Explanation:
Okay, so I need to find the vertical asymptotes and the horizontal asymptote of the function f(x) = (2x)/(x² - 5x - 14), and then add up their values a + b + c. Let me start by recalling what vertical and horizontal asymptotes are.

Vertical asymptotes occur where the denominator of a rational function is zero, provided that the numerator isn't also zero at those points. So first, I need to factor the denominator and find its roots. The denominator here is x² - 5x - 14. Let me try to factor that quadratic.

Looking for two numbers that multiply to -14 and add up to -5. Let me think... factors of -14 could be -7 and 2 because (-7) * 2 = -14, and -7 + 2 = -5. Yes, that works. So the denominator factors as (x - 7)(x + 2). Therefore, the denominator is zero when x = 7 or x = -2. So those are the vertical asymptotes. Therefore, a and b are -2 and 7? Wait, but the problem says vertical asymptotes are x = a and x = b. So I need to check if the numerator is zero at these points. Let me check.

The numerator is 2x. If x = 7, then numerator is 14, which is not zero. If x = -2, numerator is -4, which is also not zero. So both are vertical asymptotes. Therefore, a and b are -2 and 7. So a + b would be (-2) + 7 = 5. Wait, but the problem says a and b are the vertical asymptotes, so maybe they are ordered? But since addition is commutative, it doesn't matter. So a + b is 5.

Now, moving on to the horizontal asymptote. For horizontal asymptotes of rational functions, the rules depend on the degrees of the numerator and denominator. The numerator here is degree 1 (since it's 2x), and the denominator is degree 2 (since it's x² - 5x -14). When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. Wait, so is that the case here? Let me confirm.

Yes, if the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. So c is 0. Therefore, a + b + c would be 5 + 0 = 5. Wait, but let me make sure I didn't skip any steps here. Let me check again.

First, vertical asymptotes: denominator factors to (x - 7)(x + 2), so vertical asymptotes at x = 7 and x = -2. So a and b are -2 and 7. So a + b is 5. Then horizontal asymptote: since numerator degree is 1, denominator degree is 2, so horizontal asymptote is y = 0. Therefore, c = 0. Therefore, a + b + c = 5 + 0 = 5. So the answer is 5? Wait, but let me check again if there's something wrong here.

Wait, maybe I made a mistake with the horizontal asymptote? Let me recall the rules again. For horizontal asymptotes:

- If the degree of the numerator is less than the degree of the denominator, horizontal asymptote is y = 0.

- If the degree of the numerator is equal to the degree of the denominator, horizontal asymptote is y = leading coefficient of numerator / leading coefficient of denominator.

- If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote (there might be an oblique asymptote).

In this case, numerator degree is 1, denominator degree is 2. So yes, horizontal asymptote is y = 0. Therefore, c = 0. Therefore, a + b + c = (-2) + 7 + 0 = 5. So the answer is 5? Wait, but let me check if maybe the question is expecting something else. Maybe I made a mistake in factoring the denominator?

Let me check the denominator again: x² - 5x -14. Let me try factoring again. Looking for two numbers that multiply to -14 and add to -5. Let me think. Let me list the factors of -14:

1 and -14: 1 + (-14) = -13

-1 and 14: -1 + 14 = 13

2 and -7: 2 + (-7) = -5. Yes, that's correct. So the factors are (x + 2)(x - 7). So that's correct. Therefore, vertical asymptotes at x = -2 and x = 7. So that's correct.

Alternatively, maybe I can check by solving x² - 5x -14 = 0. Using quadratic formula: x = [5 ± sqrt(25 + 56)] / 2 = [5 ± sqrt(81)] / 2 = [5 ± 9]/2. Therefore, x = (5 + 9)/2 = 14/2 = 7, and x = (5 - 9)/2 = -4/2 = -2. So that's correct. So vertical asymptotes at x = -2 and x = 7. So a + b is 5. Then horizontal asymptote is y = 0. Therefore, c = 0. Therefore, a + b + c = 5 + 0 = 5. So the answer is 5? Wait, but maybe I need to check if there's a horizontal asymptote?

Alternatively, maybe I can compute the limit as x approaches infinity of f(x). Let me do that. For f(x) = (2x)/(x² - 5x -14). Divide numerator and denominator by x²:

Numerator: 2x / x² = 2/x

Denominator: (x² - 5x -14)/x² = 1 - 5/x - 14/x²

Therefore, f(x) = (2/x) / (1 - 5/x - 14/x²). As x approaches infinity, 2/x approaches 0, 5/x approaches 0, 14/x² approaches 0. Therefore, the limit is 0 / 1 = 0. Therefore, horizontal asymptote is y = 0. So c = 0. Therefore, a + b + c = (-2) + 7 + 0 = 5. So the answer is 5. Therefore, I think that is correct.

But let me check again if there's any chance that the horizontal asymptote is not zero. For example, if the degrees were equal, but here numerator is degree 1, denominator is degree 2. So yes, horizontal asymptote is y = 0. So I think that is correct.

Alternatively, maybe the question is expecting me to consider something else? For example, sometimes people confuse horizontal asymptotes with oblique asymptotes. But in this case, since the degree of the numerator is less than the denominator, there is no oblique asymptote, only horizontal. So I think that is correct.

Therefore, the answer is 5. Therefore, a + b + c = 5. Therefore, the answer is \boxed{5}.

Wait, but let me check once more. Maybe the problem says "horizontal asymptote y = c", so c is 0. So adding a + b + c. If a and b are the vertical asymptotes, which are -2 and 7, so their sum is 5, and c is 0. So 5 + 0 = 5. So yes, that's correct. Therefore, I think the answer is 5.

But just to be thorough, let me check if there's any other possible horizontal asymptote. For example, sometimes when the degrees are different, but in this case, the degree of numerator is 1, denominator is 2. So horizontal asymptote is y = 0. So that's correct. Therefore, I think my answer is correct.

**Final Answer**
\boxed{5}

To solve the problem, we analyze the function:

$$
f(x) = \frac{2x}{x^2 - 5x - 14}
$$

---

### Step 1: Find the Vertical Asymptotes

Vertical asymptotes occur where the **denominator is zero**, provided the numerator is **not zero** at those points.

Factor the denominator:

$$
x^2 - 5x - 14 = (x - 7)(x + 2)
$$

So, the denominator is zero when:

$$
x = 7 \quad \text{or} \quad x = -2
$$

Now, check the numerator at these points:

- At $ x = 7 $, the numerator is $ 2(7) = 14 \neq 0 $
- At $ x = -2 $, the numerator is $ 2(-2) = -4 \neq 0 $

Since the numerator is not zero at these points, both are **vertical asymptotes**.

Let $ a = -2 $ and $ b = 7 $, so:

$$
a + b = -2 + 7 = 5
$$

---

### Step 2: Find the Horizontal Asymptote

To find the horizontal asymptote, compare the degrees of the numerator and denominator:

- Degree of numerator: 1
- Degree of denominator: 2

Since the degree of the numerator is **less than** the degree of the denominator, the horizontal asymptote is:

$$
y = 0
$$

Let $ c = 0 $

---

### Step 3: Compute $ a + b + c $

$$
a + b + c = (-2) + 7 + 0 = 5
$$

---

### Final Answer

$$
\boxed{5}
$$

[Assistant]
Explanation:
Okay, so I need to find the vertical asymptotes and the horizontal asymptote of the function f(x) = (2x)/(x² - 5x - 14), and then add up their values a + b + c. Let me start by recalling what vertical and horizontal asymptotes are.

Vertical asymptotes occur where the denominator of a rational function is zero, provided that the numerator isn't also zero at those points. So first, I need to factor the denominator and find its roots. The denominator here is x² - 5x - 14. Let me try to factor that quadratic.

Looking for two numbers that multiply to -14 and add up to -5. Let me think... factors of -14 could be -7 and 2 because (-7) * 2 = -14, and -7 + 2 = -5. Yes, that works. So the denominator factors as (x - 7)(x + 2). Therefore, the denominator is zero when x = 7 or x = -2. So those are the vertical asymptotes. Therefore, a and b are -2 and 7? Wait, but the problem says vertical asymptotes are x = a and x = b. So I need to check if the numerator is zero at these points. Let me check.

The numerator is 2x. If x = 7, then numerator is 14, which is not zero. If x = -2, numerator is -4, which is also not zero. So both are vertical asymptotes. Therefore, a and b are -2 and 7. So a + b would be (-2) + 7 = 5. Wait, but the problem says a and b are the vertical asymptotes, so maybe they are ordered? But since addition is commutative, it doesn't matter. So a + b is 5.

Now, moving on to the horizontal asymptote. For horizontal asymptotes of rational functions, the rules depend on the degrees of the numerator and denominator. The numerator here is degree 1 (since it's 2x), and the denominator is degree 2 (since it's x² - 5x -14). When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. Wait, so is that the case here? Let me confirm.

Yes, if the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. So c is 0. Therefore, a + b + c would be 5 + 0 = 5. Wait, but let me make sure I didn't skip any steps here. Let me check again.

First, vertical asymptotes: denominator factors to (x - 7)(x + 2), so vertical asymptotes at x = 7 and x = -2. So a and b are -2 and 7. So a + b is 5. Then horizontal asymptote: since numerator degree is 1, denominator degree is 2, so horizontal asymptote is y = 0. Therefore, c = 0. Therefore, a + b + c = 5 + 0 = 5. So the answer is 5? Wait, but let me check again if there's something wrong here.

Wait, maybe I made a mistake with the horizontal asymptote? Let me recall the rules again. For horizontal asymptotes:

- If the degree of the numerator is less than the degree of the denominator, horizontal asymptote is y = 0.

- If the degree of the numerator is equal to the degree of the denominator, horizontal asymptote is y = leading coefficient of numerator / leading coefficient of denominator.

- If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote (there might be an oblique asymptote).

In this case, numerator degree is 1, denominator degree is 2. So yes, horizontal asymptote is y = 0. Therefore, c = 0. Therefore, a + b + c = (-2) + 7 + 0 = 5. So the answer is 5? Wait, but let me check if maybe the question is expecting something else. Maybe I made a mistake in factoring the denominator?

Let me check the denominator again: x² - 5x -14. Let me try factoring again. Looking for two numbers that multiply to -14 and add to -5. Let me think. Let me list the factors of -14:

1 and -14: 1 + (-14) = -13

-1 and 14: -1 + 14 = 13

2 and -7: 2 + (-7) = -5. Yes, that's correct. So the factors are (x + 2)(x - 7). So that's correct. Therefore, vertical asymptotes at x = -2 and x = 7. So that's correct.

Alternatively, maybe I can check by solving x² - 5x -14 = 0. Using quadratic formula: x = [5 ± sqrt(25 + 56)] / 2 = [5 ± sqrt(81)] / 2 = [5 ± 9]/2. Therefore, x = (5 + 9)/2 = 14/2 = 7, and x = (5 - 9)/2 = -4/2 = -2. So that's correct. So vertical asymptotes at x = -2 and x = 7. So a + b is 5. Then horizontal asymptote is y = 0. Therefore, c = 0. Therefore, a + b + c = 5 + 0 = 5. So the answer is 5? Wait, but maybe I need to check if there's a horizontal asymptote?

Alternatively, maybe I can compute the limit as x approaches infinity of f(x). Let me do that. For f(x) = (2x)/(x² - 5x -14). Divide numerator and denominator by x²:

Numerator: 2x / x² = 2/x

Denominator: (x² - 5x -14)/x² = 1 - 5/x - 14/x²

Therefore, f(x) = (2/x) / (1 - 5/x - 14/x²). As x approaches infinity, 2/x approaches 0, 5/x approaches 0, 14/x² approaches 0. Therefore, the limit is 0 / 1 = 0. Therefore, horizontal asymptote is y = 0. So c = 0. Therefore, a + b + c = (-2) + 7 + 0 = 5. So the answer is 5. Therefore, I think that is correct.

But let me check again if there's any chance that the horizontal asymptote is not zero. For example, if the degrees were equal, but here numerator is degree 1, denominator is degree 2. So yes, horizontal asymptote is y = 0. So I think that is correct.

Alternatively, maybe the question is expecting me to consider something else? For example, sometimes people confuse horizontal asymptotes with oblique asymptotes. But in this case, since the degree of the numerator is less than the denominator, there is no oblique asymptote, only horizontal. So I think that is correct.

Therefore, the answer is 5. Therefore, a + b + c = 5. Therefore, the answer is \boxed{5}.

Wait, but let me check once more. Maybe the problem says "horizontal asymptote y = c", so c is 0. So adding a + b + c. If a and b are the vertical asymptotes, which are -2 and 7, so their sum is 5, and c is 0. So 5 + 0 = 5. So yes, that's correct. Therefore, I think the answer is 5.

But just to be thorough, let me check if there's any other possible horizontal asymptote. For example, sometimes when the degrees are different, but in this case, the degree of numerator is 1, denominator is 2. So horizontal asymptote is y = 0. So that's correct. Therefore, I think my answer is correct.

**Final Answer**
\boxed{5}

To solve the problem, we analyze the function:

$$
f(x) = \frac{2x}{x^2 - 5x - 14}
$$

---

### Step 1: Find the Vertical Asymptotes

Vertical asymptotes occur where the **denominator is zero**, provided the numerator is **not zero** at those points.

Factor the denominator:

$$
x^2 - 5x - 14 = (x - 7)(x + 2)
$$

So, the denominator is zero when:

$$
x = 7 \quad \text{or} \quad x = -2
$$

Now, check the numerator at these points:

- At $ x = 7 $, the numerator is $ 2(7) = 14 \neq 0 $
- At $ x = -2 $, the numerator is $ 2(-2) = -4 \neq 0 $

Since the numerator is not zero at these points, both are **vertical asymptotes**.

Let $ a = -2 $ and $ b = 7 $, so:

$$
a + b = -2 + 7 = 5
$$

---

### Step 2: Find the Horizontal Asymptote

To find the horizontal asymptote, compare the degrees of the numerator and denominator:

- Degree of numerator: 1
- Degree of denominator: 2

Since the degree of the numerator is **less than** the degree of the denominator, the horizontal asymptote is:

$$
y = 0
$$

Let $ c = 0 $

---

### Step 3: Compute $ a + b + c $

$$
a + b + c = (-2) + 7 + 0 = 5
$$

---

### Final Answer

$$
\boxed{5}
$$

Solution:
Vertical asymptotes occur at values of $x$ where the denominator is 0.  We can factor the denominator into $(x-7)(x+2)$, so the denominator equals 0 when $x=7$ or $x=-2$. Those $x$-values are where our vertical asymptotes are located.

For horizontal asymptotes, we look at the degree of $x$ in the numerator and the denominator. The degree of the numerator is 1, and the degree of the denominator is 2, so the denominator grows faster than the numerator for large values of $x$, and the function approaches the horizontal asymptote $y=0$. We can also see that when we divide $x$ out of the numerator and denominator, we get \[\frac{2x}{x^2 - 5x - 14} = \frac{\frac{2x}{x}}{\frac{x^2-5x-14}{x}}=\frac{2}{x-5-\frac{14}{x}}.\]As $x$ approaches infinity or negative infinity, the expression approaches 0.

So, our answer is $7 + (-2) + 0 = \boxed{5}$.


Explanation:
Okay, let's try to figure out this problem. So, there are seven bags of gold coins, each with the same number of coins. Then I find an extra bag with 53 coins, making it eight bags total. I redistribute all the coins so that all eight bags have the same number, and I end up with more than 200 coins. The question is asking for the smallest number of coins I could have had before finding the 53-coin bag.

First, let me parse the problem again to make sure I understand. Initially, there are seven bags, each with the same number of coins. Let's call that number x. So, the total number of coins before finding the 53-coin bag is 7x. Then I find a bag with 53 coins, so the total becomes 7x + 53. Then I redistribute all these coins into eight bags, each with the same number of coins. Let me call that number y. So, the total number of coins after redistribution is 8y.

But since all coins are redistributed, 7x + 53 must equal 8y. Also, the problem says that after redistribution, the number of coins is more than 200. So, 8y > 200. Therefore, y > 25. Since y has to be an integer (you can't have a fraction of a coin), y must be at least 26.

But we need to find the smallest possible original number of coins, which is 7x. So, we need to find the smallest x such that 7x + 53 is divisible by 8, and 8y = 7x + 53 > 200.

So, let me write that down as equations:

1. 7x + 53 = 8y
2. 8y > 200 => y > 25
3. Find the smallest x such that these are satisfied.

So, first, let's find the possible values of y. Since y must be an integer greater than 25, the smallest possible y is 26. Let's check if 8*26 = 208. Then, 7x + 53 = 208 => 7x = 208 - 53 = 155. Then x = 155 / 7 ≈ 22.14. But x must be an integer because you can't have a fraction of a coin in a bag. So, 155 isn't divisible by 7. Therefore, y=26 is not possible.

Next, try y=27. Then 8*27=216. Then 7x + 53 = 216 => 7x = 216 - 53 = 163. 163 divided by 7 is approximately 23.28... Not an integer. Not valid.

y=28: 8*28=224. 224 - 53 = 171. 171 /7 = 24.428... Not integer.

y=29: 8*29=232. 232 -53=179. 179/7 ≈ 25.57. Not integer.

y=30: 8*30=240. 240 -53=187. 187/7=26.714... Not integer.

y=31: 8*31=248. 248 -53=195. 195/7≈27.857... Not integer.

y=32: 8*32=256. 256 -53=203. 203 divided by 7 is 29. So, x=29.

Wait, 7*29 is 203. Then 203 +53=256, which is 8*32. So that works. And 8y=256, which is more than 200. So, is this the minimal x? Wait, but maybe there is a smaller y? Wait, we tried y from 26 up to 32. But maybe there's a smaller x? Wait, but x is 29 here. But maybe there is a smaller x? Let me check if there is a smaller y that gives integer x.

Wait, but we started from y=26 upwards. So, if y=26 gives x≈22.14, which is not integer. Then y=27 gives x≈23.28, not integer. y=28 gives x≈24.42, not integer. y=29 gives x≈25.57, not integer. y=30 gives x≈26.71, not integer. y=31 gives x≈27.85, not integer. y=32 gives x=29. So, that seems like the first possible integer value. Therefore, x=29. Therefore, original number of coins was 7x=203.

But wait, the problem says "you have more than 200 coins" after redistribution. So, 8y>200. So, y>25. So, y=26 is the first possible. But since 7x +53 must be divisible by 8, so 7x ≡ -53 mod 8. Let me think if there's another way to approach this problem with modular arithmetic.

Let me try that. So, 7x + 53 ≡ 0 mod 8. Therefore, 7x ≡ -53 mod 8. Let me compute -53 mod 8. Since 53 divided by 8 is 6*8=48, 53-48=5. So, 53 ≡5 mod8. Therefore, -53 ≡ -5 mod8, which is equivalent to 3 mod8 (since -5 +8=3). Therefore, 7x ≡3 mod8.

So, 7x ≡3 mod8. Let me solve for x. Since 7 and 8 are coprime, we can find the inverse of 7 mod8. The inverse of 7 mod8 is a number m such that 7m ≡1 mod8. Let's see: 7*7=49≡1 mod8. So, inverse of 7 mod8 is 7. Therefore, multiplying both sides by 7: x ≡ 3*7 mod8 => x≡21 mod8. 21 divided by 8 is 2*8=16, 21-16=5. Therefore, x≡5 mod8. Therefore, x=8k +5 for some integer k≥0.

Therefore, the original number of coins is 7x=7*(8k +5)=56k +35. So, the original number of coins is 56k +35. Now, after adding 53 coins, total coins is 56k +35 +53=56k +88. Which must be equal to 8y. Therefore, 56k +88=8y => divide both sides by 8: 7k +11 = y.

Also, we know that 8y >200 => y>25. Since y=7k +11, then 7k +11 >25 => 7k>14 =>k>2. Therefore, k≥3.

So, the smallest k is 3. Therefore, original number of coins is 56k +35. For k=3: 56*3 +35=168 +35=203. Which matches the previous result.

So, if k=3, original coins is 203. Then adding 53 gives 256, which is 8*32. And 256>200.

But the question is asking for the smallest number of coins before finding the 53-coin bag. So, 203? But let me check if there is a smaller k. Wait, k must be ≥3. Because k=2 would give y=7*2 +11=25, but y must be greater than 25. So, y=25 is not allowed. Therefore, k=3 is the minimal. Therefore, original coins is 203.

But let me check if there is a smaller x. Wait, x=8k +5. For k=0, x=5, original coins 35. Then adding 53 gives 88, which is 8*11. But 8*11=88, which is not more than 200. So, invalid. For k=1, x=13, original coins 91. Adding 53 gives 144, which is 8*18. Still less than 200. For k=2, x=21, original coins 147. Adding 53 gives 200. Which is equal to 8*25. But the problem says "you have more than 200 coins". Therefore, 200 is not allowed. Therefore, k=2 gives total coins 200, which is not more than 200. Therefore, k must be at least 3. Therefore, original coins is 203.

Therefore, the answer is 203.

But let me check again. Let me check for k=3: original coins 7x=203. Then adding 53 gives 256. Divided by 8 is 32. 32*8=256. Which is more than 200. So that works.

Is there a smaller number? Let me think. Suppose there is another solution where k is smaller? But according to our modular arithmetic, x must be congruent to 5 mod8, so x=5,13,21,29,... So original coins would be 35, 91, 147, 203,... So, 35, 91, 147, 203.

For 35: total coins after adding 53 is 88. Which is 8*11. 88 is less than 200. Not allowed.

For 91: total coins after adding 53 is 144. 144 is 8*18. Still less than 200.

For 147: total coins after adding 53 is 200. Which is 8*25. But the problem says "more than 200", so 200 is not allowed. Therefore, the next one is 203.

Therefore, 203 is the minimal.

But let me check if there is another way. Maybe my modular approach missed something? Let me think again.

We had 7x +53 ≡0 mod8. Which led us to x≡5 mod8. Therefore, x=8k +5. Therefore, original coins is 7*(8k +5)=56k +35.

So, the total coins after adding 53 is 56k +35 +53=56k +88. Which is 8*(7k +11). Therefore, y=7k +11.

And since y>25, 7k +11>25 => 7k>14 =>k>2. Therefore, k≥3.

Therefore, the minimal k is 3, leading to original coins 56*3 +35=203.

Therefore, the answer is 203.

But let me check if there is a case where maybe the original number of coins is less than 203, but somehow the total coins after adding 53 is more than 200. For example, suppose original coins is 147. Then adding 53 gives 200. Which is not more than 200. If original coins is 147 + something? Wait, but according to our previous logic, the next possible original coins is 203.

Alternatively, maybe there is another solution where 7x +53 is divisible by 8 and 8y>200, but with smaller x?

Wait, let me think. Let me check for x=29. Original coins is 7*29=203.

But suppose there is a smaller x. Let me check x=5,13,21,29,...

For x=5: total coins 7*5 +53=35+53=88. 88/8=11. 11*8=88. Which is less than 200.

x=13: 7*13=91. 91+53=144. 144/8=18. Still less than 200.

x=21: 7*21=147. 147+53=200. 200/8=25. Which is exactly 25. But the problem says "more than 200 coins". So, 200 is not allowed.

x=29: 7*29=203. 203+53=256. 256/8=32. Which is more than 200.

So, yes, 203 is the minimal.

Therefore, the answer is 203.

But let me check if there is another way to get a smaller number. For example, maybe there is a different congruence? Wait, maybe I made a mistake in solving the congruence. Let me check again.

Original equation: 7x +53 ≡0 mod8.

So, 7x ≡ -53 mod8.

As before, 53 divided by 8 is 6*8=48, remainder 5. So 53≡5 mod8. Therefore, -53≡-5 mod8. Which is equivalent to 3 mod8.

So, 7x≡3 mod8.

Then, solving for x: multiply both sides by inverse of 7 mod8.

Since 7*7=49≡1 mod8, inverse of 7 is 7.

Therefore, x≡3*7=21≡5 mod8.

Yes, that's correct. So x≡5 mod8. Therefore, x=8k +5.

Therefore, original coins is 7x=56k +35.

So, the minimal k where 8y=56k +88>200.

Wait, 56k +88>200.

So, 56k>112.

Therefore, k>2.

So, k≥3.

Therefore, minimal k=3.

Therefore, original coins is 56*3 +35=168 +35=203.

Yes, that's correct.

Therefore, the answer is 203.

But let me check once more. Suppose someone says, what if there is a different way to distribute? But the problem says that you successfully redistribute all the coins so that all eight bags have the same number. Therefore, the total number of coins must be divisible by 8. So, 7x +53 must be divisible by 8.

Therefore, the only way is the above.

Alternatively, maybe there is a mistake in assuming that the original number of coins is 7x? Let me check.

The problem says: "You have seven bags of gold coins. Each bag has the same number of gold coins." So, yes, original total is 7x. Then you find a bag of 53 coins. So, total becomes 7x +53. Then you redistribute into eight bags with same number. So, 7x +53 must be divisible by 8.

Therefore, the equations are correct.

Therefore, the answer is 203.

But let me check with k=3, original coins is 203.

Original seven bags: 203 coins. So each bag has 203 /7=29 coins.

Then adding 53 coins: total coins is 203 +53=256.

Divided into eight bags: 256 /8=32 coins per bag.

Yes, that works.

And 256 is more than 200.

Is there a smaller number? Let me think. Suppose someone says, maybe the original number of coins is less than 203, but when adding 53, the total is more than 200. For example, suppose original coins is 147. Then adding 53 gives 200. Which is not more than 200. If original coins is 147 + something? But according to our previous logic, the next possible is 203.

Alternatively, maybe there is another value of x not congruent to 5 mod8? But according to the congruence, x must be congruent to 5 mod8. Because 7x ≡3 mod8 implies x≡5 mod8. Therefore, there are no other solutions.

Therefore, 203 is indeed the minimal.

Therefore, the answer is 203.

**Final Answer**
The smallest number of coins you could have had before finding the bag of 53 coins is \boxed{203}.

To solve the problem, we begin by defining the key variables and constraints:

- Let the original number of coins in each of the 7 bags be $ x $.
- Therefore, the total number of coins before finding the 53-coin bag is $ 7x $.
- After finding the 53-coin bag, the total becomes $ 7x + 53 $.
- These coins are then redistributed into 8 bags, each with the same number of coins, say $ y $.
- So, the total number of coins after redistribution is $ 8y $, and we are told that $ 8y > 200 $, which implies $ y > 25 $.

---

### Step 1: Set up the equation

We know that the total number of coins before and after redistribution must be the same:

$$
7x + 53 = 8y
$$

We are to find the **smallest** value of $ x $ such that this equation holds and $ y > 25 $.

---

### Step 2: Use modular arithmetic

We can rewrite the equation as:

$$
7x \equiv -53 \pmod{8}
$$

Since $ 53 \equiv 5 \pmod{8} $, we have:

$$
7x \equiv -5 \pmod{8} \Rightarrow 7x \equiv 3 \pmod{8}
$$

Now, we solve for $ x $ modulo 8. Since 7 and 8 are coprime, we can find the modular inverse of 7 modulo 8. The inverse of 7 modulo 8 is 7 (since $ 7 \times 7 = 49 \equiv 1 \pmod{8} $).

So:

$$
x \equiv 3 \times 7 \equiv 21 \equiv 5 \pmod{8}
$$

Thus, $ x = 8k + 5 $ for some integer $ k \geq 0 $.

---

### Step 3: Express total coins in terms of $ k $

Substitute $ x = 8k + 5 $ into the total coins:

$$
7x + 53 = 7(8k + 5) + 53 = 56k + 35 + 53 = 56k + 88
$$

This must be divisible by 8, and we are told that the total number of coins after redistribution is more than 200:

$$
8y = 56k + 88 > 200 \Rightarrow 56k + 88 > 200 \Rightarrow 56k > 112 \Rightarrow k > 2
$$

So the smallest integer $ k $ satisfying this is $ k = 3 $.

---

### Step 4: Compute the original number of coins

Now, plug $ k = 3 $ into the expression for the original number of coins:

$$
7x = 7(8k + 5) = 7(8 \cdot 3 + 5) = 7(24 + 5) = 7 \cdot 29 = 203
$$

---

### Step 5: Verify the solution

- Original coins: $ 7x = 203 $
- After adding 53: $ 203 + 53 = 256 $
- Divided into 8 bags: $ 256 \div 8 = 32 $, which is an integer.
- $ 256 > 200 $, satisfying the condition.

---

### Final Answer

$$
\boxed{203}
$$

[Assistant]
Explanation:
Okay, let's try to figure out this problem. So, there are seven bags of gold coins, each with the same number of coins. Then I find an extra bag with 53 coins, making it eight bags total. I redistribute all the coins so that all eight bags have the same number, and I end up with more than 200 coins. The question is asking for the smallest number of coins I could have had before finding the 53-coin bag.

First, let me parse the problem again to make sure I understand. Initially, there are seven bags, each with the same number of coins. Let's call that number x. So, the total number of coins before finding the 53-coin bag is 7x. Then I find a bag with 53 coins, so the total becomes 7x + 53. Then I redistribute all these coins into eight bags, each with the same number of coins. Let me call that number y. So, the total number of coins after redistribution is 8y.

But since all coins are redistributed, 7x + 53 must equal 8y. Also, the problem says that after redistribution, the number of coins is more than 200. So, 8y > 200. Therefore, y > 25. Since y has to be an integer (you can't have a fraction of a coin), y must be at least 26.

But we need to find the smallest possible original number of coins, which is 7x. So, we need to find the smallest x such that 7x + 53 is divisible by 8, and 8y = 7x + 53 > 200.

So, let me write that down as equations:

1. 7x + 53 = 8y
2. 8y > 200 => y > 25
3. Find the smallest x such that these are satisfied.

So, first, let's find the possible values of y. Since y must be an integer greater than 25, the smallest possible y is 26. Let's check if 8*26 = 208. Then, 7x + 53 = 208 => 7x = 208 - 53 = 155. Then x = 155 / 7 ≈ 22.14. But x must be an integer because you can't have a fraction of a coin in a bag. So, 155 isn't divisible by 7. Therefore, y=26 is not possible.

Next, try y=27. Then 8*27=216. Then 7x + 53 = 216 => 7x = 216 - 53 = 163. 163 divided by 7 is approximately 23.28... Not an integer. Not valid.

y=28: 8*28=224. 224 - 53 = 171. 171 /7 = 24.428... Not integer.

y=29: 8*29=232. 232 -53=179. 179/7 ≈ 25.57. Not integer.

y=30: 8*30=240. 240 -53=187. 187/7=26.714... Not integer.

y=31: 8*31=248. 248 -53=195. 195/7≈27.857... Not integer.

y=32: 8*32=256. 256 -53=203. 203 divided by 7 is 29. So, x=29.

Wait, 7*29 is 203. Then 203 +53=256, which is 8*32. So that works. And 8y=256, which is more than 200. So, is this the minimal x? Wait, but maybe there is a smaller y? Wait, we tried y from 26 up to 32. But maybe there's a smaller x? Wait, but x is 29 here. But maybe there is a smaller x? Let me check if there is a smaller y that gives integer x.

Wait, but we started from y=26 upwards. So, if y=26 gives x≈22.14, which is not integer. Then y=27 gives x≈23.28, not integer. y=28 gives x≈24.42, not integer. y=29 gives x≈25.57, not integer. y=30 gives x≈26.71, not integer. y=31 gives x≈27.85, not integer. y=32 gives x=29. So, that seems like the first possible integer value. Therefore, x=29. Therefore, original number of coins was 7x=203.

But wait, the problem says "you have more than 200 coins" after redistribution. So, 8y>200. So, y>25. So, y=26 is the first possible. But since 7x +53 must be divisible by 8, so 7x ≡ -53 mod 8. Let me think if there's another way to approach this problem with modular arithmetic.

Let me try that. So, 7x + 53 ≡ 0 mod 8. Therefore, 7x ≡ -53 mod 8. Let me compute -53 mod 8. Since 53 divided by 8 is 6*8=48, 53-48=5. So, 53 ≡5 mod8. Therefore, -53 ≡ -5 mod8, which is equivalent to 3 mod8 (since -5 +8=3). Therefore, 7x ≡3 mod8.

So, 7x ≡3 mod8. Let me solve for x. Since 7 and 8 are coprime, we can find the inverse of 7 mod8. The inverse of 7 mod8 is a number m such that 7m ≡1 mod8. Let's see: 7*7=49≡1 mod8. So, inverse of 7 mod8 is 7. Therefore, multiplying both sides by 7: x ≡ 3*7 mod8 => x≡21 mod8. 21 divided by 8 is 2*8=16, 21-16=5. Therefore, x≡5 mod8. Therefore, x=8k +5 for some integer k≥0.

Therefore, the original number of coins is 7x=7*(8k +5)=56k +35. So, the original number of coins is 56k +35. Now, after adding 53 coins, total coins is 56k +35 +53=56k +88. Which must be equal to 8y. Therefore, 56k +88=8y => divide both sides by 8: 7k +11 = y.

Also, we know that 8y >200 => y>25. Since y=7k +11, then 7k +11 >25 => 7k>14 =>k>2. Therefore, k≥3.

So, the smallest k is 3. Therefore, original number of coins is 56k +35. For k=3: 56*3 +35=168 +35=203. Which matches the previous result.

So, if k=3, original coins is 203. Then adding 53 gives 256, which is 8*32. And 256>200.

But the question is asking for the smallest number of coins before finding the 53-coin bag. So, 203? But let me check if there is a smaller k. Wait, k must be ≥3. Because k=2 would give y=7*2 +11=25, but y must be greater than 25. So, y=25 is not allowed. Therefore, k=3 is the minimal. Therefore, original coins is 203.

But let me check if there is a smaller x. Wait, x=8k +5. For k=0, x=5, original coins 35. Then adding 53 gives 88, which is 8*11. But 8*11=88, which is not more than 200. So, invalid. For k=1, x=13, original coins 91. Adding 53 gives 144, which is 8*18. Still less than 200. For k=2, x=21, original coins 147. Adding 53 gives 200. Which is equal to 8*25. But the problem says "you have more than 200 coins". Therefore, 200 is not allowed. Therefore, k=2 gives total coins 200, which is not more than 200. Therefore, k must be at least 3. Therefore, original coins is 203.

Therefore, the answer is 203.

But let me check again. Let me check for k=3: original coins 7x=203. Then adding 53 gives 256. Divided by 8 is 32. 32*8=256. Which is more than 200. So that works.

Is there a smaller number? Let me think. Suppose there is another solution where k is smaller? But according to our modular arithmetic, x must be congruent to 5 mod8, so x=5,13,21,29,... So original coins would be 35, 91, 147, 203,... So, 35, 91, 147, 203.

For 35: total coins after adding 53 is 88. Which is 8*11. 88 is less than 200. Not allowed.

For 91: total coins after adding 53 is 144. 144 is 8*18. Still less than 200.

For 147: total coins after adding 53 is 200. Which is 8*25. But the problem says "more than 200", so 200 is not allowed. Therefore, the next one is 203.

Therefore, 203 is the minimal.

But let me check if there is another way. Maybe my modular approach missed something? Let me think again.

We had 7x +53 ≡0 mod8. Which led us to x≡5 mod8. Therefore, x=8k +5. Therefore, original coins is 7*(8k +5)=56k +35.

So, the total coins after adding 53 is 56k +35 +53=56k +88. Which is 8*(7k +11). Therefore, y=7k +11.

And since y>25, 7k +11>25 => 7k>14 =>k>2. Therefore, k≥3.

Therefore, the minimal k is 3, leading to original coins 56*3 +35=203.

Therefore, the answer is 203.

But let me check if there is a case where maybe the original number of coins is less than 203, but somehow the total coins after adding 53 is more than 200. For example, suppose original coins is 147. Then adding 53 gives 200. Which is not more than 200. If original coins is 147 + something? Wait, but according to our previous logic, the next possible original coins is 203.

Alternatively, maybe there is another solution where 7x +53 is divisible by 8 and 8y>200, but with smaller x?

Wait, let me think. Let me check for x=29. Original coins is 7*29=203.

But suppose there is a smaller x. Let me check x=5,13,21,29,...

For x=5: total coins 7*5 +53=35+53=88. 88/8=11. 11*8=88. Which is less than 200.

x=13: 7*13=91. 91+53=144. 144/8=18. Still less than 200.

x=21: 7*21=147. 147+53=200. 200/8=25. Which is exactly 25. But the problem says "more than 200 coins". So, 200 is not allowed.

x=29: 7*29=203. 203+53=256. 256/8=32. Which is more than 200.

So, yes, 203 is the minimal.

Therefore, the answer is 203.

But let me check if there is another way to get a smaller number. For example, maybe there is a different congruence? Wait, maybe I made a mistake in solving the congruence. Let me check again.

Original equation: 7x +53 ≡0 mod8.

So, 7x ≡ -53 mod8.

As before, 53 divided by 8 is 6*8=48, remainder 5. So 53≡5 mod8. Therefore, -53≡-5 mod8. Which is equivalent to 3 mod8.

So, 7x≡3 mod8.

Then, solving for x: multiply both sides by inverse of 7 mod8.

Since 7*7=49≡1 mod8, inverse of 7 is 7.

Therefore, x≡3*7=21≡5 mod8.

Yes, that's correct. So x≡5 mod8. Therefore, x=8k +5.

Therefore, original coins is 7x=56k +35.

So, the minimal k where 8y=56k +88>200.

Wait, 56k +88>200.

So, 56k>112.

Therefore, k>2.

So, k≥3.

Therefore, minimal k=3.

Therefore, original coins is 56*3 +35=168 +35=203.

Yes, that's correct.

Therefore, the answer is 203.

But let me check once more. Suppose someone says, what if there is a different way to distribute? But the problem says that you successfully redistribute all the coins so that all eight bags have the same number. Therefore, the total number of coins must be divisible by 8. So, 7x +53 must be divisible by 8.

Therefore, the only way is the above.

Alternatively, maybe there is a mistake in assuming that the original number of coins is 7x? Let me check.

The problem says: "You have seven bags of gold coins. Each bag has the same number of gold coins." So, yes, original total is 7x. Then you find a bag of 53 coins. So, total becomes 7x +53. Then you redistribute into eight bags with same number. So, 7x +53 must be divisible by 8.

Therefore, the equations are correct.

Therefore, the answer is 203.

But let me check with k=3, original coins is 203.

Original seven bags: 203 coins. So each bag has 203 /7=29 coins.

Then adding 53 coins: total coins is 203 +53=256.

Divided into eight bags: 256 /8=32 coins per bag.

Yes, that works.

And 256 is more than 200.

Is there a smaller number? Let me think. Suppose someone says, maybe the original number of coins is less than 203, but when adding 53, the total is more than 200. For example, suppose original coins is 147. Then adding 53 gives 200. Which is not more than 200. If original coins is 147 + something? But according to our previous logic, the next possible is 203.

Alternatively, maybe there is another value of x not congruent to 5 mod8? But according to the congruence, x must be congruent to 5 mod8. Because 7x ≡3 mod8 implies x≡5 mod8. Therefore, there are no other solutions.

Therefore, 203 is indeed the minimal.

Therefore, the answer is 203.

**Final Answer**
The smallest number of coins you could have had before finding the bag of 53 coins is \boxed{203}.

To solve the problem, we begin by defining the key variables and constraints:

- Let the original number of coins in each of the 7 bags be $ x $.
- Therefore, the total number of coins before finding the 53-coin bag is $ 7x $.
- After finding the 53-coin bag, the total becomes $ 7x + 53 $.
- These coins are then redistributed into 8 bags, each with the same number of coins, say $ y $.
- So, the total number of coins after redistribution is $ 8y $, and we are told that $ 8y > 200 $, which implies $ y > 25 $.

---

### Step 1: Set up the equation

We know that the total number of coins before and after redistribution must be the same:

$$
7x + 53 = 8y
$$

We are to find the **smallest** value of $ x $ such that this equation holds and $ y > 25 $.

---

### Step 2: Use modular arithmetic

We can rewrite the equation as:

$$
7x \equiv -53 \pmod{8}
$$

Since $ 53 \equiv 5 \pmod{8} $, we have:

$$
7x \equiv -5 \pmod{8} \Rightarrow 7x \equiv 3 \pmod{8}
$$

Now, we solve for $ x $ modulo 8. Since 7 and 8 are coprime, we can find the modular inverse of 7 modulo 8. The inverse of 7 modulo 8 is 7 (since $ 7 \times 7 = 49 \equiv 1 \pmod{8} $).

So:

$$
x \equiv 3 \times 7 \equiv 21 \equiv 5 \pmod{8}
$$

Thus, $ x = 8k + 5 $ for some integer $ k \geq 0 $.

---

### Step 3: Express total coins in terms of $ k $

Substitute $ x = 8k + 5 $ into the total coins:

$$
7x + 53 = 7(8k + 5) + 53 = 56k + 35 + 53 = 56k + 88
$$

This must be divisible by 8, and we are told that the total number of coins after redistribution is more than 200:

$$
8y = 56k + 88 > 200 \Rightarrow 56k + 88 > 200 \Rightarrow 56k > 112 \Rightarrow k > 2
$$

So the smallest integer $ k $ satisfying this is $ k = 3 $.

---

### Step 4: Compute the original number of coins

Now, plug $ k = 3 $ into the expression for the original number of coins:

$$
7x = 7(8k + 5) = 7(8 \cdot 3 + 5) = 7(24 + 5) = 7 \cdot 29 = 203
$$

---

### Step 5: Verify the solution

- Original coins: $ 7x = 203 $
- After adding 53: $ 203 + 53 = 256 $
- Divided into 8 bags: $ 256 \div 8 = 32 $, which is an integer.
- $ 256 > 200 $, satisfying the condition.

---

### Final Answer

$$
\boxed{203}
$$

Solution: If there are $b$ gold coins in each of the original 7 bags, then $7b+53$ is divisible by 8. In other words, $7b + 53 \equiv 0 \pmod{8}$. Since $53 \equiv 5 \pmod{8}$ and $7 \equiv -1 \pmod{8}$, we have that $-b \equiv -5 \pmod{8}$. Multiplying both sides by $-1$, we get that $b \equiv 5 \pmod{8}$. Now, we want $7b + 53 > 200$, so as a result, $b > \frac{200-53}{7} \implies b > 21$. Therefore, we want an integer greater than 21 which leaves a remainder of 5 when divided by 8. The least such integer is 29, so you had $29 \cdot 7 = \boxed{203}$ coins before finding the bag of 53 coins.


Explanation:
Okay, so I need to simplify this expression: the square root of (2.5 squared minus 0.7 squared) divided by (2.7 minus 2.5). Let me write that down first to make sure I have it right.

So, the expression is √(2.5² - 0.7²) divided by (2.7 - 2.5). Let me check if I copied that correctly. Yeah, that seems right. Alright, let me think about how to approach this.

First, maybe I should compute the numerator and the denominator separately. Let me start with the numerator: √(2.5² - 0.7²). Hmm, that looks like a difference of squares. Remember, a² - b² factors into (a - b)(a + b). Maybe I can use that here? Let me try that.

So, if I let a = 2.5 and b = 0.7, then the numerator becomes √[(a - b)(a + b)]. Let me compute that.

First, compute a - b: 2.5 - 0.7. Let me do that subtraction. 2.5 minus 0.7. Well, 2.5 minus 0.5 is 2.0, and then minus another 0.2 is 1.8. So, 2.5 - 0.7 = 1.8.

Then, a + b is 2.5 + 0.7. Let me add those. 2.5 + 0.7. 2 + 0.5 + 0.7 is 2 + 1.2 = 3.2. Wait, no. Wait, 2.5 + 0.7. Let me think again. 2.5 + 0.5 is 3.0, and then + 0.2 more is 3.2. Yes, that's right. So, a + b = 3.2.

Therefore, the numerator simplifies to √(1.8 * 3.2). Let me compute 1.8 multiplied by 3.2. Hmm, 1.8 * 3.2. Let me do that multiplication.

First, maybe convert them to fractions? Or maybe decimal multiplication. Let me try decimal.

1.8 * 3.2. Let me think of it as (1 + 0.8) * (3 + 0.2). But maybe that's more complicated. Alternatively, multiply 1.8 * 3.2.

Well, 1.8 * 3 = 5.4, and 1.8 * 0.2 = 0.36. So adding those together: 5.4 + 0.36 = 5.76. So, 1.8 * 3.2 = 5.76. Therefore, the numerator is √5.76.

Now, what's the square root of 5.76? Let me recall. I know that 2.4 squared is 5.76 because 2.4 * 2.4. Let me check: 2 * 2 = 4, 2 * 0.4 = 0.8, 0.4 * 2 = 0.8, and 0.4 * 0.4 = 0.16. So adding those up: 4 + 0.8 + 0.8 + 0.16 = 4 + 1.6 + 0.16 = 5.76. Yes! So √5.76 = 2.4.

So the numerator simplifies to 2.4.

Now, the denominator is 2.7 - 2.5. Let me compute that. 2.7 minus 2.5. That's 0.2.

So now, the entire expression simplifies to 2.4 divided by 0.2. Let me compute that.

Dividing by 0.2 is the same as multiplying by 5, because 1/0.2 = 5. So 2.4 * 5. Let me calculate that.

2 * 5 = 10, and 0.4 * 5 = 2. So 10 + 2 = 12. Therefore, 2.4 divided by 0.2 is 12.

Wait, so the entire expression simplifies to 12? Let me check my steps again to make sure I didn't make a mistake.

First, numerator: √(2.5² - 0.7²). I used difference of squares, which is correct. Then, 2.5 - 0.7 = 1.8, and 2.5 + 0.7 = 3.2. Multiplying those gives 1.8 * 3.2. Let me verify that again.

1.8 * 3.2. Let me do it another way. 1.8 * 3.2. Multiply 1.8 by 3.2. Let me convert them to fractions. 1.8 is 9/5, and 3.2 is 16/5. So multiplying 9/5 * 16/5 = (9*16)/25. 9*16 is 144, so 144/25. Then, √(144/25) is 12/5, which is 2.4. Yes, that's correct. So that part checks out.

Then denominator: 2.7 - 2.5 = 0.2. Correct. Then 2.4 divided by 0.2. 2.4 divided by 0.2. Let me think of it as 2.4 divided by 0.2. Since 0.2 is 1/5, dividing by 1/5 is multiplying by 5. 2.4 * 5 = 12. Correct.

So, seems like the answer is 12. But let me check if there's another way to approach this problem, just to confirm.

Alternatively, maybe compute the numerator directly without factoring. Let me try that.

Compute 2.5 squared. 2.5 squared is 6.25. Then 0.7 squared is 0.49. So subtracting those: 6.25 - 0.49. Let me compute that.

6.25 minus 0.49. 6.25 - 0.49. Let me subtract 0.50 from 6.25, which would be 5.75, but since I subtracted an extra 0.01, I need to add 0.01 back. So 5.75 + 0.01 = 5.76. So that's the same as before. Then square root of 5.76 is 2.4. Then denominator is 0.2. So 2.4 / 0.2 = 12. Yep, same result.

So that seems consistent.

Alternatively, maybe convert all decimals to fractions to see if that helps. Let me try that approach.

First, 2.5 is equal to 5/2, 0.7 is 7/10, 2.7 is 27/10.

So, numerator: √[(5/2)^2 - (7/10)^2]. Let me compute that.

(5/2)^2 is 25/4. (7/10)^2 is 49/100. So subtracting those: 25/4 - 49/100. To subtract these fractions, find a common denominator. The least common denominator of 4 and 100 is 100.

Convert 25/4 to 625/100 (since 25*25=625 and 4*25=100). Then subtract 49/100: 625/100 - 49/100 = 576/100.

So the numerator is √(576/100). The square root of 576 is 24, and square root of 100 is 10, so √(576/100) = 24/10 = 2.4.

Denominator: 2.7 - 2.5 is 27/10 - 5/2. Convert 5/2 to 25/10. So 27/10 - 25/10 = 2/10 = 1/5 = 0.2.

Therefore, the entire expression is (24/10) divided by (1/5) = (24/10) * (5/1) = (24*5)/10 = 120/10 = 12.

Same result again. So that's reassuring.

Alternatively, maybe there's a way to see this without calculating? Let me think.

Wait, the original expression is √(a² - b²)/(c - a), where a = 2.5, b = 0.7, c = 2.7.

But I don't know if that helps. Maybe not.

Alternatively, maybe notice that 2.7 is 2.5 + 0.2. So denominator is 0.2.

But I think the steps I did are sufficient.

Wait, but let me check if there's any possible mistake in the calculation. For example, when I did the difference of squares, I assumed that √(a² - b²) is √[(a - b)(a + b)], which is correct. Then, I calculated a - b and a + b correctly.

Yes, 2.5 - 0.7 is 1.8, and 2.5 + 0.7 is 3.2. Then multiplying them gives 5.76, square root is 2.4. Then denominator is 0.2.

So 2.4 divided by 0.2. Let me think of 2.4 divided by 0.2. If I move the decimal one place, both numerator and denominator by 10, that becomes 24 divided by 2, which is 12. Yep, that's another way to see it.

So, all methods lead to 12. Therefore, I think the answer is 12.

But just to be thorough, let me check with decimal approximations.

Compute numerator: √(2.5² - 0.7²).

2.5 squared is 6.25. 0.7 squared is 0.49. 6.25 - 0.49 = 5.76. Square root of 5.76 is 2.4.

Denominator: 2.7 - 2.5 = 0.2.

So 2.4 divided by 0.2. Let me do this division.

Dividing 2.4 by 0.2.

Well, 0.2 goes into 2.4 how many times?

Since 0.2 * 10 = 2.0, so 0.2 * 12 = 2.4. Therefore, 12.

Yep, that's correct.

Alternatively, if I use a calculator, but since I don't have one here, but all my manual calculations confirm that.

Therefore, I think the simplified value of the expression is 12.

**Final Answer**
\boxed{12}

To simplify the expression:

$$
\frac{\sqrt{2.5^2 - 0.7^2}}{2.7 - 2.5}
$$

we proceed step by step.

---

### **Step 1: Simplify the Numerator**

The numerator is:

$$
\sqrt{2.5^2 - 0.7^2}
$$

This is a **difference of squares**, which can be factored as:

$$
\sqrt{(2.5 - 0.7)(2.5 + 0.7)}
$$

Now compute the individual terms:

- $2.5 - 0.7 = 1.8$
- $2.5 + 0.7 = 3.2$

So the numerator becomes:

$$
\sqrt{1.8 \times 3.2}
$$

Now compute the product:

$$
1.8 \times 3.2 = 5.76
$$

Then take the square root:

$$
\sqrt{5.76} = 2.4
$$

---

### **Step 2: Simplify the Denominator**

The denominator is:

$$
2.7 - 2.5 = 0.2
$$

---

### **Step 3: Combine the Results**

Now the expression becomes:

$$
\frac{2.4}{0.2}
$$

Dividing by 0.2 is the same as multiplying by 5:

$$
2.4 \div 0.2 = 2.4 \times 5 = 12
$$

---

### **Final Answer**

$$
\boxed{12}
$$

[Assistant]
Explanation:
Okay, so I need to simplify this expression: the square root of (2.5 squared minus 0.7 squared) divided by (2.7 minus 2.5). Let me write that down first to make sure I have it right.

So, the expression is √(2.5² - 0.7²) divided by (2.7 - 2.5). Let me check if I copied that correctly. Yeah, that seems right. Alright, let me think about how to approach this.

First, maybe I should compute the numerator and the denominator separately. Let me start with the numerator: √(2.5² - 0.7²). Hmm, that looks like a difference of squares. Remember, a² - b² factors into (a - b)(a + b). Maybe I can use that here? Let me try that.

So, if I let a = 2.5 and b = 0.7, then the numerator becomes √[(a - b)(a + b)]. Let me compute that.

First, compute a - b: 2.5 - 0.7. Let me do that subtraction. 2.5 minus 0.7. Well, 2.5 minus 0.5 is 2.0, and then minus another 0.2 is 1.8. So, 2.5 - 0.7 = 1.8.

Then, a + b is 2.5 + 0.7. Let me add those. 2.5 + 0.7. 2 + 0.5 + 0.7 is 2 + 1.2 = 3.2. Wait, no. Wait, 2.5 + 0.7. Let me think again. 2.5 + 0.5 is 3.0, and then + 0.2 more is 3.2. Yes, that's right. So, a + b = 3.2.

Therefore, the numerator simplifies to √(1.8 * 3.2). Let me compute 1.8 multiplied by 3.2. Hmm, 1.8 * 3.2. Let me do that multiplication.

First, maybe convert them to fractions? Or maybe decimal multiplication. Let me try decimal.

1.8 * 3.2. Let me think of it as (1 + 0.8) * (3 + 0.2). But maybe that's more complicated. Alternatively, multiply 1.8 * 3.2.

Well, 1.8 * 3 = 5.4, and 1.8 * 0.2 = 0.36. So adding those together: 5.4 + 0.36 = 5.76. So, 1.8 * 3.2 = 5.76. Therefore, the numerator is √5.76.

Now, what's the square root of 5.76? Let me recall. I know that 2.4 squared is 5.76 because 2.4 * 2.4. Let me check: 2 * 2 = 4, 2 * 0.4 = 0.8, 0.4 * 2 = 0.8, and 0.4 * 0.4 = 0.16. So adding those up: 4 + 0.8 + 0.8 + 0.16 = 4 + 1.6 + 0.16 = 5.76. Yes! So √5.76 = 2.4.

So the numerator simplifies to 2.4.

Now, the denominator is 2.7 - 2.5. Let me compute that. 2.7 minus 2.5. That's 0.2.

So now, the entire expression simplifies to 2.4 divided by 0.2. Let me compute that.

Dividing by 0.2 is the same as multiplying by 5, because 1/0.2 = 5. So 2.4 * 5. Let me calculate that.

2 * 5 = 10, and 0.4 * 5 = 2. So 10 + 2 = 12. Therefore, 2.4 divided by 0.2 is 12.

Wait, so the entire expression simplifies to 12? Let me check my steps again to make sure I didn't make a mistake.

First, numerator: √(2.5² - 0.7²). I used difference of squares, which is correct. Then, 2.5 - 0.7 = 1.8, and 2.5 + 0.7 = 3.2. Multiplying those gives 1.8 * 3.2. Let me verify that again.

1.8 * 3.2. Let me do it another way. 1.8 * 3.2. Multiply 1.8 by 3.2. Let me convert them to fractions. 1.8 is 9/5, and 3.2 is 16/5. So multiplying 9/5 * 16/5 = (9*16)/25. 9*16 is 144, so 144/25. Then, √(144/25) is 12/5, which is 2.4. Yes, that's correct. So that part checks out.

Then denominator: 2.7 - 2.5 = 0.2. Correct. Then 2.4 divided by 0.2. 2.4 divided by 0.2. Let me think of it as 2.4 divided by 0.2. Since 0.2 is 1/5, dividing by 1/5 is multiplying by 5. 2.4 * 5 = 12. Correct.

So, seems like the answer is 12. But let me check if there's another way to approach this problem, just to confirm.

Alternatively, maybe compute the numerator directly without factoring. Let me try that.

Compute 2.5 squared. 2.5 squared is 6.25. Then 0.7 squared is 0.49. So subtracting those: 6.25 - 0.49. Let me compute that.

6.25 minus 0.49. 6.25 - 0.49. Let me subtract 0.50 from 6.25, which would be 5.75, but since I subtracted an extra 0.01, I need to add 0.01 back. So 5.75 + 0.01 = 5.76. So that's the same as before. Then square root of 5.76 is 2.4. Then denominator is 0.2. So 2.4 / 0.2 = 12. Yep, same result.

So that seems consistent.

Alternatively, maybe convert all decimals to fractions to see if that helps. Let me try that approach.

First, 2.5 is equal to 5/2, 0.7 is 7/10, 2.7 is 27/10.

So, numerator: √[(5/2)^2 - (7/10)^2]. Let me compute that.

(5/2)^2 is 25/4. (7/10)^2 is 49/100. So subtracting those: 25/4 - 49/100. To subtract these fractions, find a common denominator. The least common denominator of 4 and 100 is 100.

Convert 25/4 to 625/100 (since 25*25=625 and 4*25=100). Then subtract 49/100: 625/100 - 49/100 = 576/100.

So the numerator is √(576/100). The square root of 576 is 24, and square root of 100 is 10, so √(576/100) = 24/10 = 2.4.

Denominator: 2.7 - 2.5 is 27/10 - 5/2. Convert 5/2 to 25/10. So 27/10 - 25/10 = 2/10 = 1/5 = 0.2.

Therefore, the entire expression is (24/10) divided by (1/5) = (24/10) * (5/1) = (24*5)/10 = 120/10 = 12.

Same result again. So that's reassuring.

Alternatively, maybe there's a way to see this without calculating? Let me think.

Wait, the original expression is √(a² - b²)/(c - a), where a = 2.5, b = 0.7, c = 2.7.

But I don't know if that helps. Maybe not.

Alternatively, maybe notice that 2.7 is 2.5 + 0.2. So denominator is 0.2.

But I think the steps I did are sufficient.

Wait, but let me check if there's any possible mistake in the calculation. For example, when I did the difference of squares, I assumed that √(a² - b²) is √[(a - b)(a + b)], which is correct. Then, I calculated a - b and a + b correctly.

Yes, 2.5 - 0.7 is 1.8, and 2.5 + 0.7 is 3.2. Then multiplying them gives 5.76, square root is 2.4. Then denominator is 0.2.

So 2.4 divided by 0.2. Let me think of 2.4 divided by 0.2. If I move the decimal one place, both numerator and denominator by 10, that becomes 24 divided by 2, which is 12. Yep, that's another way to see it.

So, all methods lead to 12. Therefore, I think the answer is 12.

But just to be thorough, let me check with decimal approximations.

Compute numerator: √(2.5² - 0.7²).

2.5 squared is 6.25. 0.7 squared is 0.49. 6.25 - 0.49 = 5.76. Square root of 5.76 is 2.4.

Denominator: 2.7 - 2.5 = 0.2.

So 2.4 divided by 0.2. Let me do this division.

Dividing 2.4 by 0.2.

Well, 0.2 goes into 2.4 how many times?

Since 0.2 * 10 = 2.0, so 0.2 * 12 = 2.4. Therefore, 12.

Yep, that's correct.

Alternatively, if I use a calculator, but since I don't have one here, but all my manual calculations confirm that.

Therefore, I think the simplified value of the expression is 12.

**Final Answer**
\boxed{12}

To simplify the expression:

$$
\frac{\sqrt{2.5^2 - 0.7^2}}{2.7 - 2.5}
$$

we proceed step by step.

---

### **Step 1: Simplify the Numerator**

The numerator is:

$$
\sqrt{2.5^2 - 0.7^2}
$$

This is a **difference of squares**, which can be factored as:

$$
\sqrt{(2.5 - 0.7)(2.5 + 0.7)}
$$

Now compute the individual terms:

- $2.5 - 0.7 = 1.8$
- $2.5 + 0.7 = 3.2$

So the numerator becomes:

$$
\sqrt{1.8 \times 3.2}
$$

Now compute the product:

$$
1.8 \times 3.2 = 5.76
$$

Then take the square root:

$$
\sqrt{5.76} = 2.4
$$

---

### **Step 2: Simplify the Denominator**

The denominator is:

$$
2.7 - 2.5 = 0.2
$$

---

### **Step 3: Combine the Results**

Now the expression becomes:

$$
\frac{2.4}{0.2}
$$

Dividing by 0.2 is the same as multiplying by 5:

$$
2.4 \div 0.2 = 2.4 \times 5 = 12
$$

---

### **Final Answer**

$$
\boxed{12}
$$

Solution:
We have  \begin{align*}
\frac{\sqrt{2.5^2 - 0.7^2}}{2.7-2.5} &= \frac{\sqrt{6.25 - 0.49}}{2.7-2.5} = \frac{\sqrt{5.76}}{0.2} = \frac{\sqrt{576/100}}{0.2}\\
&= \frac{\sqrt{576}/\sqrt{100}}{0.2} = \frac{24/10}{0.2} = \frac{2.4}{0.2} = \boxed{12}.\end{align*}


Explanation:
Okay, so I need to find the ordered triple (p, q, r) such that the polynomial x³ - 3x² + 4x - 1 is a factor of x⁹ + p x⁶ + q x³ + r. Hmm, let me think about how to approach this.

First, since x³ - 3x² + 4x - 1 is a factor of the ninth-degree polynomial, that means when we divide the ninth-degree polynomial by the cubic polynomial, there should be no remainder. Alternatively, maybe I can express the ninth-degree polynomial as (x³ - 3x² + 4x - 1) multiplied by some other polynomial. Let me denote the ninth-degree polynomial as P(x) = x⁹ + p x⁶ + q x³ + r.

So, if x³ - 3x² + 4x - 1 is a factor, then there exists some polynomial Q(x) such that P(x) = (x³ - 3x² + 4x - 1) * Q(x). Now, since P(x) is a ninth-degree polynomial, Q(x) must be a sixth-degree polynomial. Let me write Q(x) as a general sixth-degree polynomial: Q(x) = a x⁶ + b x⁵ + c x⁴ + d x³ + e x² + f x + g. Then, multiplying this by the cubic polynomial should give me P(x).

But multiplying out all these terms might be tedious. Maybe there's a smarter way. Let me think. Since the given polynomial is cubic, and the ninth-degree polynomial is x⁹ + ... which is similar to (x³)^3. Wait, x⁹ is (x³)^3. So maybe there's a connection here. Let me check if the ninth-degree polynomial can be expressed as (x³ - 3x² + 4x - 1) multiplied by something related to x³?

Alternatively, maybe I can use polynomial division or the concept of roots. If x³ - 3x² + 4x - 1 is a factor, then all roots of x³ - 3x² + 4x - 1 are also roots of P(x). Let me denote the roots of the cubic polynomial as α, β, γ. Then, P(α) = 0, P(β) = 0, P(γ) = 0. But P(x) is x⁹ + p x⁶ + q x³ + r. Let me note that x⁹ can be written as (x³)^3, so maybe substituting x³ into the polynomial?

Wait, if I let y = x³, then P(x) becomes y³ + p y² + q y + r. So, if α is a root of the cubic polynomial, then α³ - 3α² + 4α - 1 = 0. Therefore, α³ = 3α² - 4α + 1. Maybe I can use this to express higher powers of α in terms of lower powers. Since P(α) = α⁹ + p α⁶ + q α³ + r = 0. So, if I can express α⁹ and α⁶ in terms of lower powers using the relation α³ = 3α² - 4α + 1, then I can substitute them into P(α) and set the equation to zero. Since this must hold for all roots α, β, γ, the coefficients of the resulting equation should be zero.

This seems promising. Let me try this approach. Let me first compute α³, α⁶, α⁹ in terms of α², α, and constants.

Starting with α³ = 3α² - 4α + 1.

Now, let's compute α⁶. Since α³ = 3α² - 4α + 1, then α⁶ = (α³)² = (3α² - 4α + 1)². Let me expand that:

(3α² - 4α + 1)² = 9α⁴ - 24α³ + (16α² + 6α²) + ... Wait, maybe I should do it step by step.

Wait, (a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc. So here, a = 3α², b = -4α, c = 1. Therefore:

(3α²)² + (-4α)² + (1)² + 2*(3α²*(-4α) + 3α²*1 + (-4α)*1)

Calculating each term:

(3α²)² = 9α⁴

(-4α)² = 16α²

(1)² = 1

Then cross terms:

2*( -12α³ + 3α² - 4α )

So altogether:

9α⁴ + 16α² + 1 + 2*(-12α³ + 3α² - 4α) = 9α⁴ + 16α² + 1 - 24α³ + 6α² - 8α

Combine like terms:

9α⁴ - 24α³ + (16α² + 6α²) + (-8α) + 1 = 9α⁴ - 24α³ + 22α² - 8α + 1

But now, we need to express α⁴ in terms of lower powers. Since α³ = 3α² - 4α + 1, multiply both sides by α to get α⁴ = 3α³ - 4α² + α. Then substitute α³ again:

α⁴ = 3*(3α² - 4α + 1) - 4α² + α = 9α² - 12α + 3 - 4α² + α = (9α² - 4α²) + (-12α + α) + 3 = 5α² - 11α + 3

So, substitute α⁴ = 5α² - 11α + 3 into the expression for α⁶:

9α⁴ - 24α³ + 22α² - 8α + 1 becomes:

9*(5α² - 11α + 3) - 24α³ + 22α² - 8α + 1

Compute each term:

9*(5α² - 11α + 3) = 45α² - 99α + 27

Then subtract 24α³: -24α³

Then add 22α²: +22α²

Then subtract 8α: -8α

Then add 1: +1

So combining all terms:

-24α³ + (45α² + 22α²) + (-99α - 8α) + (27 + 1)

Which is:

-24α³ + 67α² - 107α + 28

Now, we need to replace α³ with 3α² - 4α + 1:

-24*(3α² - 4α + 1) + 67α² - 107α + 28

Compute that:

-72α² + 96α - 24 + 67α² - 107α + 28

Combine like terms:

(-72α² + 67α²) + (96α - 107α) + (-24 + 28) = (-5α²) + (-11α) + 4

So α⁶ simplifies to -5α² - 11α + 4.

Okay, so α⁶ = -5α² - 11α + 4.

Now, moving on to α⁹. Since α⁶ is known, α⁹ = α³ * α⁶. Let me compute that.

We have α³ = 3α² - 4α + 1 and α⁶ = -5α² - 11α + 4. Therefore:

α⁹ = α³ * α⁶ = (3α² - 4α + 1)*(-5α² - 11α + 4)

Let me expand this product:

First, multiply 3α² by each term in the second polynomial:

3α²*(-5α²) = -15α⁴

3α²*(-11α) = -33α³

3α²*4 = 12α²

Then, multiply -4α by each term:

-4α*(-5α²) = 20α³

-4α*(-11α) = 44α²

-4α*4 = -16α

Then, multiply 1 by each term:

1*(-5α²) = -5α²

1*(-11α) = -11α

1*4 = 4

Now, combine all these terms:

-15α⁴ -33α³ + 12α² + 20α³ + 44α² -16α -5α² -11α + 4

Combine like terms:

For α⁴: -15α⁴

For α³: -33α³ + 20α³ = -13α³

For α²: 12α² + 44α² -5α² = (12 + 44 -5)α² = 51α²

For α terms: -16α -11α = -27α

Constants: +4

So altogether:

-15α⁴ -13α³ + 51α² -27α + 4

Now, again, we need to express α⁴ and α³ in terms of lower powers. We already have α³ = 3α² - 4α + 1 and α⁴ = 5α² - 11α + 3.

Substituting these into the expression:

First, replace α⁴ with 5α² - 11α + 3:

-15*(5α² - 11α + 3) -13α³ + 51α² -27α + 4

Compute each term:

-15*(5α² - 11α + 3) = -75α² + 165α - 45

Then -13α³: -13*(3α² - 4α + 1) = -39α² + 52α -13

Then the rest: +51α² -27α + 4

Now combine all these:

-75α² + 165α - 45 -39α² + 52α -13 + 51α² -27α + 4

Combine like terms:

For α²: -75α² -39α² + 51α² = (-75 -39 +51)α² = (-63)α²

For α terms: 165α + 52α -27α = (165 + 52 -27)α = 190α

Constants: -45 -13 + 4 = -54

So α⁹ simplifies to -63α² + 190α -54.

Okay, so now we have expressions for α³, α⁶, and α⁹ in terms of α², α, and constants. Now, going back to P(α) = α⁹ + p α⁶ + q α³ + r = 0.

Substituting the expressions we found:

α⁹ = -63α² + 190α -54

p α⁶ = p*(-5α² -11α +4)

q α³ = q*(3α² -4α +1)

r = r

So adding all these together:

(-63α² + 190α -54) + p*(-5α² -11α +4) + q*(3α² -4α +1) + r = 0

Let me expand this:

-63α² + 190α -54 -5p α² -11p α +4p +3q α² -4q α + q + r = 0

Now, combine like terms:

For α² terms: (-63 -5p +3q) α²

For α terms: (190 -11p -4q) α

For constants: (-54 +4p + q + r)

Therefore, the entire expression is:

[(-63 -5p +3q) α² + (190 -11p -4q) α + (-54 +4p + q + r)] = 0

Since α is a root of the cubic polynomial x³ - 3x² + 4x -1, which is irreducible over the rationals (I think?), so α is not a rational number, and the minimal polynomial is degree 3. Therefore, the coefficients of α², α, and the constant term must all be zero. Therefore, we have the following system of equations:

1. Coefficient of α²: -63 -5p +3q = 0

2. Coefficient of α: 190 -11p -4q = 0

3. Constant term: -54 +4p + q + r = 0

So, we have three equations with three variables p, q, r. Let me write them again:

Equation (1): -5p + 3q = 63

Equation (2): -11p -4q = -190

Equation (3): 4p + q + r = 54

So, first, let me solve equations (1) and (2) for p and q.

Let me write equations (1) and (2):

Equation (1): -5p + 3q = 63

Equation (2): -11p -4q = -190

Let me solve this system. Let me use the elimination method. Let me multiply equation (1) by 4 and equation (2) by 3 to make the coefficients of q opposites?

Wait, equation (1) has 3q and equation (2) has -4q. Let me find a common multiple. Let me multiply equation (1) by 4 and equation (2) by 3:

Equation (1)*4: -20p + 12q = 252

Equation (2)*3: -33p -12q = -570

Now, adding these two equations:

(-20p -33p) + (12q -12q) = 252 -570

Which is:

-53p = -318

Therefore, p = (-318)/(-53) = 318/53. Let me check if this reduces. 53*6 = 318, so p = 6.

Wait, 53*6 is 318? Let me check: 50*6=300, 3*6=18, so 300+18=318. Yes. So p = 6.

Now, substitute p = 6 into equation (1):

-5*(6) + 3q = 63 => -30 + 3q = 63 => 3q = 63 + 30 = 93 => q = 93 / 3 = 31.

So q = 31.

Now, check equation (2) with p=6 and q=31:

-11*(6) -4*(31) = -66 -124 = -190. Which matches the right-hand side. Perfect.

Now, moving to equation (3): 4p + q + r = 54. Substitute p=6, q=31:

4*6 + 31 + r = 24 + 31 + r = 55 + r = 54 => r = 54 - 55 = -1.

Therefore, r = -1.

So the ordered triple is (p, q, r) = (6, 31, -1). Let me check if this makes sense.

But let me verify this result by another method to be sure. Maybe by considering that the ninth-degree polynomial is (x³ - 3x² + 4x -1) multiplied by some polynomial. Since P(x) = x⁹ + p x⁶ + q x³ + r, which is similar to (x³)^3 + p(x³)^2 + q(x³) + r. So if we let y = x³, then P(x) = y³ + p y² + q y + r. If the cubic polynomial is a factor, then y³ + p y² + q y + r must be divisible by the original cubic polynomial evaluated at y = x³? Wait, no. Wait, the original cubic is x³ - 3x² + 4x -1. But if we have P(x) = (x³ - 3x² + 4x -1) * Q(x). If we can write Q(x) as something that when multiplied by the cubic gives P(x). But since P(x) is x⁹ + p x⁶ + q x³ + r, which is similar to (x³)^3 + p(x³)^2 + q(x³) + r. So if we think of Q(x) as a polynomial in x³, maybe?

Alternatively, perhaps Q(x) is a cubic polynomial? Wait, no. Since P(x) is degree 9, and the cubic is degree 3, then Q(x) must be degree 6. But maybe there's a pattern here. Let me think.

Alternatively, since P(x) is a polynomial in x³, maybe Q(x) is also a polynomial in x³? Let me check. Suppose Q(x) = x^6 + a x^3 + b. Then multiplying by the cubic:

(x³ - 3x² + 4x -1)(x^6 + a x^3 + b) = x^9 + a x^6 + b x³ - 3x^8 - 3a x^5 - 3b x² + 4x^7 + 4a x^4 + 4b x - x^6 - a x³ - b.

But this would result in terms with x^9, x^8, x^7, x^6, x^5, x^4, x³, x², x, and constants. However, P(x) is x^9 + p x^6 + q x³ + r, which has no x^8, x^7, x^5, x^4, x², or x terms. Therefore, the coefficients of these terms must be zero. Therefore, we can set up equations for the coefficients of x^8, x^7, x^5, x^4, x², x to be zero. Let me try this approach.

Let me denote Q(x) = x^6 + a x^3 + b. Then, multiplying by the cubic:

(x³ - 3x² + 4x -1)(x^6 + a x^3 + b) = x^9 + a x^6 + b x³ - 3x^8 - 3a x^5 - 3b x² + 4x^7 + 4a x^4 + 4b x - x^6 - a x³ - b.

Now, combine like terms:

x^9 + (a x^6 - x^6) + (-3x^8) + (4x^7) + (-3a x^5) + (4a x^4) + (b x³ - a x³) + (-3b x²) + (4b x) + (-b)

Simplify:

x^9 + (a - 1)x^6 - 3x^8 + 4x^7 - 3a x^5 + 4a x^4 + (b - a)x³ - 3b x² + 4b x - b

Now, since P(x) is x^9 + p x^6 + q x³ + r, the coefficients of x^8, x^7, x^5, x^4, x², x must be zero. Therefore:

Coefficient of x^8: -3 = 0? Wait, that can't be. So this suggests that my assumption that Q(x) is x^6 + a x^3 + b is wrong because there is a -3x^8 term which cannot be canceled. Therefore, my initial assumption about the form of Q(x) is incorrect. Therefore, Q(x) cannot be just a polynomial in x³. Therefore, this approach may not work unless we include more terms.

Alternatively, maybe Q(x) is a polynomial that when multiplied by the cubic gives only terms in x^9, x^6, x³, and constants. Therefore, the coefficients of x^8, x^7, etc., must be zero. Therefore, we need to find Q(x) such that when multiplied by the cubic, all the unwanted terms cancel out. This seems complicated, but maybe we can use the previous result where we found p=6, q=31, r=-1. Let me check if with these values, the polynomial P(x) = x^9 + 6x^6 + 31x³ -1 is divisible by the cubic.

Alternatively, since we already solved the equations using the root substitution method, and got p=6, q=31, r=-1, maybe that is correct. However, to verify, let me check if the polynomial P(x) = x^9 + 6x^6 + 31x³ -1 is divisible by x³ - 3x² + 4x -1.

Alternatively, since we know that if α is a root of the cubic, then P(α) = 0. Let me check with the values we found. Earlier, we found that P(α) = 0 when p=6, q=31, r=-1. Therefore, that should be correct. But let me check with actual numbers. Let me pick a root of the cubic and see if P(α) = 0. However, since the cubic is x³ - 3x² + 4x -1, it's not easy to find its roots explicitly. Alternatively, maybe I can perform polynomial division.

Alternatively, since we have the expressions for α³, α⁶, α⁹, let me compute P(α) with p=6, q=31, r=-1:

P(α) = α⁹ + 6α⁶ + 31α³ -1.

From earlier, we had:

α³ = 3α² -4α +1

α⁶ = -5α² -11α +4

α⁹ = -63α² +190α -54

Therefore, substituting:

P(α) = (-63α² +190α -54) + 6*(-5α² -11α +4) + 31*(3α² -4α +1) -1

Compute each term:

First term: -63α² +190α -54

Second term: 6*(-5α² -11α +4) = -30α² -66α +24

Third term: 31*(3α² -4α +1) = 93α² -124α +31

Fourth term: -1

Now, adding all together:

(-63α² -30α² +93α²) + (190α -66α -124α) + (-54 +24 +31 -1)

Calculating each part:

For α²: (-63 -30 +93)α² = 0α²

For α: (190 -66 -124)α = 0α

Constants: (-54 +24 +31 -1) = (-54 + 24) + (31 -1) = (-30) + 30 = 0

So P(α) = 0 + 0 + 0 = 0. Perfect! Therefore, with p=6, q=31, r=-1, P(α)=0. Therefore, the values are correct. Therefore, the ordered triple is (6, 31, -1).

But just to be thorough, let me check if there's another way to get this result. For example, maybe using the fact that if the cubic divides the ninth-degree polynomial, then the ninth-degree polynomial is the cubic raised to the third power, but adjusted by some coefficients. Wait, but (x³ - 3x² + 4x -1)^3 would be a ninth-degree polynomial, but it's not necessarily the same as x⁹ + p x⁶ + q x³ + r. However, maybe there's a relation. Let me compute (x³ - 3x² + 4x -1)^3 and see if it can be expressed in terms of x⁹ + ...

But expanding (x³ - 3x² + 4x -1)^3 would be tedious, but maybe we can use the fact that if we let y = x³, then the cubic is y - 3x² + 4x -1, but that might not help. Alternatively, maybe not. Alternatively, since we have P(x) = x⁹ + p x⁶ + q x³ + r, which is similar to (x³)^3 + p(x³)^2 + q(x³) + r. If we think of this as a polynomial in x³, then if the cubic polynomial divides P(x), then perhaps the cubic polynomial divides the polynomial in y = x³? Wait, but the cubic polynomial is in x, not in y. So unless there's some relation between the two.

Alternatively, maybe if we consider that if the cubic divides P(x), then the cubic divides P(x) when considered as a polynomial in x. However, since P(x) is a polynomial in x³, maybe there's a connection. But I think the previous method is sufficient and verified. Therefore, I think the answer is (6, 31, -1).

But just to make sure, let me check with another root. Suppose I take β, another root of the cubic. Then P(β) should also be zero. But since the equations were derived for any root, and we checked for α, it should hold for all roots. Therefore, the answer is correct.

Therefore, the ordered triple is (6, 31, -1).

**Final Answer**
The ordered triple is \boxed{(6, 31, -1)}.

To solve the problem, we are given that the polynomial
$$
x^3 - 3x^2 + 4x - 1
$$
is a factor of
$$
x^9 + p x^6 + q x^3 + r.
$$
We are to find the ordered triple $(p, q, r)$.

---

### Step 1: Understand the Structure

Let the cubic polynomial be denoted as:
$$
f(x) = x^3 - 3x^2 + 4x - 1.
$$
Let the ninth-degree polynomial be:
$$
P(x) = x^9 + p x^6 + q x^3 + r.
$$

Since $f(x)$ is a factor of $P(x)$, it must divide $P(x)$ exactly. That is, there exists a polynomial $Q(x)$ such that:
$$
P(x) = f(x) \cdot Q(x).
$$
Since $P(x)$ is degree 9 and $f(x)$ is degree 3, $Q(x)$ must be degree 6.

---

### Step 2: Use Root Substitution

Let $\alpha$ be a root of $f(x)$. Then:
$$
\alpha^3 = 3\alpha^2 - 4\alpha + 1.
$$
We can use this to express higher powers of $\alpha$ in terms of lower powers. This allows us to compute $\alpha^6$ and $\alpha^9$ in terms of $\alpha^2$, $\alpha$, and constants.

#### Compute $\alpha^6$:
$$
\alpha^6 = (\alpha^3)^2 = (3\alpha^2 - 4\alpha + 1)^2 = -5\alpha^2 - 11\alpha + 4.
$$

#### Compute $\alpha^9$:
$$
\alpha^9 = \alpha^3 \cdot \alpha^6 = (3\alpha^2 - 4\alpha + 1)(-5\alpha^2 - 11\alpha + 4) = -63\alpha^2 + 190\alpha - 54.
$$

---

### Step 3: Plug into $P(\alpha)$

We now substitute into the expression for $P(x)$:
$$
P(\alpha) = \alpha^9 + p \alpha^6 + q \alpha^3 + r.
$$
Substitute the expressions:
$$
P(\alpha) = (-63\alpha^2 + 190\alpha - 54) + p(-5\alpha^2 - 11\alpha + 4) + q(3\alpha^2 - 4\alpha + 1) + r.
$$

Group like terms:
- Coefficient of $\alpha^2$: $-63 - 5p + 3q$
- Coefficient of $\alpha$: $190 - 11p - 4q$
- Constant term: $-54 + 4p + q + r$

Since $\alpha$ is a root of $f(x)$, and $f(x)$ is irreducible over the rationals, the only way for $P(\alpha) = 0$ is if all coefficients are zero:
$$
\begin{cases}
-63 - 5p + 3q = 0 \\
190 - 11p - 4q = 0 \\
-54 + 4p + q + r = 0
\end{cases}
$$

---

### Step 4: Solve the System

From the first equation:
$$
-5p + 3q = 63 \quad \text{(1)}
$$

From the second:
$$
-11p - 4q = -190 \quad \text{(2)}
$$

Multiply (1) by 4 and (2) by 3:
$$
\begin{aligned}
-20p + 12q &= 252 \\
-33p - 12q &= -570
\end{aligned}
$$

Add the two equations:
$$
-53p = -318 \Rightarrow p = 6
$$

Substitute $p = 6$ into (1):
$$
-5(6) + 3q = 63 \Rightarrow -30 + 3q = 63 \Rightarrow q = 31
$$

Now substitute $p = 6$, $q = 31$ into the third equation:
$$
-54 + 4(6) + 31 + r = 0 \Rightarrow -54 + 24 + 31 + r = 0 \Rightarrow r = -1
$$

---

### Final Answer

$$
\boxed{(6, 31, -1)}
$$

[Assistant]
Explanation:
Okay, so I need to find the ordered triple (p, q, r) such that the polynomial x³ - 3x² + 4x - 1 is a factor of x⁹ + p x⁶ + q x³ + r. Hmm, let me think about how to approach this.

First, since x³ - 3x² + 4x - 1 is a factor of the ninth-degree polynomial, that means when we divide the ninth-degree polynomial by the cubic polynomial, there should be no remainder. Alternatively, maybe I can express the ninth-degree polynomial as (x³ - 3x² + 4x - 1) multiplied by some other polynomial. Let me denote the ninth-degree polynomial as P(x) = x⁹ + p x⁶ + q x³ + r.

So, if x³ - 3x² + 4x - 1 is a factor, then there exists some polynomial Q(x) such that P(x) = (x³ - 3x² + 4x - 1) * Q(x). Now, since P(x) is a ninth-degree polynomial, Q(x) must be a sixth-degree polynomial. Let me write Q(x) as a general sixth-degree polynomial: Q(x) = a x⁶ + b x⁵ + c x⁴ + d x³ + e x² + f x + g. Then, multiplying this by the cubic polynomial should give me P(x).

But multiplying out all these terms might be tedious. Maybe there's a smarter way. Let me think. Since the given polynomial is cubic, and the ninth-degree polynomial is x⁹ + ... which is similar to (x³)^3. Wait, x⁹ is (x³)^3. So maybe there's a connection here. Let me check if the ninth-degree polynomial can be expressed as (x³ - 3x² + 4x - 1) multiplied by something related to x³?

Alternatively, maybe I can use polynomial division or the concept of roots. If x³ - 3x² + 4x - 1 is a factor, then all roots of x³ - 3x² + 4x - 1 are also roots of P(x). Let me denote the roots of the cubic polynomial as α, β, γ. Then, P(α) = 0, P(β) = 0, P(γ) = 0. But P(x) is x⁹ + p x⁶ + q x³ + r. Let me note that x⁹ can be written as (x³)^3, so maybe substituting x³ into the polynomial?

Wait, if I let y = x³, then P(x) becomes y³ + p y² + q y + r. So, if α is a root of the cubic polynomial, then α³ - 3α² + 4α - 1 = 0. Therefore, α³ = 3α² - 4α + 1. Maybe I can use this to express higher powers of α in terms of lower powers. Since P(α) = α⁹ + p α⁶ + q α³ + r = 0. So, if I can express α⁹ and α⁶ in terms of lower powers using the relation α³ = 3α² - 4α + 1, then I can substitute them into P(α) and set the equation to zero. Since this must hold for all roots α, β, γ, the coefficients of the resulting equation should be zero.

This seems promising. Let me try this approach. Let me first compute α³, α⁶, α⁹ in terms of α², α, and constants.

Starting with α³ = 3α² - 4α + 1.

Now, let's compute α⁶. Since α³ = 3α² - 4α + 1, then α⁶ = (α³)² = (3α² - 4α + 1)². Let me expand that:

(3α² - 4α + 1)² = 9α⁴ - 24α³ + (16α² + 6α²) + ... Wait, maybe I should do it step by step.

Wait, (a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc. So here, a = 3α², b = -4α, c = 1. Therefore:

(3α²)² + (-4α)² + (1)² + 2*(3α²*(-4α) + 3α²*1 + (-4α)*1)

Calculating each term:

(3α²)² = 9α⁴

(-4α)² = 16α²

(1)² = 1

Then cross terms:

2*( -12α³ + 3α² - 4α )

So altogether:

9α⁴ + 16α² + 1 + 2*(-12α³ + 3α² - 4α) = 9α⁴ + 16α² + 1 - 24α³ + 6α² - 8α

Combine like terms:

9α⁴ - 24α³ + (16α² + 6α²) + (-8α) + 1 = 9α⁴ - 24α³ + 22α² - 8α + 1

But now, we need to express α⁴ in terms of lower powers. Since α³ = 3α² - 4α + 1, multiply both sides by α to get α⁴ = 3α³ - 4α² + α. Then substitute α³ again:

α⁴ = 3*(3α² - 4α + 1) - 4α² + α = 9α² - 12α + 3 - 4α² + α = (9α² - 4α²) + (-12α + α) + 3 = 5α² - 11α + 3

So, substitute α⁴ = 5α² - 11α + 3 into the expression for α⁶:

9α⁴ - 24α³ + 22α² - 8α + 1 becomes:

9*(5α² - 11α + 3) - 24α³ + 22α² - 8α + 1

Compute each term:

9*(5α² - 11α + 3) = 45α² - 99α + 27

Then subtract 24α³: -24α³

Then add 22α²: +22α²

Then subtract 8α: -8α

Then add 1: +1

So combining all terms:

-24α³ + (45α² + 22α²) + (-99α - 8α) + (27 + 1)

Which is:

-24α³ + 67α² - 107α + 28

Now, we need to replace α³ with 3α² - 4α + 1:

-24*(3α² - 4α + 1) + 67α² - 107α + 28

Compute that:

-72α² + 96α - 24 + 67α² - 107α + 28

Combine like terms:

(-72α² + 67α²) + (96α - 107α) + (-24 + 28) = (-5α²) + (-11α) + 4

So α⁶ simplifies to -5α² - 11α + 4.

Okay, so α⁶ = -5α² - 11α + 4.

Now, moving on to α⁹. Since α⁶ is known, α⁹ = α³ * α⁶. Let me compute that.

We have α³ = 3α² - 4α + 1 and α⁶ = -5α² - 11α + 4. Therefore:

α⁹ = α³ * α⁶ = (3α² - 4α + 1)*(-5α² - 11α + 4)

Let me expand this product:

First, multiply 3α² by each term in the second polynomial:

3α²*(-5α²) = -15α⁴

3α²*(-11α) = -33α³

3α²*4 = 12α²

Then, multiply -4α by each term:

-4α*(-5α²) = 20α³

-4α*(-11α) = 44α²

-4α*4 = -16α

Then, multiply 1 by each term:

1*(-5α²) = -5α²

1*(-11α) = -11α

1*4 = 4

Now, combine all these terms:

-15α⁴ -33α³ + 12α² + 20α³ + 44α² -16α -5α² -11α + 4

Combine like terms:

For α⁴: -15α⁴

For α³: -33α³ + 20α³ = -13α³

For α²: 12α² + 44α² -5α² = (12 + 44 -5)α² = 51α²

For α terms: -16α -11α = -27α

Constants: +4

So altogether:

-15α⁴ -13α³ + 51α² -27α + 4

Now, again, we need to express α⁴ and α³ in terms of lower powers. We already have α³ = 3α² - 4α + 1 and α⁴ = 5α² - 11α + 3.

Substituting these into the expression:

First, replace α⁴ with 5α² - 11α + 3:

-15*(5α² - 11α + 3) -13α³ + 51α² -27α + 4

Compute each term:

-15*(5α² - 11α + 3) = -75α² + 165α - 45

Then -13α³: -13*(3α² - 4α + 1) = -39α² + 52α -13

Then the rest: +51α² -27α + 4

Now combine all these:

-75α² + 165α - 45 -39α² + 52α -13 + 51α² -27α + 4

Combine like terms:

For α²: -75α² -39α² + 51α² = (-75 -39 +51)α² = (-63)α²

For α terms: 165α + 52α -27α = (165 + 52 -27)α = 190α

Constants: -45 -13 + 4 = -54

So α⁹ simplifies to -63α² + 190α -54.

Okay, so now we have expressions for α³, α⁶, and α⁹ in terms of α², α, and constants. Now, going back to P(α) = α⁹ + p α⁶ + q α³ + r = 0.

Substituting the expressions we found:

α⁹ = -63α² + 190α -54

p α⁶ = p*(-5α² -11α +4)

q α³ = q*(3α² -4α +1)

r = r

So adding all these together:

(-63α² + 190α -54) + p*(-5α² -11α +4) + q*(3α² -4α +1) + r = 0

Let me expand this:

-63α² + 190α -54 -5p α² -11p α +4p +3q α² -4q α + q + r = 0

Now, combine like terms:

For α² terms: (-63 -5p +3q) α²

For α terms: (190 -11p -4q) α

For constants: (-54 +4p + q + r)

Therefore, the entire expression is:

[(-63 -5p +3q) α² + (190 -11p -4q) α + (-54 +4p + q + r)] = 0

Since α is a root of the cubic polynomial x³ - 3x² + 4x -1, which is irreducible over the rationals (I think?), so α is not a rational number, and the minimal polynomial is degree 3. Therefore, the coefficients of α², α, and the constant term must all be zero. Therefore, we have the following system of equations:

1. Coefficient of α²: -63 -5p +3q = 0

2. Coefficient of α: 190 -11p -4q = 0

3. Constant term: -54 +4p + q + r = 0

So, we have three equations with three variables p, q, r. Let me write them again:

Equation (1): -5p + 3q = 63

Equation (2): -11p -4q = -190

Equation (3): 4p + q + r = 54

So, first, let me solve equations (1) and (2) for p and q.

Let me write equations (1) and (2):

Equation (1): -5p + 3q = 63

Equation (2): -11p -4q = -190

Let me solve this system. Let me use the elimination method. Let me multiply equation (1) by 4 and equation (2) by 3 to make the coefficients of q opposites?

Wait, equation (1) has 3q and equation (2) has -4q. Let me find a common multiple. Let me multiply equation (1) by 4 and equation (2) by 3:

Equation (1)*4: -20p + 12q = 252

Equation (2)*3: -33p -12q = -570

Now, adding these two equations:

(-20p -33p) + (12q -12q) = 252 -570

Which is:

-53p = -318

Therefore, p = (-318)/(-53) = 318/53. Let me check if this reduces. 53*6 = 318, so p = 6.

Wait, 53*6 is 318? Let me check: 50*6=300, 3*6=18, so 300+18=318. Yes. So p = 6.

Now, substitute p = 6 into equation (1):

-5*(6) + 3q = 63 => -30 + 3q = 63 => 3q = 63 + 30 = 93 => q = 93 / 3 = 31.

So q = 31.

Now, check equation (2) with p=6 and q=31:

-11*(6) -4*(31) = -66 -124 = -190. Which matches the right-hand side. Perfect.

Now, moving to equation (3): 4p + q + r = 54. Substitute p=6, q=31:

4*6 + 31 + r = 24 + 31 + r = 55 + r = 54 => r = 54 - 55 = -1.

Therefore, r = -1.

So the ordered triple is (p, q, r) = (6, 31, -1). Let me check if this makes sense.

But let me verify this result by another method to be sure. Maybe by considering that the ninth-degree polynomial is (x³ - 3x² + 4x -1) multiplied by some polynomial. Since P(x) = x⁹ + p x⁶ + q x³ + r, which is similar to (x³)^3 + p(x³)^2 + q(x³) + r. So if we let y = x³, then P(x) = y³ + p y² + q y + r. If the cubic polynomial is a factor, then y³ + p y² + q y + r must be divisible by the original cubic polynomial evaluated at y = x³? Wait, no. Wait, the original cubic is x³ - 3x² + 4x -1. But if we have P(x) = (x³ - 3x² + 4x -1) * Q(x). If we can write Q(x) as something that when multiplied by the cubic gives P(x). But since P(x) is x⁹ + p x⁶ + q x³ + r, which is similar to (x³)^3 + p(x³)^2 + q(x³) + r. So if we think of Q(x) as a polynomial in x³, maybe?

Alternatively, perhaps Q(x) is a cubic polynomial? Wait, no. Since P(x) is degree 9, and the cubic is degree 3, then Q(x) must be degree 6. But maybe there's a pattern here. Let me think.

Alternatively, since P(x) is a polynomial in x³, maybe Q(x) is also a polynomial in x³? Let me check. Suppose Q(x) = x^6 + a x^3 + b. Then multiplying by the cubic:

(x³ - 3x² + 4x -1)(x^6 + a x^3 + b) = x^9 + a x^6 + b x³ - 3x^8 - 3a x^5 - 3b x² + 4x^7 + 4a x^4 + 4b x - x^6 - a x³ - b.

But this would result in terms with x^9, x^8, x^7, x^6, x^5, x^4, x³, x², x, and constants. However, P(x) is x^9 + p x^6 + q x³ + r, which has no x^8, x^7, x^5, x^4, x², or x terms. Therefore, the coefficients of these terms must be zero. Therefore, we can set up equations for the coefficients of x^8, x^7, x^5, x^4, x², x to be zero. Let me try this approach.

Let me denote Q(x) = x^6 + a x^3 + b. Then, multiplying by the cubic:

(x³ - 3x² + 4x -1)(x^6 + a x^3 + b) = x^9 + a x^6 + b x³ - 3x^8 - 3a x^5 - 3b x² + 4x^7 + 4a x^4 + 4b x - x^6 - a x³ - b.

Now, combine like terms:

x^9 + (a x^6 - x^6) + (-3x^8) + (4x^7) + (-3a x^5) + (4a x^4) + (b x³ - a x³) + (-3b x²) + (4b x) + (-b)

Simplify:

x^9 + (a - 1)x^6 - 3x^8 + 4x^7 - 3a x^5 + 4a x^4 + (b - a)x³ - 3b x² + 4b x - b

Now, since P(x) is x^9 + p x^6 + q x³ + r, the coefficients of x^8, x^7, x^5, x^4, x², x must be zero. Therefore:

Coefficient of x^8: -3 = 0? Wait, that can't be. So this suggests that my assumption that Q(x) is x^6 + a x^3 + b is wrong because there is a -3x^8 term which cannot be canceled. Therefore, my initial assumption about the form of Q(x) is incorrect. Therefore, Q(x) cannot be just a polynomial in x³. Therefore, this approach may not work unless we include more terms.

Alternatively, maybe Q(x) is a polynomial that when multiplied by the cubic gives only terms in x^9, x^6, x³, and constants. Therefore, the coefficients of x^8, x^7, etc., must be zero. Therefore, we need to find Q(x) such that when multiplied by the cubic, all the unwanted terms cancel out. This seems complicated, but maybe we can use the previous result where we found p=6, q=31, r=-1. Let me check if with these values, the polynomial P(x) = x^9 + 6x^6 + 31x³ -1 is divisible by the cubic.

Alternatively, since we already solved the equations using the root substitution method, and got p=6, q=31, r=-1, maybe that is correct. However, to verify, let me check if the polynomial P(x) = x^9 + 6x^6 + 31x³ -1 is divisible by x³ - 3x² + 4x -1.

Alternatively, since we know that if α is a root of the cubic, then P(α) = 0. Let me check with the values we found. Earlier, we found that P(α) = 0 when p=6, q=31, r=-1. Therefore, that should be correct. But let me check with actual numbers. Let me pick a root of the cubic and see if P(α) = 0. However, since the cubic is x³ - 3x² + 4x -1, it's not easy to find its roots explicitly. Alternatively, maybe I can perform polynomial division.

Alternatively, since we have the expressions for α³, α⁶, α⁹, let me compute P(α) with p=6, q=31, r=-1:

P(α) = α⁹ + 6α⁶ + 31α³ -1.

From earlier, we had:

α³ = 3α² -4α +1

α⁶ = -5α² -11α +4

α⁹ = -63α² +190α -54

Therefore, substituting:

P(α) = (-63α² +190α -54) + 6*(-5α² -11α +4) + 31*(3α² -4α +1) -1

Compute each term:

First term: -63α² +190α -54

Second term: 6*(-5α² -11α +4) = -30α² -66α +24

Third term: 31*(3α² -4α +1) = 93α² -124α +31

Fourth term: -1

Now, adding all together:

(-63α² -30α² +93α²) + (190α -66α -124α) + (-54 +24 +31 -1)

Calculating each part:

For α²: (-63 -30 +93)α² = 0α²

For α: (190 -66 -124)α = 0α

Constants: (-54 +24 +31 -1) = (-54 + 24) + (31 -1) = (-30) + 30 = 0

So P(α) = 0 + 0 + 0 = 0. Perfect! Therefore, with p=6, q=31, r=-1, P(α)=0. Therefore, the values are correct. Therefore, the ordered triple is (6, 31, -1).

But just to be thorough, let me check if there's another way to get this result. For example, maybe using the fact that if the cubic divides the ninth-degree polynomial, then the ninth-degree polynomial is the cubic raised to the third power, but adjusted by some coefficients. Wait, but (x³ - 3x² + 4x -1)^3 would be a ninth-degree polynomial, but it's not necessarily the same as x⁹ + p x⁶ + q x³ + r. However, maybe there's a relation. Let me compute (x³ - 3x² + 4x -1)^3 and see if it can be expressed in terms of x⁹ + ...

But expanding (x³ - 3x² + 4x -1)^3 would be tedious, but maybe we can use the fact that if we let y = x³, then the cubic is y - 3x² + 4x -1, but that might not help. Alternatively, maybe not. Alternatively, since we have P(x) = x⁹ + p x⁶ + q x³ + r, which is similar to (x³)^3 + p(x³)^2 + q(x³) + r. If we think of this as a polynomial in x³, then if the cubic polynomial divides P(x), then perhaps the cubic polynomial divides the polynomial in y = x³? Wait, but the cubic polynomial is in x, not in y. So unless there's some relation between the two.

Alternatively, maybe if we consider that if the cubic divides P(x), then the cubic divides P(x) when considered as a polynomial in x. However, since P(x) is a polynomial in x³, maybe there's a connection. But I think the previous method is sufficient and verified. Therefore, I think the answer is (6, 31, -1).

But just to make sure, let me check with another root. Suppose I take β, another root of the cubic. Then P(β) should also be zero. But since the equations were derived for any root, and we checked for α, it should hold for all roots. Therefore, the answer is correct.

Therefore, the ordered triple is (6, 31, -1).

**Final Answer**
The ordered triple is \boxed{(6, 31, -1)}.

To solve the problem, we are given that the polynomial
$$
x^3 - 3x^2 + 4x - 1
$$
is a factor of
$$
x^9 + p x^6 + q x^3 + r.
$$
We are to find the ordered triple $(p, q, r)$.

---

### Step 1: Understand the Structure

Let the cubic polynomial be denoted as:
$$
f(x) = x^3 - 3x^2 + 4x - 1.
$$
Let the ninth-degree polynomial be:
$$
P(x) = x^9 + p x^6 + q x^3 + r.
$$

Since $f(x)$ is a factor of $P(x)$, it must divide $P(x)$ exactly. That is, there exists a polynomial $Q(x)$ such that:
$$
P(x) = f(x) \cdot Q(x).
$$
Since $P(x)$ is degree 9 and $f(x)$ is degree 3, $Q(x)$ must be degree 6.

---

### Step 2: Use Root Substitution

Let $\alpha$ be a root of $f(x)$. Then:
$$
\alpha^3 = 3\alpha^2 - 4\alpha + 1.
$$
We can use this to express higher powers of $\alpha$ in terms of lower powers. This allows us to compute $\alpha^6$ and $\alpha^9$ in terms of $\alpha^2$, $\alpha$, and constants.

#### Compute $\alpha^6$:
$$
\alpha^6 = (\alpha^3)^2 = (3\alpha^2 - 4\alpha + 1)^2 = -5\alpha^2 - 11\alpha + 4.
$$

#### Compute $\alpha^9$:
$$
\alpha^9 = \alpha^3 \cdot \alpha^6 = (3\alpha^2 - 4\alpha + 1)(-5\alpha^2 - 11\alpha + 4) = -63\alpha^2 + 190\alpha - 54.
$$

---

### Step 3: Plug into $P(\alpha)$

We now substitute into the expression for $P(x)$:
$$
P(\alpha) = \alpha^9 + p \alpha^6 + q \alpha^3 + r.
$$
Substitute the expressions:
$$
P(\alpha) = (-63\alpha^2 + 190\alpha - 54) + p(-5\alpha^2 - 11\alpha + 4) + q(3\alpha^2 - 4\alpha + 1) + r.
$$

Group like terms:
- Coefficient of $\alpha^2$: $-63 - 5p + 3q$
- Coefficient of $\alpha$: $190 - 11p - 4q$
- Constant term: $-54 + 4p + q + r$

Since $\alpha$ is a root of $f(x)$, and $f(x)$ is irreducible over the rationals, the only way for $P(\alpha) = 0$ is if all coefficients are zero:
$$
\begin{cases}
-63 - 5p + 3q = 0 \\
190 - 11p - 4q = 0 \\
-54 + 4p + q + r = 0
\end{cases}
$$

---

### Step 4: Solve the System

From the first equation:
$$
-5p + 3q = 63 \quad \text{(1)}
$$

From the second:
$$
-11p - 4q = -190 \quad \text{(2)}
$$

Multiply (1) by 4 and (2) by 3:
$$
\begin{aligned}
-20p + 12q &= 252 \\
-33p - 12q &= -570
\end{aligned}
$$

Add the two equations:
$$
-53p = -318 \Rightarrow p = 6
$$

Substitute $p = 6$ into (1):
$$
-5(6) + 3q = 63 \Rightarrow -30 + 3q = 63 \Rightarrow q = 31
$$

Now substitute $p = 6$, $q = 31$ into the third equation:
$$
-54 + 4(6) + 31 + r = 0 \Rightarrow -54 + 24 + 31 + r = 0 \Rightarrow r = -1
$$

---

### Final Answer

$$
\boxed{(6, 31, -1)}
$$

Solution:
Let $\alpha$ be a root of $x^3 - 3x^2 + 4x - 1 = 0,$ so $\alpha^3 = 3 \alpha^2 - 4 \alpha + 1.$  Then
\[\alpha^4 = 3 \alpha^3 - 4 \alpha^2 + \alpha = 3 (3 \alpha^2 - 4 \alpha + 1) - 4 \alpha^2 + \alpha = 5 \alpha^2 - 11 \alpha + 3.\]Hence,
\begin{align*}
\alpha^6 &= (3 \alpha^2 - 4 \alpha + 1)^2 \\
&= 9 \alpha^4 - 24 \alpha^3 + 22 \alpha^2 - 8 \alpha + 1 \\
&= 9 (5 \alpha^2 - 11 \alpha + 3) - 24 (3 \alpha^2 - 4 \alpha + 1) + 22 \alpha^2 - 8 \alpha + 1 \\
&= -5 \alpha^2 - 11 \alpha + 4,
\end{align*}and
\begin{align*}
\alpha^9 &= \alpha^3 \cdot \alpha^6 \\
&= (3 \alpha^2 - 4 \alpha + 1)(-5 \alpha^2 - 11 \alpha + 4) \\
&= -15 \alpha^4 - 13 \alpha^3 + 51 \alpha^2 - 27 \alpha + 4 \\
&= -15 (5 \alpha^2 - 11 \alpha + 3) - 13 (3 \alpha^2 - 4 \alpha + 1) + 51 \alpha^2 - 27 \alpha + 4 \\
&= -63 \alpha^2 + 190 \alpha - 54.
\end{align*}Then
\begin{align*}
\alpha^9 + p \alpha^6 + q \alpha^3 + r &= (-63 \alpha^2 + 190 \alpha - 54) + p (-5 \alpha^2 - 11 \alpha + 4) + q (3 \alpha^2 - 4 \alpha + 1) + r \\
&= (-5p + 3q - 63) \alpha^2 + (-11p - 4q + 190) \alpha + (4p + q + r - 54).
\end{align*}We want this to reduce to 0, so we set
\begin{align*}
-5p + 3q &= 63, \\
11p + 4q &= 190, \\
4p + q + r &= 54.
\end{align*}Solving, we find $(p,q,r) = \boxed{(6,31,-1)}.$   For these values, $\alpha^9 + p \alpha^6 + q \alpha^3 + r$ reduces to 0 for any root $\alpha$ of $x^3 - 3x^2 + 4x - 1,$ so $x^9 + px^6 + qx^3 + r$ will be divisible by $x^3 - 3x^2 + 4x - 1.$