity doesn't matter.

Alternatively, maybe there's a mistake in considering that 1 and -1 can be roots? Let me check with another example. Suppose I want to have 1 as a root. Let me take the polynomial $2x^4 + 0x^3 + 0x^2 + 0x + 1$. Wait, substituting x=1 gives 2 + 0 + 0 + 0 + 1 = 3 ≠ 0. So, that polynomial doesn't have 1 as a root. But earlier, I thought that if I set a_3 + a_2 + a_1 = -3, then 1 is a root. For example, if I set a_3 = -3, a_2 = 0, a_1 = 0, then the polynomial is 2x^4 - 3x^3 + 0x^2 + 0x + 1. Let me check x=1: 2 - 3 + 0 + 0 + 1 = 0. Yes, that works. So, that polynomial does have 1 as a root.

Therefore, there are polynomials where 1 is a root, and others where -1 is a root, etc. Therefore, all four are possible.

Therefore, I think the answer is 4. However, I want to check if maybe the question is from a source where the answer is different. Wait, but since I don't have that context, I need to rely on my reasoning.

Alternatively, maybe there's a mistake in considering that 1/2 and -1/2 are possible. Let me check with another example. Let me take the polynomial $2x^4 + 0x^3 + 0x^2 + (-9)x + 1$. Wait, earlier I had that for x=1/2, the equation reduces to $a_3 + 2a_2 + 4a_1 = -9$. If I set a_3 = -9, a_2 = 0, a_1 = 0, then the polynomial is $2x^4 -9x^3 + 0x^2 + 0x + 1$. Let me check x=1/2:

2*(1/2)^4 -9*(1/2)^3 + 0 + 0 + 1 = 2*(1/16) -9*(1/8) + 1 = 1/8 - 9/8 + 1 = (-8/8) + 1 = -1 + 1 = 0. Yes, that works.

Similarly, for x=-1/2, take the polynomial $2x^4 + 9x^3 + 0x^2 + 0x + 1$. Substitute x=-1/2:

2*(1/16) + 9*(-1/8) + 0 + 0 + 1 = 1/8 - 9/8 + 1 = (-8/8) + 1 = -1 + 1 = 0. Correct.

Therefore, all four roots are possible. Therefore, the number of different possible rational roots is 4.

But wait, hold on. Wait, the problem says "the number of different possible rational roots of this polynomial". If the polynomial is fixed, then the number of rational roots depends on the coefficients. But since the coefficients are not fixed, but can be any integers, the question is asking for the number of distinct rational numbers that can be roots of such a polynomial. Therefore, the answer is 4.

Alternatively, maybe the question is in the style of "how many rational roots can this polynomial have?" but that would be different. For example, a quartic can have up to 4 rational roots, but since the coefficients are arbitrary, maybe it can have all four? But the question is not about the maximum number of rational roots, but the number of different possible rational roots.

Therefore, I think the answer is 4. However, I want to check once more.

Let me think of another approach. Suppose we consider all possible rational roots, which are ±1, ±1/2. Each of these can be achieved by some polynomial with integer coefficients. Therefore, the set of possible rational roots is { -1, -1/2, 1/2, 1 }, which has 4 elements. Therefore, the answer is 4.

Therefore, I think the answer is 4. However, I recall that sometimes problems like this might have a different answer. Wait, let me check with another example. Suppose someone says that maybe 1/2 cannot be a root because of the leading coefficient? But no, we already constructed a polynomial where 1/2 is a root.

Alternatively, maybe the question is from a source where the answer is different. Wait, but given the information, I think my reasoning is correct.

Therefore, I think 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**,
- $ q $ divides the **leading coefficient**.

In our case:
- The **constant term** is $ 1 $, so $ p \in \{ \pm1 \} $,
- The **leading coefficient** is $ 2 $, so $ q \in \{ \pm1, \pm2 \} $.

Thus, the possible rational roots are:
$$
\frac{p}{q} = \pm1, \pm\frac{1}{2}.
$$

So, the **set of possible rational roots** is:
$$
\left\{ -1, -\frac{1}{2}, \frac{1}{2}, 1 \right\}.
$$

---

### Step 2: Determine if Each Root is Possible

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

#### 1. $ x = 1 $:
Substitute into the polynomial:
$$
2(1)^4 + a_3(1)^3 + a_2(1)^2 + a_1(1) + 1 = 0 \Rightarrow 2 + a_3 + a_2 + a_1 + 1 = 0 \Rightarrow a_3 + a_2 + a_1 = -3.
$$
This is achievable with integer values (e.g., $ a_3 = -3, a_2 = 0, a_1 = 0 $).

#### 2. $ x = -1 $:
$$
2(-1)^4 + a_3(-1)^3 + a_2(-1)^2 + a_1(-1) + 1 = 0 \Rightarrow 2 - a_3 + a_2 - a_1 + 1 = 0 \Rightarrow a_3 - a_2 + a_1 = 3.
$$
This is also achievable (e.g., $ a_3 = 3, a_2 = 0, a_1 = 0 $).

#### 3. $ x = \frac{1}{2} $:
Substitute:
$$
2\left(\frac{1}{2}\right)^4 + a_3\left(\frac{1}{2}\right)^3 + a_2\left(\frac{1}{2}\right)^2 + a_1\left(\frac{1}{2}\right) + 1 = 0.
$$
Multiply through by 8 to eliminate denominators:
$$
1 + a_3 + 2a_2 + 4a_1 + 8 = 0 \Rightarrow a_3 + 2a_2 + 4a_1 = -9.
$$
This is achievable (e.g., $ a_3 = -9, a_2 = 0, a_1 = 0 $).

#### 4. $ x = -\frac{1}{2} $:
Substitute:
$$
2\left(-\frac{1}{2}\right)^4 + a_3\left(-\frac{1}{2}\right)^3 + a_2\left(-\frac{1}{2}\right)^2 + a_1\left(-\frac{1}{2}\right) + 1 = 0.
$$
Multiply through by 8:
$$
1 - a_3 + 2a_2 - 4a_1 + 8 = 0 \Rightarrow a_3 - 2a_2 + 4a_1 = 9.
$$
This is also achievable (e.g., $ a_3 = 9, a_2 = 0, a_1 = 0 $).

---

### Step 3: Conclusion

All four rational numbers $ -1, -\frac{1}{2}, \frac{1}{2}, 1 $ are **distinct** and **each can be a root** of some polynomial of the given form with integer coefficients. Therefore, 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 the 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? Let me check. If $p$ can be $\pm1$ and $q$ can be $\pm1$, $\pm2$, then the possible fractions are:

- $1/1 = 1$
- $-1/1 = -1$
- $1/2$
- $-1/2$

So that's four different 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 polynomial is given with integer coefficients, but the coefficients $a_3$, $a_2$, $a_1$ are variables here. So, the question is asking for the number of different possible rational roots that could exist for such a polynomial. In other words, considering all possible integer values of $a_3$, $a_2$, $a_1$, how many distinct rational roots can this polynomial have?

But actually, the Rational Root Theorem gives all possible candidates for rational roots, regardless of the coefficients. However, not all of these candidates may actually be roots for some specific polynomial. But the question is asking for the number of different possible rational roots, meaning how many different rational numbers could be roots for some polynomial of this form.

Wait, but maybe I need to think differently. Let me re-read the question: "Find the number of different possible rational roots of this polynomial." So, given that the polynomial has integer coefficients, and the form given, how many different rational roots can it have? But since the coefficients are not fixed, but can be any integers, the question is about the possible rational roots that could exist for some such polynomial.

But then, according to the Rational Root Theorem, the possible rational roots are limited to $\pm1$, $\pm\frac{1}{2}$. So, regardless of the coefficients, any rational root must be one of these four. Therefore, the number of different possible rational roots is 4? But maybe some of these are not possible? For example, can 1 be a root? Let me check.

Suppose we want 1 to be a root. Then substituting x=1 into the polynomial:

$2(1)^4 + a_3(1)^3 + a_2(1)^2 + a_1(1) + 1 = 0$

Which simplifies to:

$2 + a_3 + a_2 + a_1 + 1 = 0$

So, $a_3 + a_2 + a_1 = -3$. Since $a_3$, $a_2$, $a_1$ 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. So 1 is possible.

Similarly, check for -1:

Substitute x = -1:

$2(-1)^4 + a_3(-1)^3 + a_2(-1)^2 + a_1(-1) + 1 = 0$

Calculates to:

$2(1) + a_3(-1) + a_2(1) + a_1(-1) + 1 = 0$

Which is:

$2 - a_3 + a_2 - a_1 + 1 = 0$

Simplify:

$3 - a_3 + a_2 - a_1 = 0$ => $ -a_3 + a_2 - a_1 = -3 $ => $a_3 - a_2 + a_1 = 3$. Again, since coefficients are integers, this is possible. 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. So -1 is possible.

Now check 1/2. Let me substitute x = 1/2 into the polynomial:

$2(1/2)^4 + a_3(1/2)^3 + a_2(1/2)^2 + a_1(1/2) + 1 = 0$

Calculating each term:

$2*(1/16) = 1/8$

$a_3*(1/8)$

$a_2*(1/4)$

$a_1*(1/2)$

Plus 1. So total:

$1/8 + a_3/8 + a_2/4 + a_1/2 + 1 = 0$

Multiply all terms by 8 to eliminate denominators:

$1 + a_3 + 2a_2 + 4a_1 + 8 = 0$

Which simplifies to:

$a_3 + 2a_2 + 4a_1 + 9 = 0$

So $a_3 + 2a_2 + 4a_1 = -9$. Since $a_3$, $a_2$, $a_1$ are integers, this is possible. For example, set $a_3 = -9$, $a_2 = 0$, $a_1 = 0$. Then the polynomial becomes $2x^4 -9x^3 + 0x^2 + 0x + 1 = 0$, which has 1/2 as a root. So 1/2 is possible.

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

$2(-1/2)^4 + a_3(-1/2)^3 + a_2(-1/2)^2 + a_1(-1/2) + 1 = 0$

Calculating each term:

$2*(1/16) = 1/8$

$a_3*(-1/8)$

$a_2*(1/4)$

$a_1*(-1/2)$

Plus 1. So total:

$1/8 - a_3/8 + a_2/4 - a_1/2 + 1 = 0$

Multiply all terms by 8:

$1 - a_3 + 2a_2 - 4a_1 + 8 = 0$

Simplify:

$9 - a_3 + 2a_2 - 4a_1 = 0$ => $ -a_3 + 2a_2 - 4a_1 = -9 $ => $a_3 - 2a_2 + 4a_1 = 9$. Again, possible with integer coefficients. For example, set $a_3 = 9$, $a_2 = 0$, $a_1 = 0$. Then the polynomial is $2x^4 + 9x^3 + 0x^2 + 0x + 1 = 0$, which has -1/2 as a root. So -1/2 is possible.

Therefore, all four candidates are possible. Therefore, the number of different possible rational roots is 4. But wait, the question says "different possible rational roots". But maybe some of these are duplicates? For example, is there any overlap? The four roots are 1, -1, 1/2, -1/2. These are all distinct. So that's four different roots.

But wait, hold on. Let me think again. The question says "the number of different possible rational roots of this polynomial". But maybe there's a case where some of these roots cannot be achieved? But we just saw that each of them can be achieved by choosing appropriate coefficients. Therefore, the answer is 4?

But wait, let me check again. Wait, the polynomial is of 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. So, for example, maybe some of these roots can't be roots for some reason? But we already saw that each of them can be a root for some polynomial with integer coefficients. Therefore, the answer is 4?

But wait, I remember sometimes there are restrictions. For example, if a polynomial has a rational root, then it can be factored as (x - r) times another polynomial. But since the coefficients are integers, maybe there are some constraints? But since we can choose coefficients to make any of these roots, then they are possible.

Alternatively, maybe the question is asking for the number of possible rational roots that the polynomial can have, considering that the coefficients are integers. But since the coefficients are arbitrary integers, then the possible rational roots are exactly those given by the Rational Root Theorem, which are four. Therefore, the answer is 4.

But wait, let me check again. Let me think if there's a case where even though the Rational Root Theorem says that possible roots are those four, maybe some of them cannot be roots? For example, suppose someone says that maybe 1/2 cannot be a root? But we already constructed a polynomial where 1/2 is a root. Similarly for others. Therefore, all four are possible.

Alternatively, maybe the question is trickier? Let me think again. The polynomial is given as $2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1$. So, the constant term is 1, leading coefficient is 2. So, the possible rational roots are ±1, ±1/2. So, four possibilities. Therefore, the answer is 4.

But wait, maybe there's a mistake here. Let me check with another example. Suppose we want to have both 1 and -1 as roots. Is that possible? Let me try. If x=1 and x=-1 are both roots, then the polynomial can be factored as (x - 1)(x + 1)(quadratic). Let me compute that.

Let me suppose that the polynomial factors as (x^2 - 1)(2x^2 + bx + c). Then expanding this:

(x^2 - 1)(2x^2 + bx + c) = 2x^4 + bx^3 + cx^2 - 2x^2 - bx - c = 2x^4 + bx^3 + (c - 2)x^2 - bx - c.

Comparing with the original polynomial $2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1$, we have:

- The constant term is -c = 1 => c = -1
- The coefficient of x is -b = a_1
- The coefficient of x^2 is (c - 2) = a_2 => (-1 - 2) = -3 = a_2
- The coefficient of x^3 is b = a_3

Therefore, if we set c = -1, then the constant term is 1. Then, for example, we can choose b to be any integer, so a_3 = b, a_1 = -b, a_2 = -3. Therefore, such a polynomial exists. For example, take b = 0: then the polynomial is (x^2 - 1)(2x^2 - 1) = 2x^4 - x^2 - 2x^2 + 1 = 2x^4 - 3x^2 + 1. Which is 2x^4 + 0x^3 - 3x^2 + 0x + 1. So, this polynomial has roots at 1, -1, 1/√2, -1/√2. Wait, but 1/√2 is irrational. However, the rational roots are 1 and -1. So, even though we have a polynomial with both 1 and -1 as roots, but the other roots are irrational. So, in this case, the polynomial has two rational roots.

But the question is not about how many roots the polynomial can have, but how many different possible rational roots exist for such polynomials. So, even if some polynomials have multiple rational roots, the question is about the total number of distinct rational numbers that can be roots of some polynomial of this form.

Therefore, since we have shown that 1, -1, 1/2, -1/2 can each be roots of some polynomial with integer coefficients, and there are no other possible rational roots (as per the Rational Root Theorem), then the answer is 4.

But let me check if there's a possibility of other rational roots. Suppose someone says, what about 2? Wait, according to the Rational Root Theorem, the possible rational roots are ±1, ±1/2. Because the constant term is 1, so p must divide 1, and leading coefficient is 2, so q divides 2. Therefore, p/q can only be ±1, ±1/2. So, 2 is not a possible rational root. Similarly, 1/3 is not possible. Therefore, there are only four possible rational roots.

Therefore, the answer is 4. But wait, let me check again. Wait, the problem says "the number of different possible rational roots of this polynomial". So, maybe the question is not about all possible rational roots that can exist for any such polynomial, but rather, for a specific polynomial with integer coefficients of that form, how many different rational roots can it have? But the problem says "Find the number of different possible rational roots of this polynomial." The wording is a bit ambiguous.

But given that the coefficients are not fixed, but are variables (since they are denoted as $a_3$, $a_2$, $a_1$), the question is likely asking for the number of distinct rational numbers that could be roots of such a polynomial. Therefore, the answer is 4.

But let me check again. Suppose someone says that maybe some of these roots cannot be achieved? For example, can 1/2 be a root? As we saw earlier, yes. Similarly for others. Therefore, all four are possible.

Alternatively, maybe there's a mistake in my reasoning. Let me think again. Suppose we have a polynomial with integer coefficients, and we want to know the possible rational roots. The Rational Root Theorem says that any rational root p/q must have p dividing the constant term and q dividing the leading coefficient. Therefore, in this case, the possible rational roots are ±1, ±1/2. Therefore, there are four possible rational roots. Therefore, the answer is 4.

But maybe the question is trickier? For example, maybe some of these roots cannot be roots because of the degree? For example, if the polynomial is of degree 4, can it have, say, 1 and 1/2 as roots? Let me check. Suppose we want both 1 and 1/2 to be roots. Then, the polynomial would have factors (x - 1)(x - 1/2). Let me multiply that with another quadratic to get a quartic.

Let me compute (x - 1)(x - 1/2) = x^2 - (3/2)x + 1/2. Multiply by 2 to eliminate fractions: 2x^2 - 3x + 1. Then, multiply by another quadratic, say, (ax^2 + bx + c). Then the polynomial would be (2x^2 - 3x + 1)(ax^2 + bx + c). Let me expand this:

= 2a x^4 + (2b - 3a)x^3 + (2c - 3b + a)x^2 + (-3c + b)x + c

Comparing with the original polynomial $2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1$, we have:

- Leading coefficient: 2a = 2 => a = 1
- Constant term: c = 1
- Coefficient of x^3: 2b - 3a = a_3 => 2b - 3(1) = a_3 => 2b - 3 = a_3
- Coefficient of x^2: 2c - 3b + a = a_2 => 2(1) - 3b + 1 = a_2 => 3 - 3b = a_2
- Coefficient of x: -3c + b = a_1 => -3(1) + b = a_1 => b - 3 = a_1

So, if we set a = 1, c = 1, then we can choose b to be any integer. For example, let me choose b = 0. Then:

- a_3 = 2(0) - 3 = -3
- a_2 = 3 - 3(0) = 3
- a_1 = 0 - 3 = -3

Therefore, the polynomial is (2x^2 - 3x + 1)(x^2 + 0x + 1) = 2x^4 - 3x^3 + 3x^2 - 3x + 1. Let me check if x=1 is a root:

2(1)^4 - 3(1)^3 + 3(1)^2 - 3(1) + 1 = 2 - 3 + 3 - 3 + 1 = 0. Yes, 1 is a root. And x=1/2:

2*(1/2)^4 - 3*(1/2)^3 + 3*(1/2)^2 - 3*(1/2) + 1

= 2*(1/16) - 3*(1/8) + 3*(1/4) - 3*(1/2) + 1

= 1/8 - 3/8 + 3/4 - 3/2 + 1

Convert to eighths:

1/8 - 3/8 + 6/8 - 12/8 + 8/8 = (1 - 3 + 6 - 12 + 8)/8 = 0/8 = 0. So yes, 1/2 is a root. Therefore, this polynomial has both 1 and 1/2 as roots. Therefore, even though the polynomial has two rational roots, they are both among the four possible ones.

Therefore, this shows that multiple rational roots can exist, but the question is about the number of different possible rational roots, not how many roots a single polynomial can have. Therefore, since all four candidates are possible, the answer is 4.

But let me check once more if there's a possibility that some of these roots cannot be achieved. For example, can 1/2 be a root? As shown earlier, yes. Similarly for others. Therefore, the answer is 4.

But wait, hold on. Wait, the original polynomial is of degree 4. Suppose someone says that maybe some of these roots are not possible because of the leading coefficient or something else? For example, if we have a root 1/2, then the polynomial must satisfy certain conditions. But we already constructed such a polynomial.

Alternatively, maybe the question is in the style of a problem where you have to consider that even though the Rational Root Theorem gives possible candidates, some of them might not be possible because of the specific form of the polynomial. But in this case, since the polynomial is of the form with leading coefficient 2 and constant term 1, and the coefficients in between are arbitrary integers, there's no restriction that would prevent any of the four candidates from being roots.

Therefore, I think the answer is 4. However, I recall that sometimes problems like this might have a trick. Let me think again.

Wait, the polynomial is monic? No, the leading coefficient is 2. So, the Rational Root Theorem applies. The possible rational roots are ±1, ±1/2. Therefore, four possibilities.

Alternatively, maybe the question is asking for the number of possible rational roots that the polynomial can have, considering multiplicity? But the question says "different possible rational roots", so multiplicity doesn't matter.

Alternatively, maybe there's a mistake in considering that 1 and -1 can be roots? Let me check with another example. Suppose I want to have 1 as a root. Let me take the polynomial $2x^4 + 0x^3 + 0x^2 + 0x + 1$. Wait, substituting x=1 gives 2 + 0 + 0 + 0 + 1 = 3 ≠ 0. So, that polynomial doesn't have 1 as a root. But earlier, I thought that if I set a_3 + a_2 + a_1 = -3, then 1 is a root. For example, if I set a_3 = -3, a_2 = 0, a_1 = 0, then the polynomial is 2x^4 - 3x^3 + 0x^2 + 0x + 1. Let me check x=1: 2 - 3 + 0 + 0 + 1 = 0. Yes, that works. So, that polynomial does have 1 as a root.

Therefore, there are polynomials where 1 is a root, and others where -1 is a root, etc. Therefore, all four are possible.

Therefore, I think the answer is 4. However, I want to check if maybe the question is from a source where the answer is different. Wait, but since I don't have that context, I need to rely on my reasoning.

Alternatively, maybe there's a mistake in considering that 1/2 and -1/2 are possible. Let me check with another example. Let me take the polynomial $2x^4 + 0x^3 + 0x^2 + (-9)x + 1$. Wait, earlier I had that for x=1/2, the equation reduces to $a_3 + 2a_2 + 4a_1 = -9$. If I set a_3 = -9, a_2 = 0, a_1 = 0, then the polynomial is $2x^4 -9x^3 + 0x^2 + 0x + 1$. Let me check x=1/2:

2*(1/2)^4 -9*(1/2)^3 + 0 + 0 + 1 = 2*(1/16) -9*(1/8) + 1 = 1/8 - 9/8 + 1 = (-8/8) + 1 = -1 + 1 = 0. Yes, that works.

Similarly, for x=-1/2, take the polynomial $2x^4 + 9x^3 + 0x^2 + 0x + 1$. Substitute x=-1/2:

2*(1/16) + 9*(-1/8) + 0 + 0 + 1 = 1/8 - 9/8 + 1 = (-8/8) + 1 = -1 + 1 = 0. Correct.

Therefore, all four roots are possible. Therefore, the number of different possible rational roots is 4.

But wait, hold on. Wait, the problem says "the number of different possible rational roots of this polynomial". If the polynomial is fixed, then the number of rational roots depends on the coefficients. But since the coefficients are not fixed, but can be any integers, the question is asking for the number of distinct rational numbers that can be roots of such a polynomial. Therefore, the answer is 4.

Alternatively, maybe the question is in the style of "how many rational roots can this polynomial have?" but that would be different. For example, a quartic can have up to 4 rational roots, but since the coefficients are arbitrary, maybe it can have all four? But the question is not about the maximum number of rational roots, but the number of different possible rational roots.

Therefore, I think the answer is 4. However, I want to check once more.

Let me think of another approach. Suppose we consider all possible rational roots, which are ±1, ±1/2. Each of these can be achieved by some polynomial with integer coefficients. Therefore, the set of possible rational roots is { -1, -1/2, 1/2, 1 }, which has 4 elements. Therefore, the answer is 4.

Therefore, I think the answer is 4. However, I recall that sometimes problems like this might have a different answer. Wait, let me check with another example. Suppose someone says that maybe 1/2 cannot be a root because of the leading coefficient? But no, we already constructed a polynomial where 1/2 is a root.

Alternatively, maybe the question is from a source where the answer is different. Wait, but given the information, I think my reasoning is correct.

Therefore, I think 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**,
- $ q $ divides the **leading coefficient**.

In our case:
- The **constant term** is $ 1 $, so $ p \in \{ \pm1 \} $,
- The **leading coefficient** is $ 2 $, so $ q \in \{ \pm1, \pm2 \} $.

Thus, the possible rational roots are:
$$
\frac{p}{q} = \pm1, \pm\frac{1}{2}.
$$

So, the **set of possible rational roots** is:
$$
\left\{ -1, -\frac{1}{2}, \frac{1}{2}, 1 \right\}.
$$

---

### Step 2: Determine if Each Root is Possible

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

#### 1. $ x = 1 $:
Substitute into the polynomial:
$$
2(1)^4 + a_3(1)^3 + a_2(1)^2 + a_1(1) + 1 = 0 \Rightarrow 2 + a_3 + a_2 + a_1 + 1 = 0 \Rightarrow a_3 + a_2 + a_1 = -3.
$$
This is achievable with integer values (e.g., $ a_3 = -3, a_2 = 0, a_1 = 0 $).

#### 2. $ x = -1 $:
$$
2(-1)^4 + a_3(-1)^3 + a_2(-1)^2 + a_1(-1) + 1 = 0 \Rightarrow 2 - a_3 + a_2 - a_1 + 1 = 0 \Rightarrow a_3 - a_2 + a_1 = 3.
$$
This is also achievable (e.g., $ a_3 = 3, a_2 = 0, a_1 = 0 $).

#### 3. $ x = \frac{1}{2} $:
Substitute:
$$
2\left(\frac{1}{2}\right)^4 + a_3\left(\frac{1}{2}\right)^3 + a_2\left(\frac{1}{2}\right)^2 + a_1\left(\frac{1}{2}\right) + 1 = 0.
$$
Multiply through by 8 to eliminate denominators:
$$
1 + a_3 + 2a_2 + 4a_1 + 8 = 0 \Rightarrow a_3 + 2a_2 + 4a_1 = -9.
$$
This is achievable (e.g., $ a_3 = -9, a_2 = 0, a_1 = 0 $).

#### 4. $ x = -\frac{1}{2} $:
Substitute:
$$
2\left(-\frac{1}{2}\right)^4 + a_3\left(-\frac{1}{2}\right)^3 + a_2\left(-\frac{1}{2}\right)^2 + a_1\left(-\frac{1}{2}\right) + 1 = 0.
$$
Multiply through by 8:
$$
1 - a_3 + 2a_2 - 4a_1 + 8 = 0 \Rightarrow a_3 - 2a_2 + 4a_1 = 9.
$$
This is also achievable (e.g., $ a_3 = 9, a_2 = 0, a_1 = 0 $).

---

### Step 3: Conclusion

All four rational numbers $ -1, -\frac{1}{2}, \frac{1}{2}, 1 $ are **distinct** and **each can be a root** of some polynomial of the given form with integer coefficients. Therefore, 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 $7!\times 2!=\boxed{10,\!080}$.