Explanation:
Okay, so I need to figure out the value of 5 star 1, where the operation star is defined as a star b equals 9a plus 2b minus ab plus 5. Let me write that down to make sure I have it right.

The problem says: If a ⋆ b = 9a + 2b - ab + 5, what is the value of 5 ⋆ 1?

Alright, so first, I need to understand what this operation ⋆ does. It takes two numbers, a and b, and combines them using this formula. So, to find 5 ⋆ 1, I just need to substitute a = 5 and b = 1 into the formula, right?

Let me write that out step by step. Let me replace a with 5 and b with 1 in the expression 9a + 2b - ab + 5.

So substituting, we get:

9*(5) + 2*(1) - (5)*(1) + 5.

Now let me compute each term one by one.

First term: 9*5. Let me calculate that. 9 times 5 is 45.

Second term: 2*1. That's straightforward, 2*1 is 2.

Third term: -(5*1). So that's -5.

Fourth term: +5. So that's just adding 5.

So putting it all together: 45 + 2 - 5 + 5.

Let me compute that step by step.

Starting with 45 + 2. That gives me 47.

Then, subtract 5: 47 - 5 is 42.

Then, add 5: 42 + 5 is 47.

Wait, so does that mean 5 ⋆ 1 equals 47? Let me check my calculations again to make sure I didn't make a mistake.

First term: 9*5 is indeed 45. Second term: 2*1 is 2. Third term: - (5*1) is -5. Fourth term: +5.

So adding them up: 45 + 2 is 47. Then 47 - 5 is 42. Then 42 + 5 is 47. Yeah, that seems right.

But let me think again. Maybe I should check if I substituted the numbers correctly. The formula is 9a + 2b - ab + 5. So when a is 5 and b is 1, that's 9*5 + 2*1 - (5*1) + 5.

Yes, that's exactly what I did. So 45 + 2 - 5 + 5.

Alternatively, maybe I can rearrange the terms to make it easier. Let's see:

The expression is 9a + 2b - ab + 5. Let me group the terms with a and b.

But maybe that's complicating. Alternatively, maybe I can compute the terms in a different order. Let me try:

Start with 9a: 45. Then add 2b: 45 + 2 = 47. Then subtract ab: 47 - 5 = 42. Then add 5: 42 + 5 = 47. Yep, same result.

Alternatively, maybe I can think of the expression as (9a - ab) + (2b + 5). Let me try that.

So 9a - ab is a*(9 - b). So with a=5 and b=1, that would be 5*(9 - 1) = 5*8 = 40. Then 2b + 5 is 2*1 + 5 = 2 + 5 = 7. Then adding those together: 40 + 7 = 47.

Same answer. So that seems to confirm it.

Alternatively, maybe I can check if there's another way to interpret the original operation. Let me check the original problem again.

It says: a ⋆ b = 9a + 2b - ab + 5. So that's 9a + 2b - ab + 5. There's no parentheses that would change the order of operations, right? So it's just 9a plus 2b minus ab plus 5.

So, no, there's no ambiguity here. So substituting a=5 and b=1 gives 9*5 + 2*1 - 5*1 + 5.

Another way to check: maybe compute each term again.

First term: 9*5. 9*5 is 45.

Second term: 2*1 is 2.

Third term: -ab is - (5*1) = -5.

Fourth term: +5.

So adding them all: 45 + 2 is 47, minus 5 is 42, plus 5 is 47.

So, seems like 47 is the answer.

But just to be thorough, let me check if I misread the problem. Maybe the operation is defined differently? Let me check again.

Original problem: If a ⋆ b = 9a + 2b - ab + 5, what is the value of 5 ⋆ 1?

No, that's exactly as written. So the operation is 9a + 2b - ab + 5.

Alternatively, maybe there's a typo in my thinking? Let me check once more.

Wait, maybe the operation is 9a + 2b - ab + 5. So, if I write that as 9a + 2b - ab + 5, is there a possibility that the -ab is supposed to be something else? Like maybe - (ab + 5)? But no, the original problem says -ab + 5. So it's -ab + 5.

So, no, that's not the case.

Alternatively, maybe the problem is written with different grouping? For example, maybe 9a + 2b - (ab + 5)? But that would be different. But the original problem says "9a + 2b - ab + 5", so unless there are parentheses that aren't shown, which there aren't, it's just 9a + 2b - ab + 5.

So, I think my calculation is correct.

Alternatively, maybe I can test with another example. Let me try a simple case. Let me compute 0 ⋆ 0.

Using the formula: 9*0 + 2*0 - 0*0 + 5 = 0 + 0 - 0 + 5 = 5.

Alternatively, if I compute 1 ⋆ 1: 9*1 + 2*1 - 1*1 + 5 = 9 + 2 - 1 + 5 = 15.

So, seems like the formula works.

Alternatively, maybe check if there's a different interpretation. For example, maybe the operation is 9a + 2b - (ab + 5). But that would be different. Let me check.

If that were the case, then 5 ⋆ 1 would be 9*5 + 2*1 - (5*1 + 5) = 45 + 2 - (5 + 5) = 47 - 10 = 37. But that's different. However, the original problem says "9a + 2b - ab + 5", so unless there's a missing parenthesis, which there isn't, it's not that.

So, the original problem is 9a + 2b - ab + 5. So, the correct calculation is 45 + 2 - 5 + 5 = 47.

Alternatively, maybe the problem is written in a different order? For example, maybe it's 9a + 2b - (ab + 5). But again, that would require parentheses.

Alternatively, maybe the problem is written as 9a + 2b - ab + 5, which is the same as 9a + 2b - ab + 5.

So, unless there's some other interpretation, I think my answer is correct.

Alternatively, maybe I can think of the operation as combining terms. Let me see:

Let me rearrange the formula:

a ⋆ b = 9a + 2b - ab + 5.

Maybe factor something out? Let me see:

Looking at 9a - ab, that's a*(9 - b). Then 2b + 5. So, a*(9 - b) + 2b + 5.

So, if I plug in a=5 and b=1, that would be 5*(9 - 1) + 2*1 + 5 = 5*8 + 2 + 5 = 40 + 2 + 5 = 47.

Same result.

Alternatively, maybe factor differently? Let me see:

Looking at 9a + 2b - ab + 5. Maybe group terms with b:

= ( -ab + 2b ) + 9a + 5

= b*(-a + 2) + 9a + 5

So, if I plug in a=5 and b=1:

= 1*(-5 + 2) + 9*5 + 5

= 1*(-3) + 45 + 5

= -3 + 45 + 5

= 47.

Same answer again.

So, regardless of how I group the terms, I get 47.

Therefore, I think the answer is 47.

But just to make sure, let me check once more with direct substitution:

Original formula: 9a + 2b - ab + 5.

Substitute a=5, b=1:

9*5 = 45

2*1 = 2

-ab = -5*1 = -5

+5 = +5

So total is 45 + 2 - 5 + 5.

Compute 45 + 2 = 47. Then 47 - 5 = 42. Then 42 + 5 = 47.

Yep, same result.

Therefore, I think I can be confident that 5 ⋆ 1 equals 47.

**Final Answer**
The value of $5 \star 1$ is \boxed{47}.

To solve the problem, we are given a custom binary operation defined as:

$$
a \star b = 9a + 2b - ab + 5
$$

We are asked to compute the value of $ 5 \star 1 $, which means we substitute $ a = 5 $ and $ b = 1 $ into the expression.

---

### Step-by-step Substitution

Substitute $ a = 5 $ and $ b = 1 $ into the formula:

$$
5 \star 1 = 9(5) + 2(1) - (5)(1) + 5
$$

Now compute each term:

- $ 9(5) = 45 $
- $ 2(1) = 2 $
- $ -(5)(1) = -5 $
- $ +5 = 5 $

Now add them together:

$$
45 + 2 - 5 + 5 = 47
$$

---

### Final Answer

$$
\boxed{47}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the value of 5 star 1, where the operation star is defined as a star b equals 9a plus 2b minus ab plus 5. Let me write that down to make sure I have it right.

The problem says: If a ⋆ b = 9a + 2b - ab + 5, what is the value of 5 ⋆ 1?

Alright, so first, I need to understand what this operation ⋆ does. It takes two numbers, a and b, and combines them using this formula. So, to find 5 ⋆ 1, I just need to substitute a = 5 and b = 1 into the formula, right?

Let me write that out step by step. Let me replace a with 5 and b with 1 in the expression 9a + 2b - ab + 5.

So substituting, we get:

9*(5) + 2*(1) - (5)*(1) + 5.

Now let me compute each term one by one.

First term: 9*5. Let me calculate that. 9 times 5 is 45.

Second term: 2*1. That's straightforward, 2*1 is 2.

Third term: -(5*1). So that's -5.

Fourth term: +5. So that's just adding 5.

So putting it all together: 45 + 2 - 5 + 5.

Let me compute that step by step.

Starting with 45 + 2. That gives me 47.

Then, subtract 5: 47 - 5 is 42.

Then, add 5: 42 + 5 is 47.

Wait, so does that mean 5 ⋆ 1 equals 47? Let me check my calculations again to make sure I didn't make a mistake.

First term: 9*5 is indeed 45. Second term: 2*1 is 2. Third term: - (5*1) is -5. Fourth term: +5.

So adding them up: 45 + 2 is 47. Then 47 - 5 is 42. Then 42 + 5 is 47. Yeah, that seems right.

But let me think again. Maybe I should check if I substituted the numbers correctly. The formula is 9a + 2b - ab + 5. So when a is 5 and b is 1, that's 9*5 + 2*1 - (5*1) + 5.

Yes, that's exactly what I did. So 45 + 2 - 5 + 5.

Alternatively, maybe I can rearrange the terms to make it easier. Let's see:

The expression is 9a + 2b - ab + 5. Let me group the terms with a and b.

But maybe that's complicating. Alternatively, maybe I can compute the terms in a different order. Let me try:

Start with 9a: 45. Then add 2b: 45 + 2 = 47. Then subtract ab: 47 - 5 = 42. Then add 5: 42 + 5 = 47. Yep, same result.

Alternatively, maybe I can think of the expression as (9a - ab) + (2b + 5). Let me try that.

So 9a - ab is a*(9 - b). So with a=5 and b=1, that would be 5*(9 - 1) = 5*8 = 40. Then 2b + 5 is 2*1 + 5 = 2 + 5 = 7. Then adding those together: 40 + 7 = 47.

Same answer. So that seems to confirm it.

Alternatively, maybe I can check if there's another way to interpret the original operation. Let me check the original problem again.

It says: a ⋆ b = 9a + 2b - ab + 5. So that's 9a + 2b - ab + 5. There's no parentheses that would change the order of operations, right? So it's just 9a plus 2b minus ab plus 5.

So, no, there's no ambiguity here. So substituting a=5 and b=1 gives 9*5 + 2*1 - 5*1 + 5.

Another way to check: maybe compute each term again.

First term: 9*5. 9*5 is 45.

Second term: 2*1 is 2.

Third term: -ab is - (5*1) = -5.

Fourth term: +5.

So adding them all: 45 + 2 is 47, minus 5 is 42, plus 5 is 47.

So, seems like 47 is the answer.

But just to be thorough, let me check if I misread the problem. Maybe the operation is defined differently? Let me check again.

Original problem: If a ⋆ b = 9a + 2b - ab + 5, what is the value of 5 ⋆ 1?

No, that's exactly as written. So the operation is 9a + 2b - ab + 5.

Alternatively, maybe there's a typo in my thinking? Let me check once more.

Wait, maybe the operation is 9a + 2b - ab + 5. So, if I write that as 9a + 2b - ab + 5, is there a possibility that the -ab is supposed to be something else? Like maybe - (ab + 5)? But no, the original problem says -ab + 5. So it's -ab + 5.

So, no, that's not the case.

Alternatively, maybe the problem is written with different grouping? For example, maybe 9a + 2b - (ab + 5)? But that would be different. But the original problem says "9a + 2b - ab + 5", so unless there are parentheses that aren't shown, which there aren't, it's just 9a + 2b - ab + 5.

So, I think my calculation is correct.

Alternatively, maybe I can test with another example. Let me try a simple case. Let me compute 0 ⋆ 0.

Using the formula: 9*0 + 2*0 - 0*0 + 5 = 0 + 0 - 0 + 5 = 5.

Alternatively, if I compute 1 ⋆ 1: 9*1 + 2*1 - 1*1 + 5 = 9 + 2 - 1 + 5 = 15.

So, seems like the formula works.

Alternatively, maybe check if there's a different interpretation. For example, maybe the operation is 9a + 2b - (ab + 5). But that would be different. Let me check.

If that were the case, then 5 ⋆ 1 would be 9*5 + 2*1 - (5*1 + 5) = 45 + 2 - (5 + 5) = 47 - 10 = 37. But that's different. However, the original problem says "9a + 2b - ab + 5", so unless there's a missing parenthesis, which there isn't, it's not that.

So, the original problem is 9a + 2b - ab + 5. So, the correct calculation is 45 + 2 - 5 + 5 = 47.

Alternatively, maybe the problem is written in a different order? For example, maybe it's 9a + 2b - (ab + 5). But again, that would require parentheses.

Alternatively, maybe the problem is written as 9a + 2b - ab + 5, which is the same as 9a + 2b - ab + 5.

So, unless there's some other interpretation, I think my answer is correct.

Alternatively, maybe I can think of the operation as combining terms. Let me see:

Let me rearrange the formula:

a ⋆ b = 9a + 2b - ab + 5.

Maybe factor something out? Let me see:

Looking at 9a - ab, that's a*(9 - b). Then 2b + 5. So, a*(9 - b) + 2b + 5.

So, if I plug in a=5 and b=1, that would be 5*(9 - 1) + 2*1 + 5 = 5*8 + 2 + 5 = 40 + 2 + 5 = 47.

Same result.

Alternatively, maybe factor differently? Let me see:

Looking at 9a + 2b - ab + 5. Maybe group terms with b:

= ( -ab + 2b ) + 9a + 5

= b*(-a + 2) + 9a + 5

So, if I plug in a=5 and b=1:

= 1*(-5 + 2) + 9*5 + 5

= 1*(-3) + 45 + 5

= -3 + 45 + 5

= 47.

Same answer again.

So, regardless of how I group the terms, I get 47.

Therefore, I think the answer is 47.

But just to make sure, let me check once more with direct substitution:

Original formula: 9a + 2b - ab + 5.

Substitute a=5, b=1:

9*5 = 45

2*1 = 2

-ab = -5*1 = -5

+5 = +5

So total is 45 + 2 - 5 + 5.

Compute 45 + 2 = 47. Then 47 - 5 = 42. Then 42 + 5 = 47.

Yep, same result.

Therefore, I think I can be confident that 5 ⋆ 1 equals 47.

**Final Answer**
The value of $5 \star 1$ is \boxed{47}.

To solve the problem, we are given a custom binary operation defined as:

$$
a \star b = 9a + 2b - ab + 5
$$

We are asked to compute the value of $ 5 \star 1 $, which means we substitute $ a = 5 $ and $ b = 1 $ into the expression.

---

### Step-by-step Substitution

Substitute $ a = 5 $ and $ b = 1 $ into the formula:

$$
5 \star 1 = 9(5) + 2(1) - (5)(1) + 5
$$

Now compute each term:

- $ 9(5) = 45 $
- $ 2(1) = 2 $
- $ -(5)(1) = -5 $
- $ +5 = 5 $

Now add them together:

$$
45 + 2 - 5 + 5 = 47
$$

---

### Final Answer

$$
\boxed{47}
$$

Solution: From the defined function, we know that $5\star 1 = 9(5)+2(1)-(5)(1)+5= 45+2-5+5=\boxed{47}$.


Explanation:
Okay, so I need to find all the integer roots of the polynomial equation $x^4 + 5x^3 + 9x^2 - x - 14 = 0$. Hmm, let me think about how to approach this.

First, I remember that for polynomials with integer coefficients, the Rational Root Theorem can be helpful. The theorem states that any possible rational root, expressed in its lowest terms $\frac{p}{q}$, must have $p$ as a factor of the constant term and $q$ as a factor of the leading coefficient. Since we're looking for integer roots, $q$ must be 1 or -1, so the possible integer roots are just the factors of the constant term.

Looking at the given polynomial, the constant term is -14, and the leading coefficient is 1. Therefore, the possible integer roots are the factors of -14. Let me list them out: ±1, ±2, ±7, ±14.

Alright, so I need to test each of these values in the polynomial to see if they satisfy the equation. Let me start with the smaller numbers first.

Let me test x = 1. Plugging into the polynomial:

$1^4 + 5(1)^3 + 9(1)^2 - 1 - 14 = 1 + 5 + 9 - 1 - 14$. Let me compute that step by step:

1 + 5 = 6; 6 + 9 = 15; 15 - 1 = 14; 14 - 14 = 0. Oh, that equals zero! So x = 1 is a root. Great, that's one integer root.

Now let me check x = -1. Plugging in:

$(-1)^4 + 5(-1)^3 + 9(-1)^2 - (-1) - 14$. Let me compute each term:

(-1)^4 = 1; 5(-1)^3 = 5(-1) = -5; 9(-1)^2 = 9(1) = 9; -(-1) = +1; and then -14. So adding them up:

1 - 5 + 9 + 1 - 14. Let me compute step by step:

1 - 5 = -4; -4 + 9 = 5; 5 + 1 = 6; 6 - 14 = -8. Not zero. So x = -1 is not a root.

Next, let me try x = 2. Plugging in:

$2^4 + 5(2)^3 + 9(2)^2 - 2 - 14$. Calculating each term:

2^4 = 16; 5*(8) = 40; 9*(4) = 36; then -2 and -14. So total:

16 + 40 = 56; 56 + 36 = 92; 92 - 2 = 90; 90 - 14 = 76. Not zero. So x = 2 is not a root.

How about x = -2? Let's compute:

(-2)^4 + 5(-2)^3 + 9(-2)^2 - (-2) -14.

Calculating each term:

(-2)^4 = 16; 5*(-8) = -40; 9*(4) = 36; -(-2) = +2; and -14.

Adding them up: 16 - 40 = -24; -24 + 36 = 12; 12 + 2 = 14; 14 -14 = 0. Oh! That equals zero. So x = -2 is also a root. Nice, so now we have two integer roots: 1 and -2.

Moving on to x = 7. Let me check that.

7^4 + 5*(7)^3 + 9*(7)^2 - 7 -14.

Calculating each term:

7^4 is 2401; 5*(343) = 1715; 9*(49) = 441; then -7 and -14.

Adding them up: 2401 + 1715 = 4116; 4116 + 441 = 4557; 4557 - 7 = 4550; 4550 -14 = 4536. Definitely not zero. So x = 7 is not a root.

How about x = -7? Let me try that.

(-7)^4 + 5*(-7)^3 + 9*(-7)^2 - (-7) -14.

Calculating each term:

(-7)^4 = 2401; 5*(-343) = -1715; 9*(49) = 441; -(-7) = +7; and -14.

Adding them up: 2401 - 1715 = 686; 686 + 441 = 1127; 1127 + 7 = 1134; 1134 -14 = 1120. Not zero. So x = -7 is not a root.

Next, x = 14. Let me check that.

14^4 + 5*(14)^3 + 9*(14)^2 -14 -14.

This is going to be a huge number. Let me see if I can compute it step by step.

14^2 = 196; 14^3 = 14*196 = 2744; 14^4 = 14*2744 = 38416.

So:

38416 + 5*(2744) + 9*(196) -14 -14.

Calculating each term:

5*2744 = 13720; 9*196 = 1764.

Adding up: 38416 + 13720 = 52136; 52136 + 1764 = 53900; 53900 -14 = 53886; 53886 -14 = 53872. Definitely not zero. So x = 14 is not a root.

Similarly, x = -14 would be even larger in magnitude, so I can safely say it's not a root.

So far, we have found two integer roots: 1 and -2. But since the polynomial is of degree 4, there could be more roots. However, since we are only looking for integer roots, and we have already checked all possible candidates from the Rational Root Theorem, maybe there are no more integer roots? Wait, but maybe there are other integer roots that are not factors of -14? Wait, no, the Rational Root Theorem says that any rational root must be a factor of the constant term over a factor of the leading coefficient. Since the leading coefficient is 1, the only possible rational roots are the factors of -14, which we have already checked. Therefore, there are no other possible integer roots.

But just to be thorough, let me check if maybe there are multiple roots or something. Wait, but we found two roots already. Let me try to factor the polynomial using these roots.

If x = 1 and x = -2 are roots, then (x - 1) and (x + 2) are factors of the polynomial. Let me perform polynomial division or use synthetic division to factor them out.

First, let me divide the original polynomial by (x - 1). Wait, but since we found x = 1 is a root, so (x - 1) is a factor. Let me use synthetic division with root 1.

The coefficients of the polynomial are 1 (x^4), 5 (x^3), 9 (x^2), -1 (x), -14 (constant).

Set up synthetic division:

1 | 1   5   9   -1   -14
  |     1   6   15   14
  ------------------------
    1   6  15   14   0

So the result is x^3 + 6x^2 + 15x + 14.

So now, the polynomial factors as (x - 1)(x^3 + 6x^2 + 15x + 14).

Now, let me check if x = -2 is a root of this cubic polynomial. Let me plug in x = -2 into x^3 + 6x^2 + 15x + 14.

(-2)^3 + 6*(-2)^2 + 15*(-2) + 14 = -8 + 6*(4) + (-30) + 14 = -8 + 24 - 30 + 14.

Calculating step by step: -8 +24 = 16; 16 -30 = -14; -14 +14 = 0. Yes, so x = -2 is a root. Therefore, we can factor out (x + 2) from the cubic.

Let me perform synthetic division on the cubic polynomial x^3 + 6x^2 + 15x + 14 with root -2.

Coefficients: 1, 6, 15, 14

-2 | 1   6   15   14
    |     -2   -8   -14
    -------------------
      1   4   7   0

So the result is x^2 + 4x + 7.

Therefore, the original polynomial factors as (x - 1)(x + 2)(x^2 + 4x + 7).

Now, we need to check if the quadratic factor x^2 + 4x + 7 has any integer roots. Let me compute its discriminant: D = 16 - 4*1*7 = 16 - 28 = -12. Since the discriminant is negative, there are no real roots, let alone integer roots. Therefore, the only integer roots of the original polynomial are x = 1 and x = -2.

Wait, but let me just confirm that I didn't make any mistakes in the synthetic division steps. Let me check the first division again.

Original polynomial: x^4 +5x^3 +9x^2 -x -14 divided by (x - 1).

Using synthetic division with root 1:

Bring down the 1. Multiply by 1: 1. Add to next coefficient: 5 + 1 = 6. Multiply by 1: 6. Add to next coefficient: 9 + 6 = 15. Multiply by 1: 15. Add to next coefficient: -1 + 15 = 14. Multiply by 1: 14. Add to last coefficient: -14 +14 = 0. Correct. So the cubic is x^3 +6x^2 +15x +14.

Then dividing that by (x + 2), which is equivalent to root -2.

Coefficients 1, 6, 15, 14. Bring down the 1. Multiply by -2: -2. Add to 6: 4. Multiply by -2: -8. Add to 15: 7. Multiply by -2: -14. Add to 14: 0. Correct. So quadratic is x^2 +4x +7.

Yes, that seems correct. So the quadratic has no real roots, so no integer roots.

Therefore, the only integer roots are 1 and -2.

But just to be absolutely thorough, let me check if there are any other possible integer roots that I might have missed. Wait, the Rational Root Theorem says that all possible rational roots are factors of -14, which we checked. So there can't be any other integer roots.

Alternatively, maybe I can factor the original polynomial in another way? Let me see.

Alternatively, maybe I can try factoring by grouping? Let me try.

Original polynomial: x^4 +5x^3 +9x^2 -x -14.

Hmm, grouping terms. Let me see.

Looking at x^4 +5x^3 +9x^2 -x -14. Maybe group as (x^4 +5x^3) + (9x^2 -x) -14.

Factor x^3 from first group: x^3(x + 5) + x(9x -1) -14. Doesn't seem helpful.

Alternatively, maybe different grouping. Let me think.

Alternatively, since we already factored it as (x - 1)(x + 2)(x^2 +4x +7), which is the complete factorization. Since the quadratic doesn't factor further over integers, that's the end.

Alternatively, maybe check if the original polynomial can be factored into quadratics? Let me try.

Suppose we try to write it as (x^2 + ax + b)(x^2 + cx + d). Then expanding:

x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd.

Comparing coefficients with original polynomial:

Original coefficients: 1 (x^4), 5 (x^3), 9 (x^2), -1 (x), -14 (constant).

So equations:

1. a + c = 5
2. ac + b + d = 9
3. ad + bc = -1
4. bd = -14

We need integers a, b, c, d satisfying these equations.

We already know that the polynomial factors as (x - 1)(x + 2)(x^2 +4x +7). Let me check if that corresponds to such a factorization.

If we take (x - 1)(x + 2) = x^2 + x - 2. Then multiply by (x^2 +4x +7):

(x^2 + x - 2)(x^2 +4x +7) = x^4 +4x^3 +7x^2 +x^3 +4x^2 +7x -2x^2 -8x -14.

Combine like terms:

x^4 + (4x^3 +x^3) + (7x^2 +4x^2 -2x^2) + (7x -8x) -14 = x^4 +5x^3 +9x^2 -x -14. Which matches.

So if we try to factor into quadratics, we can have (x^2 + x - 2)(x^2 +4x +7).

But since we already found the roots, maybe this isn't necessary. However, just to see if there are other possible factorizations.

But since we already have the roots, and the quadratic factor doesn't have integer roots, I think we can be confident that the only integer roots are 1 and -2.

Alternatively, maybe there's another way to factor? Let me try with different b and d.

From equation 4: bd = -14. So possible integer pairs (b, d) are (1, -14), (-1, 14), (2, -7), (-2, 7), (7, -2), (-7, 2), (14, -1), (-14, 1).

Let me try each possibility.

First, let me try b = 1, d = -14. Then equation 1: a + c = 5. Equation 2: ac + 1 + (-14) = ac -13 = 9 => ac = 22. Equation 3: ad + bc = a*(-14) + c*(1) = -14a + c = -1.

So we have:

a + c = 5

ac = 22

-14a + c = -1

Let me solve these equations. From the first equation, c = 5 - a. Substitute into the third equation:

-14a + (5 - a) = -1 => -15a +5 = -1 => -15a = -6 => a = (-6)/(-15) = 2/5. Not integer. So discard this pair.

Next pair: b = -1, d = 14. Then equation 2: ac + (-1) +14 = ac +13 = 9 => ac = -4. Equation 3: ad + bc = a*14 + c*(-1) = 14a - c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

14a - (5 - a) = 14a -5 + a = 15a -5 = -1 => 15a = 4 => a = 4/15. Not integer. Disregard.

Next pair: b = 2, d = -7. Equation 2: ac + 2 + (-7) = ac -5 = 9 => ac = 14. Equation 3: ad + bc = a*(-7) + c*2 = -7a + 2c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

-7a + 2*(5 - a) = -7a +10 -2a = -9a +10 = -1 => -9a = -11 => a = 11/9. Not integer.

Next pair: b = -2, d = 7. Equation 2: ac + (-2) +7 = ac +5 = 9 => ac = 4. Equation 3: ad + bc = a*7 + c*(-2) = 7a -2c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

7a -2*(5 - a) = 7a -10 + 2a = 9a -10 = -1 => 9a = 9 => a = 1. Then c = 5 -1 = 4.

Check if ac = 1*4 = 4, which matches equation 2. So this works!

Therefore, we have a = 1, c = 4, b = -2, d = 7. Therefore, the factorization is (x^2 + ax + b)(x^2 + cx + d) = (x^2 + x -2)(x^2 +4x +7). Which is exactly what we had earlier.

So that's consistent.

Next pair: b = 7, d = -2. Equation 2: ac +7 + (-2) = ac +5 = 9 => ac = 4. Equation 3: ad + bc = a*(-2) + c*7 = -2a +7c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

-2a +7*(5 - a) = -2a +35 -7a = -9a +35 = -1 => -9a = -36 => a = 4. Then c = 5 -4 = 1.

Check ac = 4*1 = 4, which matches. So this is another valid factorization. So same as before, just swapping the quadratics.

Similarly, other pairs:

b = -7, d = 2. Equation 2: ac + (-7) +2 = ac -5 = 9 => ac = 14. Equation 3: ad + bc = a*2 + c*(-7) = 2a -7c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

2a -7*(5 - a) = 2a -35 +7a = 9a -35 = -1 => 9a = 34 => a = 34/9. Not integer.

Next pair: b = 14, d = -1. Equation 2: ac +14 + (-1) = ac +13 = 9 => ac = -4. Equation 3: ad + bc = a*(-1) + c*14 = -a +14c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

-a +14*(5 - a) = -a +70 -14a = -15a +70 = -1 => -15a = -71 => a = 71/15. Not integer.

Last pair: b = -14, d = 1. Equation 2: ac + (-14) +1 = ac -13 = 9 => ac = 22. Equation 3: ad + bc = a*1 + c*(-14) = a -14c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

a -14*(5 - a) = a -70 +14a = 15a -70 = -1 => 15a = 69 => a = 69/15 = 23/5. Not integer.

So the only valid factorizations are the ones we found earlier, which confirm that the polynomial factors into (x^2 +x -2)(x^2 +4x +7). And since x^2 +x -2 factors further into (x -1)(x +2), which gives us the integer roots 1 and -2, and the other quadratic has no real roots, we can confirm that these are the only integer roots.

Therefore, after thorough checking, the integer roots are 1 and -2.

Just to recap, I used the Rational Root Theorem to list possible integer roots, tested them, found two roots, factored the polynomial accordingly, and confirmed that the remaining quadratic doesn't have integer roots. Additionally, I tried factoring the polynomial into quadratics to verify the factorization, which also led to the same conclusion.

I think that covers all possibilities. I don't see any mistakes in my reasoning, so I'm confident that the integer roots are 1 and -2.

**Final Answer**
The integer roots are \boxed{-2} and \boxed{1}.

To find all the **integer roots** of the polynomial

$$
x^4 + 5x^3 + 9x^2 - x - 14 = 0,
$$

we begin by applying the **Rational Root Theorem**. This theorem tells us that any rational root of a polynomial with integer coefficients must be of the form $\frac{p}{q}$, where:

- $p$ is a factor of the **constant term** (here, $-14$),
- $q$ is a factor of the **leading coefficient** (here, $1$).

Since the leading coefficient is 1, the only possible rational roots are the **integer factors** of $-14$, which are:

$$
\pm 1, \pm 2, \pm 7, \pm 14.
$$

---

### Step 1: Test Possible Integer Roots

We substitute each of these values into the polynomial to check if they are roots.

- **$x = 1$:**
  $$
  1^4 + 5(1)^3 + 9(1)^2 - 1 - 14 = 1 + 5 + 9 - 1 - 14 = 0
  $$
  ✅ **Root: $x = 1$**

- **$x = -1$:**
  $$
  (-1)^4 + 5(-1)^3 + 9(-1)^2 - (-1) - 14 = 1 - 5 + 9 + 1 - 14 = -8
  $$
  ❌ Not a root

- **$x = 2$:**
  $$
  2^4 + 5(2)^3 + 9(2)^2 - 2 - 14 = 16 + 40 + 36 - 2 - 14 = 76
  $$
  ❌ Not a root

- **$x = -2$:**
  $$
  (-2)^4 + 5(-2)^3 + 9(-2)^2 - (-2) - 14 = 16 - 40 + 36 + 2 - 14 = 0
  $$
  ✅ **Root: $x = -2$**

- **$x = 7$, $x = -7$, $x = 14$, $x = -14$:**
  These values result in large positive or negative values when substituted, and none yield zero.
  ❌ Not roots

So, the only **integer roots** are $x = 1$ and $x = -2$.

---

### Step 2: Factor the Polynomial

We now factor the polynomial using the known roots. Since $x = 1$ and $x = -2$ are roots, the polynomial is divisible by $(x - 1)(x + 2)$. We perform **synthetic division** to divide the original polynomial by $(x - 1)$, then by $(x + 2)$.

#### First Division: $(x - 1)$

Divide $x^4 + 5x^3 + 9x^2 - x - 14$ by $(x - 1)$:

$$
\begin{array}{r|rrrrr}
1 & 1 & 5 & 9 & -1 & -14 \\
  &   & 1 & 6 & 15 & 14 \\
\hline
  & 1 & 6 & 15 & 14 & 0 \\
\end{array}
$$

Result: $x^3 + 6x^2 + 15x + 14$

#### Second Division: $(x + 2)$

Divide $x^3 + 6x^2 + 15x + 14$ by $(x + 2)$:

$$
\begin{array}{r|rrrr}
-2 & 1 & 6 & 15 & 14 \\
   &   & -2 & -8 & -14 \\
\hline
   & 1 & 4 & 7 & 0 \\
\end{array}
$$

Result: $x^2 + 4x + 7$

---

### Step 3: Analyze the Remaining Quadratic

Now we have:

$$
x^4 + 5x^3 + 9x^2 - x - 14 = (x - 1)(x + 2)(x^2 + 4x + 7)
$$

We now check if the quadratic $x^2 + 4x + 7$ has any **real** roots. Compute the discriminant:

$$
D = b^2 - 4ac = 16 - 28 = -12
$$

Since the discriminant is **negative**, the quadratic has **no real roots**, and thus **no integer roots**.

---

### Final Conclusion

The only **integer roots** of the polynomial are:

$$
\boxed{-2}, \boxed{1}
$$

[Assistant]
Explanation:
Okay, so I need to find all the integer roots of the polynomial equation $x^4 + 5x^3 + 9x^2 - x - 14 = 0$. Hmm, let me think about how to approach this.

First, I remember that for polynomials with integer coefficients, the Rational Root Theorem can be helpful. The theorem states that any possible rational root, expressed in its lowest terms $\frac{p}{q}$, must have $p$ as a factor of the constant term and $q$ as a factor of the leading coefficient. Since we're looking for integer roots, $q$ must be 1 or -1, so the possible integer roots are just the factors of the constant term.

Looking at the given polynomial, the constant term is -14, and the leading coefficient is 1. Therefore, the possible integer roots are the factors of -14. Let me list them out: ±1, ±2, ±7, ±14.

Alright, so I need to test each of these values in the polynomial to see if they satisfy the equation. Let me start with the smaller numbers first.

Let me test x = 1. Plugging into the polynomial:

$1^4 + 5(1)^3 + 9(1)^2 - 1 - 14 = 1 + 5 + 9 - 1 - 14$. Let me compute that step by step:

1 + 5 = 6; 6 + 9 = 15; 15 - 1 = 14; 14 - 14 = 0. Oh, that equals zero! So x = 1 is a root. Great, that's one integer root.

Now let me check x = -1. Plugging in:

$(-1)^4 + 5(-1)^3 + 9(-1)^2 - (-1) - 14$. Let me compute each term:

(-1)^4 = 1; 5(-1)^3 = 5(-1) = -5; 9(-1)^2 = 9(1) = 9; -(-1) = +1; and then -14. So adding them up:

1 - 5 + 9 + 1 - 14. Let me compute step by step:

1 - 5 = -4; -4 + 9 = 5; 5 + 1 = 6; 6 - 14 = -8. Not zero. So x = -1 is not a root.

Next, let me try x = 2. Plugging in:

$2^4 + 5(2)^3 + 9(2)^2 - 2 - 14$. Calculating each term:

2^4 = 16; 5*(8) = 40; 9*(4) = 36; then -2 and -14. So total:

16 + 40 = 56; 56 + 36 = 92; 92 - 2 = 90; 90 - 14 = 76. Not zero. So x = 2 is not a root.

How about x = -2? Let's compute:

(-2)^4 + 5(-2)^3 + 9(-2)^2 - (-2) -14.

Calculating each term:

(-2)^4 = 16; 5*(-8) = -40; 9*(4) = 36; -(-2) = +2; and -14.

Adding them up: 16 - 40 = -24; -24 + 36 = 12; 12 + 2 = 14; 14 -14 = 0. Oh! That equals zero. So x = -2 is also a root. Nice, so now we have two integer roots: 1 and -2.

Moving on to x = 7. Let me check that.

7^4 + 5*(7)^3 + 9*(7)^2 - 7 -14.

Calculating each term:

7^4 is 2401; 5*(343) = 1715; 9*(49) = 441; then -7 and -14.

Adding them up: 2401 + 1715 = 4116; 4116 + 441 = 4557; 4557 - 7 = 4550; 4550 -14 = 4536. Definitely not zero. So x = 7 is not a root.

How about x = -7? Let me try that.

(-7)^4 + 5*(-7)^3 + 9*(-7)^2 - (-7) -14.

Calculating each term:

(-7)^4 = 2401; 5*(-343) = -1715; 9*(49) = 441; -(-7) = +7; and -14.

Adding them up: 2401 - 1715 = 686; 686 + 441 = 1127; 1127 + 7 = 1134; 1134 -14 = 1120. Not zero. So x = -7 is not a root.

Next, x = 14. Let me check that.

14^4 + 5*(14)^3 + 9*(14)^2 -14 -14.

This is going to be a huge number. Let me see if I can compute it step by step.

14^2 = 196; 14^3 = 14*196 = 2744; 14^4 = 14*2744 = 38416.

So:

38416 + 5*(2744) + 9*(196) -14 -14.

Calculating each term:

5*2744 = 13720; 9*196 = 1764.

Adding up: 38416 + 13720 = 52136; 52136 + 1764 = 53900; 53900 -14 = 53886; 53886 -14 = 53872. Definitely not zero. So x = 14 is not a root.

Similarly, x = -14 would be even larger in magnitude, so I can safely say it's not a root.

So far, we have found two integer roots: 1 and -2. But since the polynomial is of degree 4, there could be more roots. However, since we are only looking for integer roots, and we have already checked all possible candidates from the Rational Root Theorem, maybe there are no more integer roots? Wait, but maybe there are other integer roots that are not factors of -14? Wait, no, the Rational Root Theorem says that any rational root must be a factor of the constant term over a factor of the leading coefficient. Since the leading coefficient is 1, the only possible rational roots are the factors of -14, which we have already checked. Therefore, there are no other possible integer roots.

But just to be thorough, let me check if maybe there are multiple roots or something. Wait, but we found two roots already. Let me try to factor the polynomial using these roots.

If x = 1 and x = -2 are roots, then (x - 1) and (x + 2) are factors of the polynomial. Let me perform polynomial division or use synthetic division to factor them out.

First, let me divide the original polynomial by (x - 1). Wait, but since we found x = 1 is a root, so (x - 1) is a factor. Let me use synthetic division with root 1.

The coefficients of the polynomial are 1 (x^4), 5 (x^3), 9 (x^2), -1 (x), -14 (constant).

Set up synthetic division:

1 | 1   5   9   -1   -14
  |     1   6   15   14
  ------------------------
    1   6  15   14   0

So the result is x^3 + 6x^2 + 15x + 14.

So now, the polynomial factors as (x - 1)(x^3 + 6x^2 + 15x + 14).

Now, let me check if x = -2 is a root of this cubic polynomial. Let me plug in x = -2 into x^3 + 6x^2 + 15x + 14.

(-2)^3 + 6*(-2)^2 + 15*(-2) + 14 = -8 + 6*(4) + (-30) + 14 = -8 + 24 - 30 + 14.

Calculating step by step: -8 +24 = 16; 16 -30 = -14; -14 +14 = 0. Yes, so x = -2 is a root. Therefore, we can factor out (x + 2) from the cubic.

Let me perform synthetic division on the cubic polynomial x^3 + 6x^2 + 15x + 14 with root -2.

Coefficients: 1, 6, 15, 14

-2 | 1   6   15   14
    |     -2   -8   -14
    -------------------
      1   4   7   0

So the result is x^2 + 4x + 7.

Therefore, the original polynomial factors as (x - 1)(x + 2)(x^2 + 4x + 7).

Now, we need to check if the quadratic factor x^2 + 4x + 7 has any integer roots. Let me compute its discriminant: D = 16 - 4*1*7 = 16 - 28 = -12. Since the discriminant is negative, there are no real roots, let alone integer roots. Therefore, the only integer roots of the original polynomial are x = 1 and x = -2.

Wait, but let me just confirm that I didn't make any mistakes in the synthetic division steps. Let me check the first division again.

Original polynomial: x^4 +5x^3 +9x^2 -x -14 divided by (x - 1).

Using synthetic division with root 1:

Bring down the 1. Multiply by 1: 1. Add to next coefficient: 5 + 1 = 6. Multiply by 1: 6. Add to next coefficient: 9 + 6 = 15. Multiply by 1: 15. Add to next coefficient: -1 + 15 = 14. Multiply by 1: 14. Add to last coefficient: -14 +14 = 0. Correct. So the cubic is x^3 +6x^2 +15x +14.

Then dividing that by (x + 2), which is equivalent to root -2.

Coefficients 1, 6, 15, 14. Bring down the 1. Multiply by -2: -2. Add to 6: 4. Multiply by -2: -8. Add to 15: 7. Multiply by -2: -14. Add to 14: 0. Correct. So quadratic is x^2 +4x +7.

Yes, that seems correct. So the quadratic has no real roots, so no integer roots.

Therefore, the only integer roots are 1 and -2.

But just to be absolutely thorough, let me check if there are any other possible integer roots that I might have missed. Wait, the Rational Root Theorem says that all possible rational roots are factors of -14, which we checked. So there can't be any other integer roots.

Alternatively, maybe I can factor the original polynomial in another way? Let me see.

Alternatively, maybe I can try factoring by grouping? Let me try.

Original polynomial: x^4 +5x^3 +9x^2 -x -14.

Hmm, grouping terms. Let me see.

Looking at x^4 +5x^3 +9x^2 -x -14. Maybe group as (x^4 +5x^3) + (9x^2 -x) -14.

Factor x^3 from first group: x^3(x + 5) + x(9x -1) -14. Doesn't seem helpful.

Alternatively, maybe different grouping. Let me think.

Alternatively, since we already factored it as (x - 1)(x + 2)(x^2 +4x +7), which is the complete factorization. Since the quadratic doesn't factor further over integers, that's the end.

Alternatively, maybe check if the original polynomial can be factored into quadratics? Let me try.

Suppose we try to write it as (x^2 + ax + b)(x^2 + cx + d). Then expanding:

x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd.

Comparing coefficients with original polynomial:

Original coefficients: 1 (x^4), 5 (x^3), 9 (x^2), -1 (x), -14 (constant).

So equations:

1. a + c = 5
2. ac + b + d = 9
3. ad + bc = -1
4. bd = -14

We need integers a, b, c, d satisfying these equations.

We already know that the polynomial factors as (x - 1)(x + 2)(x^2 +4x +7). Let me check if that corresponds to such a factorization.

If we take (x - 1)(x + 2) = x^2 + x - 2. Then multiply by (x^2 +4x +7):

(x^2 + x - 2)(x^2 +4x +7) = x^4 +4x^3 +7x^2 +x^3 +4x^2 +7x -2x^2 -8x -14.

Combine like terms:

x^4 + (4x^3 +x^3) + (7x^2 +4x^2 -2x^2) + (7x -8x) -14 = x^4 +5x^3 +9x^2 -x -14. Which matches.

So if we try to factor into quadratics, we can have (x^2 + x - 2)(x^2 +4x +7).

But since we already found the roots, maybe this isn't necessary. However, just to see if there are other possible factorizations.

But since we already have the roots, and the quadratic factor doesn't have integer roots, I think we can be confident that the only integer roots are 1 and -2.

Alternatively, maybe there's another way to factor? Let me try with different b and d.

From equation 4: bd = -14. So possible integer pairs (b, d) are (1, -14), (-1, 14), (2, -7), (-2, 7), (7, -2), (-7, 2), (14, -1), (-14, 1).

Let me try each possibility.

First, let me try b = 1, d = -14. Then equation 1: a + c = 5. Equation 2: ac + 1 + (-14) = ac -13 = 9 => ac = 22. Equation 3: ad + bc = a*(-14) + c*(1) = -14a + c = -1.

So we have:

a + c = 5

ac = 22

-14a + c = -1

Let me solve these equations. From the first equation, c = 5 - a. Substitute into the third equation:

-14a + (5 - a) = -1 => -15a +5 = -1 => -15a = -6 => a = (-6)/(-15) = 2/5. Not integer. So discard this pair.

Next pair: b = -1, d = 14. Then equation 2: ac + (-1) +14 = ac +13 = 9 => ac = -4. Equation 3: ad + bc = a*14 + c*(-1) = 14a - c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

14a - (5 - a) = 14a -5 + a = 15a -5 = -1 => 15a = 4 => a = 4/15. Not integer. Disregard.

Next pair: b = 2, d = -7. Equation 2: ac + 2 + (-7) = ac -5 = 9 => ac = 14. Equation 3: ad + bc = a*(-7) + c*2 = -7a + 2c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

-7a + 2*(5 - a) = -7a +10 -2a = -9a +10 = -1 => -9a = -11 => a = 11/9. Not integer.

Next pair: b = -2, d = 7. Equation 2: ac + (-2) +7 = ac +5 = 9 => ac = 4. Equation 3: ad + bc = a*7 + c*(-2) = 7a -2c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

7a -2*(5 - a) = 7a -10 + 2a = 9a -10 = -1 => 9a = 9 => a = 1. Then c = 5 -1 = 4.

Check if ac = 1*4 = 4, which matches equation 2. So this works!

Therefore, we have a = 1, c = 4, b = -2, d = 7. Therefore, the factorization is (x^2 + ax + b)(x^2 + cx + d) = (x^2 + x -2)(x^2 +4x +7). Which is exactly what we had earlier.

So that's consistent.

Next pair: b = 7, d = -2. Equation 2: ac +7 + (-2) = ac +5 = 9 => ac = 4. Equation 3: ad + bc = a*(-2) + c*7 = -2a +7c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

-2a +7*(5 - a) = -2a +35 -7a = -9a +35 = -1 => -9a = -36 => a = 4. Then c = 5 -4 = 1.

Check ac = 4*1 = 4, which matches. So this is another valid factorization. So same as before, just swapping the quadratics.

Similarly, other pairs:

b = -7, d = 2. Equation 2: ac + (-7) +2 = ac -5 = 9 => ac = 14. Equation 3: ad + bc = a*2 + c*(-7) = 2a -7c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

2a -7*(5 - a) = 2a -35 +7a = 9a -35 = -1 => 9a = 34 => a = 34/9. Not integer.

Next pair: b = 14, d = -1. Equation 2: ac +14 + (-1) = ac +13 = 9 => ac = -4. Equation 3: ad + bc = a*(-1) + c*14 = -a +14c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

-a +14*(5 - a) = -a +70 -14a = -15a +70 = -1 => -15a = -71 => a = 71/15. Not integer.

Last pair: b = -14, d = 1. Equation 2: ac + (-14) +1 = ac -13 = 9 => ac = 22. Equation 3: ad + bc = a*1 + c*(-14) = a -14c = -1.

With a + c = 5. So c = 5 - a. Substitute into equation 3:

a -14*(5 - a) = a -70 +14a = 15a -70 = -1 => 15a = 69 => a = 69/15 = 23/5. Not integer.

So the only valid factorizations are the ones we found earlier, which confirm that the polynomial factors into (x^2 +x -2)(x^2 +4x +7). And since x^2 +x -2 factors further into (x -1)(x +2), which gives us the integer roots 1 and -2, and the other quadratic has no real roots, we can confirm that these are the only integer roots.

Therefore, after thorough checking, the integer roots are 1 and -2.

Just to recap, I used the Rational Root Theorem to list possible integer roots, tested them, found two roots, factored the polynomial accordingly, and confirmed that the remaining quadratic doesn't have integer roots. Additionally, I tried factoring the polynomial into quadratics to verify the factorization, which also led to the same conclusion.

I think that covers all possibilities. I don't see any mistakes in my reasoning, so I'm confident that the integer roots are 1 and -2.

**Final Answer**
The integer roots are \boxed{-2} and \boxed{1}.

To find all the **integer roots** of the polynomial

$$
x^4 + 5x^3 + 9x^2 - x - 14 = 0,
$$

we begin by applying the **Rational Root Theorem**. This theorem tells us that any rational root of a polynomial with integer coefficients must be of the form $\frac{p}{q}$, where:

- $p$ is a factor of the **constant term** (here, $-14$),
- $q$ is a factor of the **leading coefficient** (here, $1$).

Since the leading coefficient is 1, the only possible rational roots are the **integer factors** of $-14$, which are:

$$
\pm 1, \pm 2, \pm 7, \pm 14.
$$

---

### Step 1: Test Possible Integer Roots

We substitute each of these values into the polynomial to check if they are roots.

- **$x = 1$:**
  $$
  1^4 + 5(1)^3 + 9(1)^2 - 1 - 14 = 1 + 5 + 9 - 1 - 14 = 0
  $$
  ✅ **Root: $x = 1$**

- **$x = -1$:**
  $$
  (-1)^4 + 5(-1)^3 + 9(-1)^2 - (-1) - 14 = 1 - 5 + 9 + 1 - 14 = -8
  $$
  ❌ Not a root

- **$x = 2$:**
  $$
  2^4 + 5(2)^3 + 9(2)^2 - 2 - 14 = 16 + 40 + 36 - 2 - 14 = 76
  $$
  ❌ Not a root

- **$x = -2$:**
  $$
  (-2)^4 + 5(-2)^3 + 9(-2)^2 - (-2) - 14 = 16 - 40 + 36 + 2 - 14 = 0
  $$
  ✅ **Root: $x = -2$**

- **$x = 7$, $x = -7$, $x = 14$, $x = -14$:**
  These values result in large positive or negative values when substituted, and none yield zero.
  ❌ Not roots

So, the only **integer roots** are $x = 1$ and $x = -2$.

---

### Step 2: Factor the Polynomial

We now factor the polynomial using the known roots. Since $x = 1$ and $x = -2$ are roots, the polynomial is divisible by $(x - 1)(x + 2)$. We perform **synthetic division** to divide the original polynomial by $(x - 1)$, then by $(x + 2)$.

#### First Division: $(x - 1)$

Divide $x^4 + 5x^3 + 9x^2 - x - 14$ by $(x - 1)$:

$$
\begin{array}{r|rrrrr}
1 & 1 & 5 & 9 & -1 & -14 \\
  &   & 1 & 6 & 15 & 14 \\
\hline
  & 1 & 6 & 15 & 14 & 0 \\
\end{array}
$$

Result: $x^3 + 6x^2 + 15x + 14$

#### Second Division: $(x + 2)$

Divide $x^3 + 6x^2 + 15x + 14$ by $(x + 2)$:

$$
\begin{array}{r|rrrr}
-2 & 1 & 6 & 15 & 14 \\
   &   & -2 & -8 & -14 \\
\hline
   & 1 & 4 & 7 & 0 \\
\end{array}
$$

Result: $x^2 + 4x + 7$

---

### Step 3: Analyze the Remaining Quadratic

Now we have:

$$
x^4 + 5x^3 + 9x^2 - x - 14 = (x - 1)(x + 2)(x^2 + 4x + 7)
$$

We now check if the quadratic $x^2 + 4x + 7$ has any **real** roots. Compute the discriminant:

$$
D = b^2 - 4ac = 16 - 28 = -12
$$

Since the discriminant is **negative**, the quadratic has **no real roots**, and thus **no integer roots**.

---

### Final Conclusion

The only **integer roots** of the polynomial are:

$$
\boxed{-2}, \boxed{1}
$$

Solution: By the Integer Root Theorem, the possible integer roots are all the divisors of 14 (including negative divisors), which are $-14,$ $-7,$ $-2,$ $-1,$ $1,$ $2,$ $7,$ and $14.$  Checking, we find that the only integer roots are $\boxed{-2,1}.$


Explanation:
Okay, so I need to figure out the remainder when 54 is divided by 6. Let me think. Hmm, modulus operations can sometimes be tricky, but I remember that the remainder is what's left after dividing the number by the modulus. So, basically, I need to divide 54 by 6 and find out what's left over.

Wait, let me recall the definition. The remainder when a number 'a' is divided by 'n' is the integer 'r' such that 0 ≤ r < n and a = qn + r, where q is the quotient. So in this case, a is 54 and n is 6. So I need to find q and r such that 54 = 6q + r, and r is between 0 and 5 inclusive.

Let me try dividing 54 by 6. Well, 6 times 9 is 54, right? 6*9=54. So if I divide 54 by 6, the quotient q is 9, and the remainder r would be 0 because 6*9 is exactly 54. So there's no remainder. Therefore, the remainder is 0.

But wait, let me check that again. Maybe I made a mistake. Let me do the division step by step. 6 goes into 54 how many times? Let's see, 6*10 is 60, which is too big. So 6*9 is 54. So yes, exactly 9 times. So 54 divided by 6 is 9 with no remainder. Therefore, the remainder is 0.

Alternatively, maybe I can think of it in terms of multiples. 6*0=0, 6*1=6, 6*2=12, 6*3=18, 6*4=24, 6*5=30, 6*6=36, 6*7=42, 6*8=48, 6*9=54. So 54 is exactly 6*9, so there's no remainder.

Another way to think about it is using the modulus operation. If I subtract multiples of 6 from 54 until I get a number less than 6. Let's try that. 54 - 6*9 = 54 - 54 = 0. So again, the remainder is 0.

Wait, but maybe I should check with smaller numbers to make sure I'm not making a mistake. For example, if I take 54 divided by 6, let's say I do 54 divided by 6. 6*9 is 54, so remainder is 0. If I take 55 divided by 6, then 6*9=54, so 55-54=1, so remainder 1. Similarly, 54 divided by 6 is exactly 9, so remainder 0.

Alternatively, maybe I can use the formula: remainder = dividend - divisor * floor(dividend / divisor). So here, dividend is 54, divisor is 6. So floor(54 / 6) is 9. Then remainder is 54 - 6*9 = 54 - 54 = 0. Yep, same result.

Is there another way to think about this? Maybe using negative numbers? Wait, no, since we're dealing with positive numbers here.

Alternatively, maybe using modular arithmetic properties. For example, 54 mod 6. Since 6 is a small number, maybe I can break down 54 into parts. Let's see, 54 is 50 + 4. 50 divided by 6 is 8*6=48, remainder 2. Then 4 divided by 6 is 0 with remainder 4. So adding the remainders? Wait, no, that's not the right way. Wait, actually, when you split the number into parts, you can take the modulus of each part and then add them? Let me check.

So 54 = 50 + 4. Then 50 mod 6 is 2, and 4 mod 6 is 4. Then 2 + 4 = 6. But 6 mod 6 is 0. So that gives 0, which matches. So that works.

Alternatively, maybe 54 divided by 6. Let me think of 54 as 6*9, so obviously, 54 is a multiple of 6, so the remainder is 0.

Wait, but maybe I should check with another method. Let me use the division algorithm. The division algorithm states that for any integers a and b (with b > 0), there exist unique integers q and r such that a = bq + r and 0 ≤ r < b.

So here, a = 54, b = 6. So we need to find q and r.

We can find q by dividing 54 by 6. 54 divided by 6 is 9, so q = 9. Then r = a - bq = 54 - 6*9 = 54 - 54 = 0.

Therefore, the remainder r is 0.

Is there any chance that the remainder could be something else? For example, if I thought that 54 divided by 6 is 9.0, but maybe I need to consider decimal places? No, because in modulus, we are dealing with integer division. The remainder is always an integer between 0 and the divisor minus one.

Alternatively, if I use a calculator to divide 54 by 6, it gives exactly 9. So there's no decimal part, so the remainder is 0.

Another way: think of the number line. Starting at 0, adding 6 each time: 0, 6, 12, 18, 24, 30, 36, 42, 48, 54. So 54 is the 9th multiple of 6. Therefore, when you divide 54 by 6, you land exactly on a multiple, so the remainder is 0.

Alternatively, if I think of 54 divided by 6, how many times does 6 go into 54? 6*9=54, so exactly 9 times. Therefore, no remainder.

Wait, maybe I can think of it in terms of congruence. 54 ≡ r mod 6, where 0 ≤ r < 6. So we need to find r such that 54 - r is divisible by 6. So 54 - r must be a multiple of 6. Since 54 is already a multiple of 6, then r must be 0.

Alternatively, if I didn't know that 54 is a multiple of 6, I could compute 54 divided by 6. Let me do that. 6 goes into 5 (the first digit) 0 times. 6 goes into 54 how many times? 6*9=54, so 9 times. So 9*6=54, so remainder is 0.

Alternatively, maybe I can use the fact that 54 is 6*9, so 54 mod 6 is 0.

Is there any possible way that the remainder could be non-zero? For example, if the number wasn't a multiple, but in this case, it is.

Wait, let me check with another example. Let's say 55 mod 6. 55 divided by 6 is 9*6=54, remainder 1. So 55 mod 6 is 1. Similarly, 54 is 55 -1, so 54 mod 6 is 1 -1 = 0? Wait, no, that's not the right way.

But if 55 mod 6 is 1, then 54 mod 6 is 55 mod 6 minus 1, but since 55 mod 6 is 1, then 1 -1 = 0, which is correct. So that works.

Alternatively, maybe using negative numbers. For example, 54 divided by 6. If I think of 54 as 6*9 + 0, so remainder 0.

Alternatively, if I use the Euclidean algorithm. Wait, the Euclidean algorithm is for finding GCDs, but maybe I can use it here? Let me see.

The Euclidean algorithm says that GCD(a, b) = GCD(b, a mod b). But here, we are just trying to find a mod b. But maybe if I consider GCD(54, 6). Well, since 54 is a multiple of 6, the GCD is 6. But I don't know if that helps here.

Alternatively, maybe I can think of 54 divided by 6. Let me write it out as long division.

6 | 54

How many times does 6 go into 5? 0 times. So 0. Then 6 goes into 54 how many times? 9 times. 9*6 is 54. Subtract 54 from 54, get 0. So the remainder is 0.

So all methods point to the remainder being 0.

Wait, but maybe I need to check if there's a different interpretation of the question? The question says "the remainder of 54 (mod 6)". So mod 6, so 54 mod 6. Which is the same as asking for the remainder when 54 is divided by 6. Which is 0.

Alternatively, maybe the question is trying to trick me? Like, maybe 54 is written in a different base? But the question doesn't specify that. So I think it's safe to assume that 54 is in base 10.

Alternatively, maybe the question is in another language, and "remainder" is being used differently? But no, in mathematics, the remainder when dividing by a number is standard.

Alternatively, maybe I need to consider negative remainders? But no, the remainder is always non-negative.

Wait, if I consider negative numbers, for example, if I have 54 divided by 6, but if I take a negative quotient, but the remainder is still defined to be non-negative.

For example, if I do 54 divided by 6, the quotient can be 9, and remainder 0. If I take quotient as 8, then remainder would be 54 - 6*8 = 54 - 48 = 6, but 6 is not less than 6, so that's invalid. So the remainder has to be less than 6. So the only valid remainder is 0.

Therefore, I think I can be confident that the remainder is 0.

But just to make sure, let me check with another approach. Let me think of 54 divided by 6.

6*10 = 60, which is larger than 54. So 6*9 = 54. So 9 times. So 54 - 6*9 = 0.

Alternatively, think of 54 divided by 6. 6*5=30, 54-30=24. 6*4=24, so total is 5+4=9. 24-24=0. So again, remainder 0.

Alternatively, think of 54 divided by 6. 6*2=12, 54-12=42. 6*7=42, so total is 2+7=9. 42-42=0.

So regardless of how I break it down, the remainder is 0.

Therefore, I think the answer is 0.

**Final Answer**
The remainder of 54 divided by 6 is \boxed{0}.

To determine the remainder of $ 54 \mod 6 $, we apply the concept of modular arithmetic, which involves finding the **remainder** when one number is divided by another.

---

### Step-by-Step Breakdown:

We are given:
- Dividend: $ 54 $
- Divisor: $ 6 $

We want to find the **remainder** when $ 54 $ is divided by $ 6 $, which is denoted as $ 54 \mod 6 $.

---

### Step 1: Perform the Division

We divide $ 54 $ by $ 6 $:

$$
54 \div 6 = 9
$$

This means that $ 6 $ fits into $ 54 $ exactly **9 times** with **no remainder**.

---

### Step 2: Use the Division Algorithm

According to the **division algorithm**, for any integers $ a $ and $ b $ (with $ b > 0 $), there exist unique integers $ q $ (quotient) and $ r $ (remainder) such that:

$$
a = bq + r \quad \text{where } 0 \leq r < b
$$

In this case:
- $ a = 54 $
- $ b = 6 $
- $ q = 9 $
- $ r = 54 - 6 \times 9 = 54 - 54 = 0 $

So, the remainder $ r = 0 $.

---

### Step 3: Confirm with Modular Arithmetic

We can also directly compute:

$$
54 \mod 6 = 0
$$

This confirms that $ 54 $ is a **multiple of 6**, and thus the remainder is **zero**.

---

### Final Answer

$$
\boxed{0}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the remainder when 54 is divided by 6. Let me think. Hmm, modulus operations can sometimes be tricky, but I remember that the remainder is what's left after dividing the number by the modulus. So, basically, I need to divide 54 by 6 and find out what's left over.

Wait, let me recall the definition. The remainder when a number 'a' is divided by 'n' is the integer 'r' such that 0 ≤ r < n and a = qn + r, where q is the quotient. So in this case, a is 54 and n is 6. So I need to find q and r such that 54 = 6q + r, and r is between 0 and 5 inclusive.

Let me try dividing 54 by 6. Well, 6 times 9 is 54, right? 6*9=54. So if I divide 54 by 6, the quotient q is 9, and the remainder r would be 0 because 6*9 is exactly 54. So there's no remainder. Therefore, the remainder is 0.

But wait, let me check that again. Maybe I made a mistake. Let me do the division step by step. 6 goes into 54 how many times? Let's see, 6*10 is 60, which is too big. So 6*9 is 54. So yes, exactly 9 times. So 54 divided by 6 is 9 with no remainder. Therefore, the remainder is 0.

Alternatively, maybe I can think of it in terms of multiples. 6*0=0, 6*1=6, 6*2=12, 6*3=18, 6*4=24, 6*5=30, 6*6=36, 6*7=42, 6*8=48, 6*9=54. So 54 is exactly 6*9, so there's no remainder.

Another way to think about it is using the modulus operation. If I subtract multiples of 6 from 54 until I get a number less than 6. Let's try that. 54 - 6*9 = 54 - 54 = 0. So again, the remainder is 0.

Wait, but maybe I should check with smaller numbers to make sure I'm not making a mistake. For example, if I take 54 divided by 6, let's say I do 54 divided by 6. 6*9 is 54, so remainder is 0. If I take 55 divided by 6, then 6*9=54, so 55-54=1, so remainder 1. Similarly, 54 divided by 6 is exactly 9, so remainder 0.

Alternatively, maybe I can use the formula: remainder = dividend - divisor * floor(dividend / divisor). So here, dividend is 54, divisor is 6. So floor(54 / 6) is 9. Then remainder is 54 - 6*9 = 54 - 54 = 0. Yep, same result.

Is there another way to think about this? Maybe using negative numbers? Wait, no, since we're dealing with positive numbers here.

Alternatively, maybe using modular arithmetic properties. For example, 54 mod 6. Since 6 is a small number, maybe I can break down 54 into parts. Let's see, 54 is 50 + 4. 50 divided by 6 is 8*6=48, remainder 2. Then 4 divided by 6 is 0 with remainder 4. So adding the remainders? Wait, no, that's not the right way. Wait, actually, when you split the number into parts, you can take the modulus of each part and then add them? Let me check.

So 54 = 50 + 4. Then 50 mod 6 is 2, and 4 mod 6 is 4. Then 2 + 4 = 6. But 6 mod 6 is 0. So that gives 0, which matches. So that works.

Alternatively, maybe 54 divided by 6. Let me think of 54 as 6*9, so obviously, 54 is a multiple of 6, so the remainder is 0.

Wait, but maybe I should check with another method. Let me use the division algorithm. The division algorithm states that for any integers a and b (with b > 0), there exist unique integers q and r such that a = bq + r and 0 ≤ r < b.

So here, a = 54, b = 6. So we need to find q and r.

We can find q by dividing 54 by 6. 54 divided by 6 is 9, so q = 9. Then r = a - bq = 54 - 6*9 = 54 - 54 = 0.

Therefore, the remainder r is 0.

Is there any chance that the remainder could be something else? For example, if I thought that 54 divided by 6 is 9.0, but maybe I need to consider decimal places? No, because in modulus, we are dealing with integer division. The remainder is always an integer between 0 and the divisor minus one.

Alternatively, if I use a calculator to divide 54 by 6, it gives exactly 9. So there's no decimal part, so the remainder is 0.

Another way: think of the number line. Starting at 0, adding 6 each time: 0, 6, 12, 18, 24, 30, 36, 42, 48, 54. So 54 is the 9th multiple of 6. Therefore, when you divide 54 by 6, you land exactly on a multiple, so the remainder is 0.

Alternatively, if I think of 54 divided by 6, how many times does 6 go into 54? 6*9=54, so exactly 9 times. Therefore, no remainder.

Wait, maybe I can think of it in terms of congruence. 54 ≡ r mod 6, where 0 ≤ r < 6. So we need to find r such that 54 - r is divisible by 6. So 54 - r must be a multiple of 6. Since 54 is already a multiple of 6, then r must be 0.

Alternatively, if I didn't know that 54 is a multiple of 6, I could compute 54 divided by 6. Let me do that. 6 goes into 5 (the first digit) 0 times. 6 goes into 54 how many times? 6*9=54, so 9 times. So 9*6=54, so remainder is 0.

Alternatively, maybe I can use the fact that 54 is 6*9, so 54 mod 6 is 0.

Is there any possible way that the remainder could be non-zero? For example, if the number wasn't a multiple, but in this case, it is.

Wait, let me check with another example. Let's say 55 mod 6. 55 divided by 6 is 9*6=54, remainder 1. So 55 mod 6 is 1. Similarly, 54 is 55 -1, so 54 mod 6 is 1 -1 = 0? Wait, no, that's not the right way.

But if 55 mod 6 is 1, then 54 mod 6 is 55 mod 6 minus 1, but since 55 mod 6 is 1, then 1 -1 = 0, which is correct. So that works.

Alternatively, maybe using negative numbers. For example, 54 divided by 6. If I think of 54 as 6*9 + 0, so remainder 0.

Alternatively, if I use the Euclidean algorithm. Wait, the Euclidean algorithm is for finding GCDs, but maybe I can use it here? Let me see.

The Euclidean algorithm says that GCD(a, b) = GCD(b, a mod b). But here, we are just trying to find a mod b. But maybe if I consider GCD(54, 6). Well, since 54 is a multiple of 6, the GCD is 6. But I don't know if that helps here.

Alternatively, maybe I can think of 54 divided by 6. Let me write it out as long division.

6 | 54

How many times does 6 go into 5? 0 times. So 0. Then 6 goes into 54 how many times? 9 times. 9*6 is 54. Subtract 54 from 54, get 0. So the remainder is 0.

So all methods point to the remainder being 0.

Wait, but maybe I need to check if there's a different interpretation of the question? The question says "the remainder of 54 (mod 6)". So mod 6, so 54 mod 6. Which is the same as asking for the remainder when 54 is divided by 6. Which is 0.

Alternatively, maybe the question is trying to trick me? Like, maybe 54 is written in a different base? But the question doesn't specify that. So I think it's safe to assume that 54 is in base 10.

Alternatively, maybe the question is in another language, and "remainder" is being used differently? But no, in mathematics, the remainder when dividing by a number is standard.

Alternatively, maybe I need to consider negative remainders? But no, the remainder is always non-negative.

Wait, if I consider negative numbers, for example, if I have 54 divided by 6, but if I take a negative quotient, but the remainder is still defined to be non-negative.

For example, if I do 54 divided by 6, the quotient can be 9, and remainder 0. If I take quotient as 8, then remainder would be 54 - 6*8 = 54 - 48 = 6, but 6 is not less than 6, so that's invalid. So the remainder has to be less than 6. So the only valid remainder is 0.

Therefore, I think I can be confident that the remainder is 0.

But just to make sure, let me check with another approach. Let me think of 54 divided by 6.

6*10 = 60, which is larger than 54. So 6*9 = 54. So 9 times. So 54 - 6*9 = 0.

Alternatively, think of 54 divided by 6. 6*5=30, 54-30=24. 6*4=24, so total is 5+4=9. 24-24=0. So again, remainder 0.

Alternatively, think of 54 divided by 6. 6*2=12, 54-12=42. 6*7=42, so total is 2+7=9. 42-42=0.

So regardless of how I break it down, the remainder is 0.

Therefore, I think the answer is 0.

**Final Answer**
The remainder of 54 divided by 6 is \boxed{0}.

To determine the remainder of $ 54 \mod 6 $, we apply the concept of modular arithmetic, which involves finding the **remainder** when one number is divided by another.

---

### Step-by-Step Breakdown:

We are given:
- Dividend: $ 54 $
- Divisor: $ 6 $

We want to find the **remainder** when $ 54 $ is divided by $ 6 $, which is denoted as $ 54 \mod 6 $.

---

### Step 1: Perform the Division

We divide $ 54 $ by $ 6 $:

$$
54 \div 6 = 9
$$

This means that $ 6 $ fits into $ 54 $ exactly **9 times** with **no remainder**.

---

### Step 2: Use the Division Algorithm

According to the **division algorithm**, for any integers $ a $ and $ b $ (with $ b > 0 $), there exist unique integers $ q $ (quotient) and $ r $ (remainder) such that:

$$
a = bq + r \quad \text{where } 0 \leq r < b
$$

In this case:
- $ a = 54 $
- $ b = 6 $
- $ q = 9 $
- $ r = 54 - 6 \times 9 = 54 - 54 = 0 $

So, the remainder $ r = 0 $.

---

### Step 3: Confirm with Modular Arithmetic

We can also directly compute:

$$
54 \mod 6 = 0
$$

This confirms that $ 54 $ is a **multiple of 6**, and thus the remainder is **zero**.

---

### Final Answer

$$
\boxed{0}
$$

Solution: $54 = 9 \cdot 6 + 0 \Rightarrow 54 \equiv \boxed{0} \pmod{6}$.


Explanation:
Okay, so I need to figure out how far a car would travel if it's going 50 miles per hour for 2 and 3/4 hours. Hmm, let me think. I remember that distance equals speed multiplied by time. So the formula is distance = speed × time. Right?

Alright, so the speed here is 50 miles per hour, and the time is 2 and 3/4 hours. Wait, 2 and 3/4 hours is a mixed number. Maybe I should convert that to an improper fraction or a decimal to make the multiplication easier? Let me try both ways and see which is simpler.

First, converting 2 3/4 hours to an improper fraction. The whole number part is 2, which is 2/1. The fractional part is 3/4. To add them together, I need a common denominator. So 2 is 8/4, right? Because 2 times 4 is 8. Then adding 8/4 and 3/4 gives me 11/4. So 2 3/4 hours is equal to 11/4 hours. Okay, that seems right.

Alternatively, if I convert it to a decimal, 3/4 is 0.75, so 2.75 hours. Either way, I can use that in the calculation. Let me try both methods to check my answer.

Starting with the improper fraction: distance = 50 mph × 11/4 hours. Let me compute that. Multiplying 50 by 11 first. 50 times 11 is... 550. Then divide by 4. So 550 divided by 4. Let me do that division. 4 goes into 550 how many times?

Well, 4 × 137 is 548, because 4 × 100 is 400, 4 × 30 is 120, 4 × 7 is 28. So 400 + 120 is 520, plus 28 is 548. Then 550 minus 548 is 2. So that would be 137 and 2/4, which simplifies to 137 and 1/2. So 137.5 miles? Wait, but the question says to express the answer as a mixed number. So 137 1/2 miles. Let me check that again.

Alternatively, using decimals: 50 mph × 2.75 hours. Let me compute that. 50 × 2 is 100, and 50 × 0.75 is 37.5. Adding them together, 100 + 37.5 is 137.5. Which is the same as 137 1/2. So that matches. So both methods give me the same answer. That seems good.

Wait, let me just make sure I didn't make any calculation errors. Let me verify the improper fraction method again. 50 multiplied by 11 is indeed 550. Divided by 4. 550 divided by 4. Let me divide 550 by 2 first, which is 275, then divide by 2 again, which is 137.5. Yep, same result. So that's 137.5, which is 137 and 1/2. So that's correct.

Alternatively, if I think about the time. 2 and 3/4 hours is 2 hours plus 45 minutes. Because 3/4 of an hour is 45 minutes. So if the car is going 50 mph, in 2 hours it would go 50 × 2 = 100 miles. Then in 45 minutes, which is 3/4 of an hour, it would go 50 × 3/4. Let me calculate that. 50 × 3 is 150, divided by 4 is 37.5. So adding 100 + 37.5 is 137.5. Same answer again. So that's consistent.

Therefore, regardless of the method I use, I get 137.5 miles, which as a mixed number is 137 1/2. So I think that's the answer. Let me just check if there's any chance I messed up the conversion from the mixed number to the improper fraction. Original time was 2 3/4 hours. So 2 is 8/4, plus 3/4 is 11/4. Yes, that's correct. So 11/4 hours. Then 50 × 11/4. 50 × 11 is 550. Divided by 4. Yep, 550/4 reduces to 275/2, which is 137 1/2. So that's correct.

Alternatively, if I think about units. The speed is in miles per hour, time is in hours, so multiplying them gives miles. So units check out. So 50 miles per hour times 2.75 hours is 137.5 miles. So that's correct.

I think I've verified it multiple ways. So I feel confident that the answer is 137 and 1/2 miles. So expressed as a mixed number, that's 137 1/2. So I should write that in the box.

**Final Answer**
The car would travel \boxed{137\frac{1}{2}} miles.

To determine how far a car would travel at a constant speed of 50 miles per hour over a period of $ 2\frac{3}{4} $ hours, we use the fundamental formula for distance:

$$
\text{Distance} = \text{Speed} \times \text{Time}
$$

---

### Step 1: Convert the Mixed Number to an Improper Fraction

The time given is $ 2\frac{3}{4} $ hours. To convert this to an improper fraction:

$$
2\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{8 + 3}{4} = \frac{11}{4}
$$

---

### Step 2: Multiply Speed by Time

Now, multiply the speed (50 miles per hour) by the time ($ \frac{11}{4} $ hours):

$$
\text{Distance} = 50 \times \frac{11}{4} = \frac{50 \times 11}{4} = \frac{550}{4}
$$

---

### Step 3: Simplify the Result

Divide 550 by 4:

$$
\frac{550}{4} = 137.5
$$

To express this as a **mixed number**, we note that:

$$
137.5 = 137 + 0.5 = 137\frac{1}{2}
$$

---

### Final Answer

$$
\boxed{137\frac{1}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how far a car would travel if it's going 50 miles per hour for 2 and 3/4 hours. Hmm, let me think. I remember that distance equals speed multiplied by time. So the formula is distance = speed × time. Right?

Alright, so the speed here is 50 miles per hour, and the time is 2 and 3/4 hours. Wait, 2 and 3/4 hours is a mixed number. Maybe I should convert that to an improper fraction or a decimal to make the multiplication easier? Let me try both ways and see which is simpler.

First, converting 2 3/4 hours to an improper fraction. The whole number part is 2, which is 2/1. The fractional part is 3/4. To add them together, I need a common denominator. So 2 is 8/4, right? Because 2 times 4 is 8. Then adding 8/4 and 3/4 gives me 11/4. So 2 3/4 hours is equal to 11/4 hours. Okay, that seems right.

Alternatively, if I convert it to a decimal, 3/4 is 0.75, so 2.75 hours. Either way, I can use that in the calculation. Let me try both methods to check my answer.

Starting with the improper fraction: distance = 50 mph × 11/4 hours. Let me compute that. Multiplying 50 by 11 first. 50 times 11 is... 550. Then divide by 4. So 550 divided by 4. Let me do that division. 4 goes into 550 how many times?

Well, 4 × 137 is 548, because 4 × 100 is 400, 4 × 30 is 120, 4 × 7 is 28. So 400 + 120 is 520, plus 28 is 548. Then 550 minus 548 is 2. So that would be 137 and 2/4, which simplifies to 137 and 1/2. So 137.5 miles? Wait, but the question says to express the answer as a mixed number. So 137 1/2 miles. Let me check that again.

Alternatively, using decimals: 50 mph × 2.75 hours. Let me compute that. 50 × 2 is 100, and 50 × 0.75 is 37.5. Adding them together, 100 + 37.5 is 137.5. Which is the same as 137 1/2. So that matches. So both methods give me the same answer. That seems good.

Wait, let me just make sure I didn't make any calculation errors. Let me verify the improper fraction method again. 50 multiplied by 11 is indeed 550. Divided by 4. 550 divided by 4. Let me divide 550 by 2 first, which is 275, then divide by 2 again, which is 137.5. Yep, same result. So that's 137.5, which is 137 and 1/2. So that's correct.

Alternatively, if I think about the time. 2 and 3/4 hours is 2 hours plus 45 minutes. Because 3/4 of an hour is 45 minutes. So if the car is going 50 mph, in 2 hours it would go 50 × 2 = 100 miles. Then in 45 minutes, which is 3/4 of an hour, it would go 50 × 3/4. Let me calculate that. 50 × 3 is 150, divided by 4 is 37.5. So adding 100 + 37.5 is 137.5. Same answer again. So that's consistent.

Therefore, regardless of the method I use, I get 137.5 miles, which as a mixed number is 137 1/2. So I think that's the answer. Let me just check if there's any chance I messed up the conversion from the mixed number to the improper fraction. Original time was 2 3/4 hours. So 2 is 8/4, plus 3/4 is 11/4. Yes, that's correct. So 11/4 hours. Then 50 × 11/4. 50 × 11 is 550. Divided by 4. Yep, 550/4 reduces to 275/2, which is 137 1/2. So that's correct.

Alternatively, if I think about units. The speed is in miles per hour, time is in hours, so multiplying them gives miles. So units check out. So 50 miles per hour times 2.75 hours is 137.5 miles. So that's correct.

I think I've verified it multiple ways. So I feel confident that the answer is 137 and 1/2 miles. So expressed as a mixed number, that's 137 1/2. So I should write that in the box.

**Final Answer**
The car would travel \boxed{137\frac{1}{2}} miles.

To determine how far a car would travel at a constant speed of 50 miles per hour over a period of $ 2\frac{3}{4} $ hours, we use the fundamental formula for distance:

$$
\text{Distance} = \text{Speed} \times \text{Time}
$$

---

### Step 1: Convert the Mixed Number to an Improper Fraction

The time given is $ 2\frac{3}{4} $ hours. To convert this to an improper fraction:

$$
2\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{8 + 3}{4} = \frac{11}{4}
$$

---

### Step 2: Multiply Speed by Time

Now, multiply the speed (50 miles per hour) by the time ($ \frac{11}{4} $ hours):

$$
\text{Distance} = 50 \times \frac{11}{4} = \frac{50 \times 11}{4} = \frac{550}{4}
$$

---

### Step 3: Simplify the Result

Divide 550 by 4:

$$
\frac{550}{4} = 137.5
$$

To express this as a **mixed number**, we note that:

$$
137.5 = 137 + 0.5 = 137\frac{1}{2}
$$

---

### Final Answer

$$
\boxed{137\frac{1}{2}}
$$

Solution: In two hours, a car traveling at $50$ mph would travel $50$ mph $\times 2$ hours $= 100$ miles. Now we find how far a car can travel in $3/4$ of an hour which is $50$ mph $\times \frac{3}{4}$ hours $ = \frac{150}{4} = 37 \frac{1}{2}$ miles. Thus, the car would travel a total of $100 + 37 \frac{1}{2}= \boxed{137 \frac{1}{2}}$ miles.


Explanation:
Okay, so I need to find the value of this product of logarithms: log base 2 of 3 times log base 3 of 4 times log base 4 of 5, and so on up to log base 7 of 8. Let me write that down to visualize it better:

log₂3 × log₃4 × log₄5 × log₅6 × log₆7 × log₇8

Hmm, I remember that there's a property of logarithms that can help simplify products of logs with consecutive bases. Let me recall... Oh yeah! The change of base formula. Wait, but isn't there something called the chain rule for logarithms? Or maybe it's called the telescoping product?

Let me think. If I have log_a b × log_b c, that's equal to log_a c, right? Wait, is that true? Let me check with numbers. Let's take log₂4 × log₄8. Log base 2 of 4 is 2, log base 4 of 8 is log₄(4×2) = log₄4 + log₄2 = 1 + 0.5 = 1.5. Multiplying them gives 2 × 1.5 = 3. On the other hand, log₂8 is 3. So yes! It works. So log_a b × log_b c = log_a c. That seems to be the case. So this is called the chain rule for logarithms? Or maybe it's a property of changing the base?

So if that's the case, then maybe this entire product telescopes? Let me try applying that property step by step.

Starting with log₂3 × log₃4. According to the property, that should be log₂4. Then multiply by log₄5, so log₂4 × log₄5 = log₂5. Then multiply by log₅6, so log₂5 × log₅6 = log₂6. Then multiply by log₆7, so log₂6 × log₆7 = log₂7. Then multiply by log₇8, so log₂7 × log₇8 = log₂8.

So the entire product simplifies to log₂8. And log base 2 of 8 is 3, since 2^3 = 8. Therefore, the value of the entire product is 3. Wait, that seems straightforward. Let me verify with another approach to make sure I didn't make a mistake.

Alternatively, I can use the change of base formula on each term. The change of base formula says that log_a b = ln b / ln a, where ln is the natural logarithm. So let me rewrite each term:

log₂3 = ln3 / ln2

log₃4 = ln4 / ln3

log₄5 = ln5 / ln4

log₅6 = ln6 / ln5

log₆7 = ln7 / ln6

log₇8 = ln8 / ln7

So multiplying all these together:

(ln3 / ln2) × (ln4 / ln3) × (ln5 / ln4) × (ln6 / ln5) × (ln7 / ln6) × (ln8 / ln7)

Now, let's see if there's cancellation here. The ln3 in the numerator of the first term cancels with the ln3 in the denominator of the second term. Similarly, ln4 in the numerator of the second term cancels with ln4 in the denominator of the third term. This pattern continues all the way through. So after cancellation, we are left with ln8 / ln2. Which is log₂8, same as before. And log₂8 is 3. So that confirms the result.

Therefore, regardless of the method, the answer is 3. Wait, but let me check once more with actual numbers. Let me compute each term numerically and multiply them to see if it's approximately 3.

First, compute each logarithm:

log₂3 ≈ 1.58496

log₃4 ≈ 1.26186

log₄5 ≈ 1.16096

log₅6 ≈ 1.11328

log₆7 ≈ 1.08605

log₇8 ≈ 1.0686

Now multiply them all together:

Start with 1.58496 × 1.26186 ≈ Let me compute that. 1.58496 × 1.26186. Let me approximate:

First, 1.5 × 1.26 = 1.89, but more accurately:

1.58496 × 1.26186 ≈ Let me use calculator steps:

Well, 1.58496 × 1.26186 ≈ Let me approximate:

Let me note that 1.58496 × 1.26186 ≈ Let me use decimal multiplication:

Approximately, 1.585 × 1.262 ≈

Multiply 1.585 × 1.262:

First compute 1.5 × 1.262 = 1.893

Then 0.085 × 1.262 ≈ 0.10727

Adding together: 1.893 + 0.10727 ≈ 2.00027. So approximately 2.0003.

So that's the first two terms. Then multiply by the third term, which is approx 1.16096:

2.0003 × 1.16096 ≈ 2 × 1.16096 = 2.32192, plus 0.0003 × 1.16096 ≈ 0.000348, so total ≈ 2.32227.

Next multiply by 1.11328:

2.32227 × 1.11328 ≈ Let me approximate:

2 × 1.11328 = 2.22656

0.32227 × 1.11328 ≈ Let me compute 0.3 × 1.11328 = 0.33398, and 0.02227 × 1.11328 ≈ approx 0.02477. So total ≈ 0.33398 + 0.02477 ≈ 0.35875. So total ≈ 2.22656 + 0.35875 ≈ 2.58531.

Next multiply by 1.08605:

2.58531 × 1.08605 ≈ Let me compute 2.58531 × 1.08 ≈ 2.58531 + 2.58531 × 0.08 ≈ 2.58531 + 0.20682 ≈ 2.79213. Then adjust for the extra 0.00605. So 2.58531 × 0.00605 ≈ approx 0.01564. So total ≈ 2.79213 + 0.01564 ≈ 2.80777.

Then multiply by 1.0686:

2.80777 × 1.0686 ≈ Let me compute 2.80777 × 1.06 ≈ 2.80777 + 2.80777 × 0.06 ≈ 2.80777 + 0.16846 ≈ 2.97623. Then 2.80777 × 0.0086 ≈ approx 0.02415. So total ≈ 2.97623 + 0.02415 ≈ 3.00038.

Wow! So the approximate value is about 3.00038, which is very close to 3. Considering that I approximated each logarithm and did some rounding, this seems to confirm that the exact value is indeed 3. So that gives me more confidence.

Therefore, the answer is 3. But let me just think again if there's another way to see this. Maybe using exponents?

Alternatively, since each term is log_{n} (n+1) where n starts at 2 and goes up to 7. Wait, but the first term is log₂3, then log₃4, etc., so yeah, each term is log_{k} (k+1) for k from 2 to 7. So the product is from k=2 to k=7 of log_k (k+1). And as we saw, this telescopes to log₂8, which is 3. So that's another way to think about it.

Alternatively, if I consider the general case: the product from k = a to k = b of log_k (k+1) equals log_a (b+1). Is that a general formula? Let me check with smaller numbers. Let's take a=2, b=2: product is log₂3, which is log₂(2+1). If a=2, b=3: product is log₂3 × log₃4 = log₂4 = log₂(3+1). So yes, seems like the general formula is that the product from k=a to k=b of log_k (k+1) equals log_a (b+1). Therefore, in our case, a=2, b=7, so log₂(7+1)=log₂8=3. So that's another way to see it.

Therefore, regardless of the approach, the answer is 3. I think that's solid. I don't see any mistakes in my reasoning. The key idea is recognizing the telescoping product through the change of base formula or the chain rule for logarithms. It's a neat property that allows the product to collapse into a single logarithm. So yeah, the answer is 3.

**Final Answer**
\boxed{3}

To evaluate the expression:

$$
\log_2{3} \cdot \log_3{4} \cdot \log_4{5} \cdot \log_5{6} \cdot \log_6{7} \cdot \log_7{8}
$$

we can utilize a key property of logarithms known as the **change of base formula** and the **telescoping behavior** of logarithmic products.

---

### Step 1: Apply the Change of Base Formula

Recall the change of base formula:

$$
\log_a b = \frac{\ln b}{\ln a}
$$

Using this, we can rewrite each term in the product:

$$
\log_2{3} = \frac{\ln 3}{\ln 2}, \quad
\log_3{4} = \frac{\ln 4}{\ln 3}, \quad
\log_4{5} = \frac{\ln 5}{\ln 4}, \quad
\log_5{6} = \frac{\ln 6}{\ln 5}, \quad
\log_6{7} = \frac{\ln 7}{\ln 6}, \quad
\log_7{8} = \frac{\ln 8}{\ln 7}
$$

Now, multiplying all these together:

$$
\left( \frac{\ln 3}{\ln 2} \right) \cdot \left( \frac{\ln 4}{\ln 3} \right) \cdot \left( \frac{\ln 5}{\ln 4} \right) \cdot \left( \frac{\ln 6}{\ln 5} \right) \cdot \left( \frac{\ln 7}{\ln 6} \right) \cdot \left( \frac{\ln 8}{\ln 7} \right)
$$

---

### Step 2: Observe the Telescoping Pattern

Notice that the numerator of each fraction cancels with the denominator of the next:

- $\ln 3$ in the numerator of the first term cancels with $\ln 3$ in the denominator of the second term.
- $\ln 4$ in the numerator of the second term cancels with $\ln 4$ in the denominator of the third term.
- This pattern continues all the way through.

After all cancellations, only the first denominator and the last numerator remain:

$$
\frac{\ln 8}{\ln 2}
$$

This is equivalent to:

$$
\log_2{8}
$$

---

### Step 3: Evaluate the Final Logarithm

$$
\log_2{8} = 3 \quad \text{since } 2^3 = 8
$$

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of this product of logarithms: log base 2 of 3 times log base 3 of 4 times log base 4 of 5, and so on up to log base 7 of 8. Let me write that down to visualize it better:

log₂3 × log₃4 × log₄5 × log₅6 × log₆7 × log₇8

Hmm, I remember that there's a property of logarithms that can help simplify products of logs with consecutive bases. Let me recall... Oh yeah! The change of base formula. Wait, but isn't there something called the chain rule for logarithms? Or maybe it's called the telescoping product?

Let me think. If I have log_a b × log_b c, that's equal to log_a c, right? Wait, is that true? Let me check with numbers. Let's take log₂4 × log₄8. Log base 2 of 4 is 2, log base 4 of 8 is log₄(4×2) = log₄4 + log₄2 = 1 + 0.5 = 1.5. Multiplying them gives 2 × 1.5 = 3. On the other hand, log₂8 is 3. So yes! It works. So log_a b × log_b c = log_a c. That seems to be the case. So this is called the chain rule for logarithms? Or maybe it's a property of changing the base?

So if that's the case, then maybe this entire product telescopes? Let me try applying that property step by step.

Starting with log₂3 × log₃4. According to the property, that should be log₂4. Then multiply by log₄5, so log₂4 × log₄5 = log₂5. Then multiply by log₅6, so log₂5 × log₅6 = log₂6. Then multiply by log₆7, so log₂6 × log₆7 = log₂7. Then multiply by log₇8, so log₂7 × log₇8 = log₂8.

So the entire product simplifies to log₂8. And log base 2 of 8 is 3, since 2^3 = 8. Therefore, the value of the entire product is 3. Wait, that seems straightforward. Let me verify with another approach to make sure I didn't make a mistake.

Alternatively, I can use the change of base formula on each term. The change of base formula says that log_a b = ln b / ln a, where ln is the natural logarithm. So let me rewrite each term:

log₂3 = ln3 / ln2

log₃4 = ln4 / ln3

log₄5 = ln5 / ln4

log₅6 = ln6 / ln5

log₆7 = ln7 / ln6

log₇8 = ln8 / ln7

So multiplying all these together:

(ln3 / ln2) × (ln4 / ln3) × (ln5 / ln4) × (ln6 / ln5) × (ln7 / ln6) × (ln8 / ln7)

Now, let's see if there's cancellation here. The ln3 in the numerator of the first term cancels with the ln3 in the denominator of the second term. Similarly, ln4 in the numerator of the second term cancels with ln4 in the denominator of the third term. This pattern continues all the way through. So after cancellation, we are left with ln8 / ln2. Which is log₂8, same as before. And log₂8 is 3. So that confirms the result.

Therefore, regardless of the method, the answer is 3. Wait, but let me check once more with actual numbers. Let me compute each term numerically and multiply them to see if it's approximately 3.

First, compute each logarithm:

log₂3 ≈ 1.58496

log₃4 ≈ 1.26186

log₄5 ≈ 1.16096

log₅6 ≈ 1.11328

log₆7 ≈ 1.08605

log₇8 ≈ 1.0686

Now multiply them all together:

Start with 1.58496 × 1.26186 ≈ Let me compute that. 1.58496 × 1.26186. Let me approximate:

First, 1.5 × 1.26 = 1.89, but more accurately:

1.58496 × 1.26186 ≈ Let me use calculator steps:

Well, 1.58496 × 1.26186 ≈ Let me approximate:

Let me note that 1.58496 × 1.26186 ≈ Let me use decimal multiplication:

Approximately, 1.585 × 1.262 ≈

Multiply 1.585 × 1.262:

First compute 1.5 × 1.262 = 1.893

Then 0.085 × 1.262 ≈ 0.10727

Adding together: 1.893 + 0.10727 ≈ 2.00027. So approximately 2.0003.

So that's the first two terms. Then multiply by the third term, which is approx 1.16096:

2.0003 × 1.16096 ≈ 2 × 1.16096 = 2.32192, plus 0.0003 × 1.16096 ≈ 0.000348, so total ≈ 2.32227.

Next multiply by 1.11328:

2.32227 × 1.11328 ≈ Let me approximate:

2 × 1.11328 = 2.22656

0.32227 × 1.11328 ≈ Let me compute 0.3 × 1.11328 = 0.33398, and 0.02227 × 1.11328 ≈ approx 0.02477. So total ≈ 0.33398 + 0.02477 ≈ 0.35875. So total ≈ 2.22656 + 0.35875 ≈ 2.58531.

Next multiply by 1.08605:

2.58531 × 1.08605 ≈ Let me compute 2.58531 × 1.08 ≈ 2.58531 + 2.58531 × 0.08 ≈ 2.58531 + 0.20682 ≈ 2.79213. Then adjust for the extra 0.00605. So 2.58531 × 0.00605 ≈ approx 0.01564. So total ≈ 2.79213 + 0.01564 ≈ 2.80777.

Then multiply by 1.0686:

2.80777 × 1.0686 ≈ Let me compute 2.80777 × 1.06 ≈ 2.80777 + 2.80777 × 0.06 ≈ 2.80777 + 0.16846 ≈ 2.97623. Then 2.80777 × 0.0086 ≈ approx 0.02415. So total ≈ 2.97623 + 0.02415 ≈ 3.00038.

Wow! So the approximate value is about 3.00038, which is very close to 3. Considering that I approximated each logarithm and did some rounding, this seems to confirm that the exact value is indeed 3. So that gives me more confidence.

Therefore, the answer is 3. But let me just think again if there's another way to see this. Maybe using exponents?

Alternatively, since each term is log_{n} (n+1) where n starts at 2 and goes up to 7. Wait, but the first term is log₂3, then log₃4, etc., so yeah, each term is log_{k} (k+1) for k from 2 to 7. So the product is from k=2 to k=7 of log_k (k+1). And as we saw, this telescopes to log₂8, which is 3. So that's another way to think about it.

Alternatively, if I consider the general case: the product from k = a to k = b of log_k (k+1) equals log_a (b+1). Is that a general formula? Let me check with smaller numbers. Let's take a=2, b=2: product is log₂3, which is log₂(2+1). If a=2, b=3: product is log₂3 × log₃4 = log₂4 = log₂(3+1). So yes, seems like the general formula is that the product from k=a to k=b of log_k (k+1) equals log_a (b+1). Therefore, in our case, a=2, b=7, so log₂(7+1)=log₂8=3. So that's another way to see it.

Therefore, regardless of the approach, the answer is 3. I think that's solid. I don't see any mistakes in my reasoning. The key idea is recognizing the telescoping product through the change of base formula or the chain rule for logarithms. It's a neat property that allows the product to collapse into a single logarithm. So yeah, the answer is 3.

**Final Answer**
\boxed{3}

To evaluate the expression:

$$
\log_2{3} \cdot \log_3{4} \cdot \log_4{5} \cdot \log_5{6} \cdot \log_6{7} \cdot \log_7{8}
$$

we can utilize a key property of logarithms known as the **change of base formula** and the **telescoping behavior** of logarithmic products.

---

### Step 1: Apply the Change of Base Formula

Recall the change of base formula:

$$
\log_a b = \frac{\ln b}{\ln a}
$$

Using this, we can rewrite each term in the product:

$$
\log_2{3} = \frac{\ln 3}{\ln 2}, \quad
\log_3{4} = \frac{\ln 4}{\ln 3}, \quad
\log_4{5} = \frac{\ln 5}{\ln 4}, \quad
\log_5{6} = \frac{\ln 6}{\ln 5}, \quad
\log_6{7} = \frac{\ln 7}{\ln 6}, \quad
\log_7{8} = \frac{\ln 8}{\ln 7}
$$

Now, multiplying all these together:

$$
\left( \frac{\ln 3}{\ln 2} \right) \cdot \left( \frac{\ln 4}{\ln 3} \right) \cdot \left( \frac{\ln 5}{\ln 4} \right) \cdot \left( \frac{\ln 6}{\ln 5} \right) \cdot \left( \frac{\ln 7}{\ln 6} \right) \cdot \left( \frac{\ln 8}{\ln 7} \right)
$$

---

### Step 2: Observe the Telescoping Pattern

Notice that the numerator of each fraction cancels with the denominator of the next:

- $\ln 3$ in the numerator of the first term cancels with $\ln 3$ in the denominator of the second term.
- $\ln 4$ in the numerator of the second term cancels with $\ln 4$ in the denominator of the third term.
- This pattern continues all the way through.

After all cancellations, only the first denominator and the last numerator remain:

$$
\frac{\ln 8}{\ln 2}
$$

This is equivalent to:

$$
\log_2{8}
$$

---

### Step 3: Evaluate the Final Logarithm

$$
\log_2{8} = 3 \quad \text{since } 2^3 = 8
$$

---

### Final Answer

$$
\boxed{3}
$$

Solution:
We use the change-of-base identity $\log_a{b}=\frac{\log{b}}{\log{a}}$ to find $$\log_2{3} \cdot \log_3{4} \cdot \log_4{5} \cdot \log_5{6} \cdot \log_6{7} \cdot \log_7{8}=
\frac{\log3}{\log2} \cdot \frac{\log4}{\log3} \cdot \frac{\log5}{\log4} \cdot \frac{\log6}{\log5} \cdot \frac{\log7}{\log6} \cdot \frac{\log8}{\log7}.$$Simplifying, we get $\frac{\log8}{\log2}=\log_2{8}=\boxed{3}$.