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" so maybe they are ordered? But since addition is commutative, it doesn't matter. So a + b is 5.

Now, moving on to 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. Correct. 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 is (-2) + 7 + 0 = 5. So the answer is 5? Wait, but let me check again if there's a possibility that the horizontal asymptote isn't zero. Maybe I need to do a more detailed analysis?

Alternatively, horizontal asymptote can be found by taking the limit as x approaches infinity of f(x). Let me compute that. For large x, f(x) ≈ (2x)/x² = 2/x, which approaches 0 as x approaches infinity. Similarly, as x approaches negative infinity, 2/x also approaches 0. Therefore, horizontal asymptote is indeed y = 0. So c = 0. Therefore, a + b + c = (-2) + 7 + 0 = 5. So the answer is 5?

Wait, but let me check if there's a mistake in factoring the denominator. Let me check the denominator again: x² -5x -14. Let me use the quadratic formula to find roots. The roots are [5 ± sqrt(25 + 56)] / 2 = [5 ± sqrt(81)] / 2 = [5 ± 9]/2. Therefore, (5 + 9)/2 = 14/2 = 7, and (5 - 9)/2 = -4/2 = -2. So yes, the roots are 7 and -2. So that's correct. So vertical asymptotes at x = -2 and x = 7. So a and b are -2 and 7. Therefore, a + b is 5. Then horizontal asymptote is y = 0, so c = 0. Therefore, total is 5 + 0 = 5. So answer is 5?

But wait, the problem says "the graph of f(x) has vertical asymptotes x = a and x = b, and horizontal asymptote y = c. Find a + b + c." So if a and b are the vertical asymptotes, then they are -2 and 7. So a + b is 5. Then c is 0. So total is 5 + 0 = 5. Therefore, the answer is 5. But let me check again if there's something wrong here.

Alternatively, maybe the horizontal asymptote isn't zero? Wait, if the degree of numerator is less than denominator, horizontal asymptote is y = 0. If they are equal, it's the ratio of leading coefficients. If numerator's degree is higher, then there's no horizontal asymptote, but maybe an oblique asymptote. But here, numerator is degree 1, denominator is degree 2, so horizontal asymptote is y = 0. So that's correct.

Alternatively, maybe I need to check if there's a hole? But holes occur when there's a common factor in numerator and denominator. Let me check if numerator and denominator have any common factors. The denominator factors as (x - 7)(x + 2). The numerator is 2x. There are no common factors, so there are no holes. Therefore, vertical asymptotes are indeed at x = -2 and x = 7, and horizontal asymptote at y = 0. Therefore, a + b + c = (-2) + 7 + 0 = 5. Therefore, the answer is 5. So I think that's correct.

But let me check once more. Maybe the problem is expecting something else? Let me check the original function again: f(x) = 2x / (x² -5x -14). Yes. So denominator is x² -5x -14. Factored as (x - 7)(x + 2). Correct. Vertical asymptotes at x = 7 and x = -2. Horizontal asymptote y = 0. Therefore, a + b + c = (-2) + 7 + 0 = 5. So answer is 5. Therefore, I think that's correct.

But just to be thorough, let me check if there's a possibility that the horizontal asymptote is something else. For example, sometimes when people are confused between horizontal and oblique asymptotes. But since the degree of numerator is less than denominator, horizontal asymptote is y = 0. If the degree was equal, then it would be the ratio of leading coefficients. For example, if the function was (2x^2)/(x^2 -5x -14), then horizontal asymptote would be y = 2/1 = 2. But in our case, numerator is degree 1, denominator degree 2, so horizontal asymptote is y = 0. So that's correct.

Alternatively, maybe the problem is expecting me to write a and b as 7 and -2, so a + b is 5, and c is 0, so total is 5. So yeah, that's correct. Therefore, I think the answer is 5.

**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**.

Thus, the vertical asymptotes are at:

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

Let $ a = -2 $, $ b = 7 $

---

### 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" so maybe they are ordered? But since addition is commutative, it doesn't matter. So a + b is 5.

Now, moving on to 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. Correct. 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 is (-2) + 7 + 0 = 5. So the answer is 5? Wait, but let me check again if there's a possibility that the horizontal asymptote isn't zero. Maybe I need to do a more detailed analysis?

Alternatively, horizontal asymptote can be found by taking the limit as x approaches infinity of f(x). Let me compute that. For large x, f(x) ≈ (2x)/x² = 2/x, which approaches 0 as x approaches infinity. Similarly, as x approaches negative infinity, 2/x also approaches 0. Therefore, horizontal asymptote is indeed y = 0. So c = 0. Therefore, a + b + c = (-2) + 7 + 0 = 5. So the answer is 5?

Wait, but let me check if there's a mistake in factoring the denominator. Let me check the denominator again: x² -5x -14. Let me use the quadratic formula to find roots. The roots are [5 ± sqrt(25 + 56)] / 2 = [5 ± sqrt(81)] / 2 = [5 ± 9]/2. Therefore, (5 + 9)/2 = 14/2 = 7, and (5 - 9)/2 = -4/2 = -2. So yes, the roots are 7 and -2. So that's correct. So vertical asymptotes at x = -2 and x = 7. So a and b are -2 and 7. Therefore, a + b is 5. Then horizontal asymptote is y = 0, so c = 0. Therefore, total is 5 + 0 = 5. So answer is 5?

But wait, the problem says "the graph of f(x) has vertical asymptotes x = a and x = b, and horizontal asymptote y = c. Find a + b + c." So if a and b are the vertical asymptotes, then they are -2 and 7. So a + b is 5. Then c is 0. So total is 5 + 0 = 5. Therefore, the answer is 5. But let me check again if there's something wrong here.

Alternatively, maybe the horizontal asymptote isn't zero? Wait, if the degree of numerator is less than denominator, horizontal asymptote is y = 0. If they are equal, it's the ratio of leading coefficients. If numerator's degree is higher, then there's no horizontal asymptote, but maybe an oblique asymptote. But here, numerator is degree 1, denominator is degree 2, so horizontal asymptote is y = 0. So that's correct.

Alternatively, maybe I need to check if there's a hole? But holes occur when there's a common factor in numerator and denominator. Let me check if numerator and denominator have any common factors. The denominator factors as (x - 7)(x + 2). The numerator is 2x. There are no common factors, so there are no holes. Therefore, vertical asymptotes are indeed at x = -2 and x = 7, and horizontal asymptote at y = 0. Therefore, a + b + c = (-2) + 7 + 0 = 5. Therefore, the answer is 5. So I think that's correct.

But let me check once more. Maybe the problem is expecting something else? Let me check the original function again: f(x) = 2x / (x² -5x -14). Yes. So denominator is x² -5x -14. Factored as (x - 7)(x + 2). Correct. Vertical asymptotes at x = 7 and x = -2. Horizontal asymptote y = 0. Therefore, a + b + c = (-2) + 7 + 0 = 5. So answer is 5. Therefore, I think that's correct.

But just to be thorough, let me check if there's a possibility that the horizontal asymptote is something else. For example, sometimes when people are confused between horizontal and oblique asymptotes. But since the degree of numerator is less than denominator, horizontal asymptote is y = 0. If the degree was equal, then it would be the ratio of leading coefficients. For example, if the function was (2x^2)/(x^2 -5x -14), then horizontal asymptote would be y = 2/1 = 2. But in our case, numerator is degree 1, denominator degree 2, so horizontal asymptote is y = 0. So that's correct.

Alternatively, maybe the problem is expecting me to write a and b as 7 and -2, so a + b is 5, and c is 0, so total is 5. So yeah, that's correct. Therefore, I think the answer is 5.

**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**.

Thus, the vertical asymptotes are at:

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

Let $ a = -2 $, $ b = 7 $

---

### 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 of coins. Also, after redistribution, the total number of coins is more than 200. The question is asking for the smallest number of coins I could have had before finding the bag of 53 coins.

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 the number of coins in each original bag "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's call the number of coins in each of the eight bags "y". Therefore, the total number of coins after redistribution is 8y.

Since all coins are redistributed, the total number of coins before and after should be the same. Therefore, 7x + 53 = 8y. Also, we know that after redistribution, the total number of coins is more than 200. So, 8y > 200. Therefore, y > 200 / 8, which is 25. So, y must be at least 26.

But we need to find the smallest number of coins before finding the 53-coin bag, 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 down the equations:

1. 7x + 53 = 8y
2. 8y > 200 => y > 25
3. Find the minimal 7x.

So, since 7x + 53 must be divisible by 8, we can write:

7x + 53 ≡ 0 mod 8

Which implies:

7x ≡ -53 mod 8

But -53 mod 8 is the same as (-53 + 64) mod 8 = 11 mod 8 = 3 mod 8. Wait, let me check that again. Let me compute 53 divided by 8. 8*6=48, so 53-48=5. Therefore, 53 ≡ 5 mod 8. Therefore, -53 ≡ -5 mod 8, which is equivalent to 3 mod 8 because -5 + 8 = 3. Therefore, 7x ≡ 3 mod 8.

So, 7x ≡ 3 mod 8. We need to solve for x here. Let me think about how to solve this congruence.

First, 7x ≡ 3 mod 8. Let me find the multiplicative inverse of 7 mod 8. Since 7 and 8 are coprime, there is an inverse. Let me find a number k such that 7k ≡ 1 mod 8. Testing k=1: 7*1=7≡7 mod8. k=7: 7*7=49≡1 mod8. Yes, because 49 divided by 8 is 6*8=48, remainder 1. Therefore, inverse of 7 mod8 is 7. Therefore, multiplying both sides of 7x ≡3 mod8 by 7:

7*7x ≡7*3 mod8 => 49x ≡21 mod8. But 49 mod8 is 1, so 1*x ≡21 mod8. 21 divided by 8 is 2*8=16, remainder 5. Therefore, x ≡5 mod8. Therefore, x=8k +5 for some integer k≥0.

Therefore, the original number of coins per bag is x=8k +5. Therefore, the total number of coins before finding the 53-coin bag is 7x=7*(8k +5)=56k +35.

Now, we also have that after adding 53 coins, the total is 56k +35 +53=56k +88. This must equal 8y, so 8y=56k +88. Therefore, y=(56k +88)/8=7k +11.

Additionally, we know that 8y >200. So, substituting y=7k +11:

8*(7k +11) >200 => 56k +88 >200 => 56k > 112 => k > 2.

Therefore, k must be greater than 2. Since k is an integer, the smallest possible value of k is 3.

Therefore, substituting k=3 into 7x=56k +35:

7x=56*3 +35=168 +35=203.

Wait, but hold on. Let me check if this satisfies the original condition. Let me check if 7x +53 is divisible by 8.

If x= (56k +35)/7= 8k +5. For k=3, x=8*3 +5=29. Therefore, original total coins is 7*29=203. Then adding 53 gives 203 +53=256. Then 256 divided by 8 is 32. So y=32. Then 8y=256, which is indeed more than 200.

But wait, the problem says "you have more than 200 coins" after redistribution. So 256 is more than 200, so that's okay. But is 203 the minimal possible? Let me check if k=2 would work.

If k=2, then k is not greater than 2, but let's check. If k=2, then 56k +88= 112 +88=200. Therefore, 8y=200, so y=25. But the problem says that after redistribution, you have more than 200 coins. Therefore, 200 is not allowed. Therefore, k must be at least 3. Therefore, the minimal total coins before finding the 53-coin bag is 203.

But wait, let me check if there is a smaller x. Because maybe there is another solution where k is smaller? Wait, we had x ≡5 mod8, so x=5,13,21,29,... So for k=0, x=5, total coins 35. Then adding 53 gives 88, which is 8*11=88. But 8y=88, so y=11. But 8y=88 is not more than 200. Similarly, k=1: x=13, total coins 91. Adding 53 gives 144. 144/8=18. Still way below 200. k=2: x=21, total coins 147. Adding 53 gives 200. Which is equal to 200, but the problem says "more than 200", so that's not allowed. Therefore, k=3 is the first one that gives 256, which is more than 200. Therefore, the minimal total coins before finding the 53-coin bag is 203.

But wait, the question says "the smallest number of coins you could have had before finding the bag of 53 coins". So that would be 7x=203. But let me check if there is another way to get a smaller total.

Wait, maybe there's another solution to the congruence 7x ≡3 mod8? Let me check.

We had 7x ≡3 mod8. Since 7 is congruent to -1 mod8, so -x ≡3 mod8 => x≡-3 mod8 => x≡5 mod8. So that's the only solution. Therefore, x must be congruent to 5 mod8. Therefore, the minimal x is 5, then 13, 21, 29,... So the corresponding total coins before adding 53 is 35, 91, 147, 203,... So as we saw, only when k=3 (i.e., x=29) do we get total coins before adding 53 as 203, which when adding 53 gives 256>200.

But let me check if there's another way. Maybe I made a mistake in the congruence? Let me check again.

Original equation: 7x +53 is divisible by 8. So 7x +53 ≡0 mod8. 53 mod8 is 5, so 7x +5 ≡0 mod8 => 7x ≡-5 mod8 => 7x≡3 mod8. Then solving 7x≡3 mod8. Since 7≡-1 mod8, so -x≡3 mod8 => x≡-3 mod8 => x≡5 mod8. So that's correct.

Therefore, x must be 5 mod8. Therefore, the minimal x is 5, but that gives total coins before adding 53 as 35. Then adding 53 gives 88, which is 8*11. But 8y=88 is 88, which is less than 200. So not acceptable. Then next is x=13, total coins 91. Adding 53 gives 144, which is 8*18. Still less than 200. Next x=21, total coins 147. Adding 53 gives 200, which is equal to 200. But the problem says "more than 200", so that's not acceptable. Then x=29, total coins 203. Adding 53 gives 256, which is 8*32. 256>200. So that works.

Therefore, the minimal total coins before finding the 53-coin bag is 203. Therefore, the answer is 203.

But let me check if there's another way to approach this problem. Maybe by considering the total coins after adding 53 must be divisible by 8 and greater than 200. So total coins after adding is 7x +53. So 7x +53 must be divisible by 8 and greater than 200. So 7x +53 >200 => 7x>147 => x>21. So x must be greater than 21. Also, since x must be such that 7x +53 is divisible by 8.

So, let me think of numbers greater than 21, such that 7x +53 is divisible by 8. Let me try x=21: 7*21 +53=147 +53=200. Which is divisible by 8? 200 divided by 8 is 25. Yes, but 200 is not more than 200. So not acceptable. Next x=22: 7*22 +53=154 +53=207. 207 divided by 8 is 25.875. Not integer. x=23: 7*23=161 +53=214. 214/8=26.75. Not integer. x=24: 7*24=168 +53=221. 221/8=27.625. Not integer. x=25: 7*25=175 +53=228. 228/8=28.5. Not integer. x=26: 7*26=182 +53=235. 235/8=29.375. Not integer. x=27: 7*27=189 +53=242. 242/8=30.25. Not integer. x=28: 7*28=196 +53=249. 249/8=31.125. Not integer. x=29: 7*29=203 +53=256. 256/8=32. Which is integer. So that works. Therefore, x=29 is the first x greater than 21 where 7x +53 is divisible by 8 and the total is 256>200. Therefore, the original total coins is 7x=203.

Therefore, the answer is 203.

But let me check if there is a smaller x than 29 that satisfies the conditions. Wait, we saw that x must be congruent to 5 mod8. So x=5,13,21,29,... So x=21 is the previous one before 29. But x=21 gives total coins 147, which when adding 53 gives 200, which is not more than 200. Therefore, 29 is the next one. Therefore, 203 is the minimal.

Alternatively, maybe there's another way where the total coins after adding 53 is a multiple of 8 greater than 200. Let me think. The total coins after adding is 7x +53. Let me denote T = 7x +53. T must be divisible by 8 and T > 200. So T is the smallest multiple of 8 greater than 200. The smallest multiple of 8 greater than 200 is 200 + (8 - 200 mod8). Let me compute 200 divided by 8 is 25, so 200 is exactly divisible by 8. Therefore, the next multiple is 208. Therefore, T must be at least 208. Therefore, 7x +53 ≥208 => 7x ≥155 => x ≥155/7 ≈22.14. Therefore, x must be at least 23.

But we also have that T must be divisible by 8. So T=208, 216, 224,... Let me check for each T starting from 208:

For T=208: 7x +53=208 => 7x=155 => x=155/7≈22.14. Not integer.

T=216: 7x=216-53=163. 163/7≈23.28. Not integer.

T=224: 7x=224-53=171. 171/7≈24.428. Not integer.

T=232: 7x=232-53=179. 179/7≈25.57. Not integer.

T=240: 7x=240-53=187. 187/7≈26.71. Not integer.

T=248: 7x=248-53=195. 195/7≈27.85. Not integer.

T=256: 7x=256-53=203. 203/7=29. Which is integer. So x=29.

Therefore, the first T that works is 256, which gives x=29. Therefore, the minimal total coins before adding the 53-coin bag is 7x=203.

Therefore, the answer is 203.

But let me check if there is a T between 208 and 256 that I might have missed. For example, T=208, 216,... up to 256. We saw that T=256 is the first one that gives integer x. Therefore, 203 is indeed the minimal.

Alternatively, maybe there's another approach. Since we know that the original total coins is 7x, and after adding 53, it's 8y. Therefore, 7x = 8y -53. Since 7x must be positive, 8y -53 >0 => y>53/8≈6.625. But we also have that 8y>200 => y>25. So y must be at least 26. Therefore, the minimal y is 26. Let's check for y=26: 8*26=208. Then 7x=208 -53=155. 155/7≈22.14. Not integer.

y=27: 8*27=216. 216-53=163. 163/7≈23.28. Not integer.

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

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

y=30: 240-53=187. 187/7≈26.71. Not integer.

y=31: 248-53=195. 195/7≈27.85. Not integer.

y=32: 256-53=203. 203/7=29. Which is integer.

Therefore, y=32 gives x=29. Therefore, the minimal total coins before adding the 53-coin bag is 7x=203.

Therefore, regardless of the approach, the answer seems to be 203.

But let me check once more if there's a smaller x. For example, if there is a T between 200 and 256 that is divisible by 8 and satisfies 7x = T -53 with x integer. But as we saw, the first T that works is 256. Therefore, 203 is the minimal.

Alternatively, maybe there's a mistake in assuming that the original number of coins per bag is x. Let me check the problem statement again.

"You have seven bags of gold coins. Each bag has the same number of gold coins. One day, you find a bag of 53 coins. You decide to redistribute the number of coins you have so that all eight bags you hold have the same number of coins. You successfully manage to redistribute all the coins, and you also note that you have more than 200 coins. What is the smallest number of coins you could have had before finding the bag of 53 coins?"

So, before finding the 53-coin bag, there were seven bags, each with x coins. Then you find a bag of 53 coins, so total coins is 7x +53. Then you redistribute into eight bags with equal number of coins. So the total coins after redistribution is 8y. Therefore, 7x +53 =8y. Also, after redistribution, you have more than 200 coins. So 8y>200. Therefore, y>25.

So, the problem is to find minimal 7x such that 7x +53 is divisible by 8 and 8y=7x +53>200. Which is exactly what we did.

Therefore, I think 203 is correct.

But let me check if there's a case where the original number of coins was less than 203 but somehow the total after adding 53 is more than 200. Wait, for example, if 7x +53>200, then 7x>147, so x>21. So x must be at least 22. But as we saw, for x=22, 7x +53=154 +53=207. Which is 207. Which is more than 200. But 207 is not divisible by 8. 207 divided by 8 is 25.875. So not integer. Therefore, not possible. Similarly, x=23 gives 7*23 +53=161 +53=214. 214/8=26.75. Not integer. x=24: 168 +53=221. 221/8=27.625. Not integer. x=25: 175 +53=228. 228/8=28.5. Not integer. x=26: 182 +53=235. 235/8=29.375. Not integer. x=27: 189 +53=242. 242/8=30.25. Not integer. x=28: 196 +53=249. 249/8=31.125. Not integer. x=29: 203 +53=256. 256/8=32. Integer. So yes, x=29 is the first x where 7x +53 is divisible by 8 and greater than 200. Therefore, 7x=203 is the minimal.

Therefore, I think the answer is 203.

But let me check once again if there's a different interpretation of the problem. For example, maybe the problem says that after redistribution, you have more than 200 coins in total. Which is what we considered. So 8y>200. So y>25. So y=26 and above.

Alternatively, maybe the problem says that after redistribution, each bag has more than 200 coins? But that would be different. But the problem says "you have more than 200 coins", which is total coins. So 8y>200.

Alternatively, maybe the problem is in the redistribution. For example, maybe you have to have each bag with the same number of coins, but you can't have fractions. Which is already considered.

Alternatively, maybe the original number of coins per bag is an integer, which it is.

Therefore, I think 203 is correct.

**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 $ x $ be the number of coins in each of the original 7 bags.
- 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 this total is **more than 200**.

---

### Step 1: Set up the equation

Since all coins are redistributed, the total number of coins remains the same:

$$
7x + 53 = 8y
$$

We are also told that:

$$
8y > 200 \Rightarrow y > 25
$$

So, $ y \geq 26 $.

---

### Step 2: Solve the congruence

We want to find the smallest $ x $ such that $ 7x + 53 $ is divisible by 8. That is:

$$
7x + 53 \equiv 0 \pmod{8}
$$

Simplify:

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

Since $ -53 \equiv 3 \pmod{8} $, we get:

$$
7x \equiv 3 \pmod{8}
$$

Now, we solve this congruence. Since 7 and 8 are coprime, 7 has a multiplicative inverse modulo 8. The inverse of 7 modulo 8 is 7 (since $ 7 \times 7 = 49 \equiv 1 \pmod{8} $).

Multiply both sides of the congruence by 7:

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

So, $ x \equiv 5 \pmod{8} $, meaning $ x = 8k + 5 $ for some integer $ k \geq 0 $.

---

### Step 3: Find the smallest valid $ x $

We now compute the total number of coins before finding the 53-coin bag:

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

We also know that:

$$
7x + 53 = 56k + 35 + 53 = 56k + 88 = 8y
$$

So:

$$
y = \frac{56k + 88}{8} = 7k + 11
$$

We are told that $ 8y > 200 $, so:

$$
8(7k + 11) > 200 \Rightarrow 56k + 88 > 200 \Rightarrow 56k > 112 \Rightarrow k > 2
$$

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

---

### Step 4: Compute the minimal total coins

For $ k = 3 $:

$$
x = 8k + 5 = 8(3) + 5 = 29
$$

$$
\text{Total coins before finding the 53-coin bag} = 7x = 7 \times 29 = 203
$$

Check that this satisfies all conditions:

- $ 7x + 53 = 203 + 53 = 256 $
- $ 256 \div 8 = 32 $, so the coins can be evenly redistributed into 8 bags.
- $ 256 > 200 $, so the total is more than 200.

---

### 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 of coins. Also, after redistribution, the total number of coins is more than 200. The question is asking for the smallest number of coins I could have had before finding the bag of 53 coins.

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 the number of coins in each original bag "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's call the number of coins in each of the eight bags "y". Therefore, the total number of coins after redistribution is 8y.

Since all coins are redistributed, the total number of coins before and after should be the same. Therefore, 7x + 53 = 8y. Also, we know that after redistribution, the total number of coins is more than 200. So, 8y > 200. Therefore, y > 200 / 8, which is 25. So, y must be at least 26.

But we need to find the smallest number of coins before finding the 53-coin bag, 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 down the equations:

1. 7x + 53 = 8y
2. 8y > 200 => y > 25
3. Find the minimal 7x.

So, since 7x + 53 must be divisible by 8, we can write:

7x + 53 ≡ 0 mod 8

Which implies:

7x ≡ -53 mod 8

But -53 mod 8 is the same as (-53 + 64) mod 8 = 11 mod 8 = 3 mod 8. Wait, let me check that again. Let me compute 53 divided by 8. 8*6=48, so 53-48=5. Therefore, 53 ≡ 5 mod 8. Therefore, -53 ≡ -5 mod 8, which is equivalent to 3 mod 8 because -5 + 8 = 3. Therefore, 7x ≡ 3 mod 8.

So, 7x ≡ 3 mod 8. We need to solve for x here. Let me think about how to solve this congruence.

First, 7x ≡ 3 mod 8. Let me find the multiplicative inverse of 7 mod 8. Since 7 and 8 are coprime, there is an inverse. Let me find a number k such that 7k ≡ 1 mod 8. Testing k=1: 7*1=7≡7 mod8. k=7: 7*7=49≡1 mod8. Yes, because 49 divided by 8 is 6*8=48, remainder 1. Therefore, inverse of 7 mod8 is 7. Therefore, multiplying both sides of 7x ≡3 mod8 by 7:

7*7x ≡7*3 mod8 => 49x ≡21 mod8. But 49 mod8 is 1, so 1*x ≡21 mod8. 21 divided by 8 is 2*8=16, remainder 5. Therefore, x ≡5 mod8. Therefore, x=8k +5 for some integer k≥0.

Therefore, the original number of coins per bag is x=8k +5. Therefore, the total number of coins before finding the 53-coin bag is 7x=7*(8k +5)=56k +35.

Now, we also have that after adding 53 coins, the total is 56k +35 +53=56k +88. This must equal 8y, so 8y=56k +88. Therefore, y=(56k +88)/8=7k +11.

Additionally, we know that 8y >200. So, substituting y=7k +11:

8*(7k +11) >200 => 56k +88 >200 => 56k > 112 => k > 2.

Therefore, k must be greater than 2. Since k is an integer, the smallest possible value of k is 3.

Therefore, substituting k=3 into 7x=56k +35:

7x=56*3 +35=168 +35=203.

Wait, but hold on. Let me check if this satisfies the original condition. Let me check if 7x +53 is divisible by 8.

If x= (56k +35)/7= 8k +5. For k=3, x=8*3 +5=29. Therefore, original total coins is 7*29=203. Then adding 53 gives 203 +53=256. Then 256 divided by 8 is 32. So y=32. Then 8y=256, which is indeed more than 200.

But wait, the problem says "you have more than 200 coins" after redistribution. So 256 is more than 200, so that's okay. But is 203 the minimal possible? Let me check if k=2 would work.

If k=2, then k is not greater than 2, but let's check. If k=2, then 56k +88= 112 +88=200. Therefore, 8y=200, so y=25. But the problem says that after redistribution, you have more than 200 coins. Therefore, 200 is not allowed. Therefore, k must be at least 3. Therefore, the minimal total coins before finding the 53-coin bag is 203.

But wait, let me check if there is a smaller x. Because maybe there is another solution where k is smaller? Wait, we had x ≡5 mod8, so x=5,13,21,29,... So for k=0, x=5, total coins 35. Then adding 53 gives 88, which is 8*11=88. But 8y=88, so y=11. But 8y=88 is not more than 200. Similarly, k=1: x=13, total coins 91. Adding 53 gives 144. 144/8=18. Still way below 200. k=2: x=21, total coins 147. Adding 53 gives 200. Which is equal to 200, but the problem says "more than 200", so that's not allowed. Therefore, k=3 is the first one that gives 256, which is more than 200. Therefore, the minimal total coins before finding the 53-coin bag is 203.

But wait, the question says "the smallest number of coins you could have had before finding the bag of 53 coins". So that would be 7x=203. But let me check if there is another way to get a smaller total.

Wait, maybe there's another solution to the congruence 7x ≡3 mod8? Let me check.

We had 7x ≡3 mod8. Since 7 is congruent to -1 mod8, so -x ≡3 mod8 => x≡-3 mod8 => x≡5 mod8. So that's the only solution. Therefore, x must be congruent to 5 mod8. Therefore, the minimal x is 5, then 13, 21, 29,... So the corresponding total coins before adding 53 is 35, 91, 147, 203,... So as we saw, only when k=3 (i.e., x=29) do we get total coins before adding 53 as 203, which when adding 53 gives 256>200.

But let me check if there's another way. Maybe I made a mistake in the congruence? Let me check again.

Original equation: 7x +53 is divisible by 8. So 7x +53 ≡0 mod8. 53 mod8 is 5, so 7x +5 ≡0 mod8 => 7x ≡-5 mod8 => 7x≡3 mod8. Then solving 7x≡3 mod8. Since 7≡-1 mod8, so -x≡3 mod8 => x≡-3 mod8 => x≡5 mod8. So that's correct.

Therefore, x must be 5 mod8. Therefore, the minimal x is 5, but that gives total coins before adding 53 as 35. Then adding 53 gives 88, which is 8*11. But 8y=88 is 88, which is less than 200. So not acceptable. Then next is x=13, total coins 91. Adding 53 gives 144, which is 8*18. Still less than 200. Next x=21, total coins 147. Adding 53 gives 200, which is equal to 200. But the problem says "more than 200", so that's not acceptable. Then x=29, total coins 203. Adding 53 gives 256, which is 8*32. 256>200. So that works.

Therefore, the minimal total coins before finding the 53-coin bag is 203. Therefore, the answer is 203.

But let me check if there's another way to approach this problem. Maybe by considering the total coins after adding 53 must be divisible by 8 and greater than 200. So total coins after adding is 7x +53. So 7x +53 must be divisible by 8 and greater than 200. So 7x +53 >200 => 7x>147 => x>21. So x must be greater than 21. Also, since x must be such that 7x +53 is divisible by 8.

So, let me think of numbers greater than 21, such that 7x +53 is divisible by 8. Let me try x=21: 7*21 +53=147 +53=200. Which is divisible by 8? 200 divided by 8 is 25. Yes, but 200 is not more than 200. So not acceptable. Next x=22: 7*22 +53=154 +53=207. 207 divided by 8 is 25.875. Not integer. x=23: 7*23=161 +53=214. 214/8=26.75. Not integer. x=24: 7*24=168 +53=221. 221/8=27.625. Not integer. x=25: 7*25=175 +53=228. 228/8=28.5. Not integer. x=26: 7*26=182 +53=235. 235/8=29.375. Not integer. x=27: 7*27=189 +53=242. 242/8=30.25. Not integer. x=28: 7*28=196 +53=249. 249/8=31.125. Not integer. x=29: 7*29=203 +53=256. 256/8=32. Which is integer. So that works. Therefore, x=29 is the first x greater than 21 where 7x +53 is divisible by 8 and the total is 256>200. Therefore, the original total coins is 7x=203.

Therefore, the answer is 203.

But let me check if there is a smaller x than 29 that satisfies the conditions. Wait, we saw that x must be congruent to 5 mod8. So x=5,13,21,29,... So x=21 is the previous one before 29. But x=21 gives total coins 147, which when adding 53 gives 200, which is not more than 200. Therefore, 29 is the next one. Therefore, 203 is the minimal.

Alternatively, maybe there's another way where the total coins after adding 53 is a multiple of 8 greater than 200. Let me think. The total coins after adding is 7x +53. Let me denote T = 7x +53. T must be divisible by 8 and T > 200. So T is the smallest multiple of 8 greater than 200. The smallest multiple of 8 greater than 200 is 200 + (8 - 200 mod8). Let me compute 200 divided by 8 is 25, so 200 is exactly divisible by 8. Therefore, the next multiple is 208. Therefore, T must be at least 208. Therefore, 7x +53 ≥208 => 7x ≥155 => x ≥155/7 ≈22.14. Therefore, x must be at least 23.

But we also have that T must be divisible by 8. So T=208, 216, 224,... Let me check for each T starting from 208:

For T=208: 7x +53=208 => 7x=155 => x=155/7≈22.14. Not integer.

T=216: 7x=216-53=163. 163/7≈23.28. Not integer.

T=224: 7x=224-53=171. 171/7≈24.428. Not integer.

T=232: 7x=232-53=179. 179/7≈25.57. Not integer.

T=240: 7x=240-53=187. 187/7≈26.71. Not integer.

T=248: 7x=248-53=195. 195/7≈27.85. Not integer.

T=256: 7x=256-53=203. 203/7=29. Which is integer. So x=29.

Therefore, the first T that works is 256, which gives x=29. Therefore, the minimal total coins before adding the 53-coin bag is 7x=203.

Therefore, the answer is 203.

But let me check if there is a T between 208 and 256 that I might have missed. For example, T=208, 216,... up to 256. We saw that T=256 is the first one that gives integer x. Therefore, 203 is indeed the minimal.

Alternatively, maybe there's another approach. Since we know that the original total coins is 7x, and after adding 53, it's 8y. Therefore, 7x = 8y -53. Since 7x must be positive, 8y -53 >0 => y>53/8≈6.625. But we also have that 8y>200 => y>25. So y must be at least 26. Therefore, the minimal y is 26. Let's check for y=26: 8*26=208. Then 7x=208 -53=155. 155/7≈22.14. Not integer.

y=27: 8*27=216. 216-53=163. 163/7≈23.28. Not integer.

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

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

y=30: 240-53=187. 187/7≈26.71. Not integer.

y=31: 248-53=195. 195/7≈27.85. Not integer.

y=32: 256-53=203. 203/7=29. Which is integer.

Therefore, y=32 gives x=29. Therefore, the minimal total coins before adding the 53-coin bag is 7x=203.

Therefore, regardless of the approach, the answer seems to be 203.

But let me check once more if there's a smaller x. For example, if there is a T between 200 and 256 that is divisible by 8 and satisfies 7x = T -53 with x integer. But as we saw, the first T that works is 256. Therefore, 203 is the minimal.

Alternatively, maybe there's a mistake in assuming that the original number of coins per bag is x. Let me check the problem statement again.

"You have seven bags of gold coins. Each bag has the same number of gold coins. One day, you find a bag of 53 coins. You decide to redistribute the number of coins you have so that all eight bags you hold have the same number of coins. You successfully manage to redistribute all the coins, and you also note that you have more than 200 coins. What is the smallest number of coins you could have had before finding the bag of 53 coins?"

So, before finding the 53-coin bag, there were seven bags, each with x coins. Then you find a bag of 53 coins, so total coins is 7x +53. Then you redistribute into eight bags with equal number of coins. So the total coins after redistribution is 8y. Therefore, 7x +53 =8y. Also, after redistribution, you have more than 200 coins. So 8y>200. Therefore, y>25.

So, the problem is to find minimal 7x such that 7x +53 is divisible by 8 and 8y=7x +53>200. Which is exactly what we did.

Therefore, I think 203 is correct.

But let me check if there's a case where the original number of coins was less than 203 but somehow the total after adding 53 is more than 200. Wait, for example, if 7x +53>200, then 7x>147, so x>21. So x must be at least 22. But as we saw, for x=22, 7x +53=154 +53=207. Which is 207. Which is more than 200. But 207 is not divisible by 8. 207 divided by 8 is 25.875. So not integer. Therefore, not possible. Similarly, x=23 gives 7*23 +53=161 +53=214. 214/8=26.75. Not integer. x=24: 168 +53=221. 221/8=27.625. Not integer. x=25: 175 +53=228. 228/8=28.5. Not integer. x=26: 182 +53=235. 235/8=29.375. Not integer. x=27: 189 +53=242. 242/8=30.25. Not integer. x=28: 196 +53=249. 249/8=31.125. Not integer. x=29: 203 +53=256. 256/8=32. Integer. So yes, x=29 is the first x where 7x +53 is divisible by 8 and greater than 200. Therefore, 7x=203 is the minimal.

Therefore, I think the answer is 203.

But let me check once again if there's a different interpretation of the problem. For example, maybe the problem says that after redistribution, you have more than 200 coins in total. Which is what we considered. So 8y>200. So y>25. So y=26 and above.

Alternatively, maybe the problem says that after redistribution, each bag has more than 200 coins? But that would be different. But the problem says "you have more than 200 coins", which is total coins. So 8y>200.

Alternatively, maybe the problem is in the redistribution. For example, maybe you have to have each bag with the same number of coins, but you can't have fractions. Which is already considered.

Alternatively, maybe the original number of coins per bag is an integer, which it is.

Therefore, I think 203 is correct.

**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 $ x $ be the number of coins in each of the original 7 bags.
- 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 this total is **more than 200**.

---

### Step 1: Set up the equation

Since all coins are redistributed, the total number of coins remains the same:

$$
7x + 53 = 8y
$$

We are also told that:

$$
8y > 200 \Rightarrow y > 25
$$

So, $ y \geq 26 $.

---

### Step 2: Solve the congruence

We want to find the smallest $ x $ such that $ 7x + 53 $ is divisible by 8. That is:

$$
7x + 53 \equiv 0 \pmod{8}
$$

Simplify:

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

Since $ -53 \equiv 3 \pmod{8} $, we get:

$$
7x \equiv 3 \pmod{8}
$$

Now, we solve this congruence. Since 7 and 8 are coprime, 7 has a multiplicative inverse modulo 8. The inverse of 7 modulo 8 is 7 (since $ 7 \times 7 = 49 \equiv 1 \pmod{8} $).

Multiply both sides of the congruence by 7:

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

So, $ x \equiv 5 \pmod{8} $, meaning $ x = 8k + 5 $ for some integer $ k \geq 0 $.

---

### Step 3: Find the smallest valid $ x $

We now compute the total number of coins before finding the 53-coin bag:

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

We also know that:

$$
7x + 53 = 56k + 35 + 53 = 56k + 88 = 8y
$$

So:

$$
y = \frac{56k + 88}{8} = 7k + 11
$$

We are told that $ 8y > 200 $, so:

$$
8(7k + 11) > 200 \Rightarrow 56k + 88 > 200 \Rightarrow 56k > 112 \Rightarrow k > 2
$$

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

---

### Step 4: Compute the minimal total coins

For $ k = 3 $:

$$
x = 8k + 5 = 8(3) + 5 = 29
$$

$$
\text{Total coins before finding the 53-coin bag} = 7x = 7 \times 29 = 203
$$

Check that this satisfies all conditions:

- $ 7x + 53 = 203 + 53 = 256 $
- $ 256 \div 8 = 32 $, so the coins can be evenly redistributed into 8 bags.
- $ 256 > 200 $, so the total is more than 200.

---

### 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 to simplify the expression under the square root. Let me try that.

So, 2.5 squared minus 0.7 squared would be (2.5 - 0.7)(2.5 + 0.7). Let me calculate those values.

First, 2.5 minus 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 that's 1.8.

Then, 2.5 plus 0.7. Let me add those. 2.5 + 0.7. 2.5 + 0.5 is 3.0, plus another 0.2 is 3.2. So that's 3.2.

So, the numerator becomes √(1.8 * 3.2). Let me compute 1.8 multiplied by 3.2. Hmm, 1.8 times 3.2. Let me think. Maybe I can do this multiplication step by step.

First, 1 times 3.2 is 3.2. Then 0.8 times 3.2. Let me compute 0.8 * 3.2. Well, 0.8 * 3 is 2.4, and 0.8 * 0.2 is 0.16. Adding those together gives 2.4 + 0.16 = 2.56. Then adding that to the previous 3.2 gives 3.2 + 2.56 = 5.76.

So, the numerator simplifies to √5.76. Now, what's the square root of 5.76? Let me think. I know that 2.4 squared is 5.76 because 2.4 * 2.4. Let me check that. 2 * 2 is 4, 2 * 0.4 is 0.8, 0.4 * 2 is another 0.8, and 0.4 * 0.4 is 0.16. Adding those up: 4 + 0.8 + 0.8 + 0.16 = 4 + 1.6 + 0.16 = 5.76. Yes, that's correct. So √5.76 is 2.4.

Okay, so the numerator simplifies to 2.4. Now, let me look at the denominator: 2.7 minus 2.5. That's straightforward. 2.7 - 2.5. Let me subtract those. 2.7 minus 2.5 is 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, right? Because 1 divided by 0.2 is 5. So 2.4 * 5. Let me calculate that.

2 * 5 is 10, and 0.4 * 5 is 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.

Starting with the numerator: √(2.5² - 0.7²). Factored as (2.5 - 0.7)(2.5 + 0.7) = 1.8 * 3.2 = 5.76. Square root of 5.76 is 2.4. Then denominator is 2.7 - 2.5 = 0.2. Then 2.4 / 0.2 = 12. That seems correct.

But let me verify the multiplication of 1.8 * 3.2 again. Maybe I can do it another way. Let me convert them to fractions to see if that helps.

1.8 is the same as 18/10, and 3.2 is 32/10. Multiplying them together: (18/10)*(32/10) = (18*32)/100. Let me compute 18*32.

Well, 10*32 = 320, 8*32 = 256. So 320 + 256 = 576. Therefore, 576/100 = 5.76. Yep, that's the same as before. So square root of 5.76 is indeed 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 multiply numerator and denominator by 10 to eliminate the decimal, that becomes 24 divided by 2, which is 12. Yep, that's correct.

Alternatively, 0.2 goes into 2.4 how many times? Since 0.2 * 10 = 2.0, so 0.2 * 12 = 2.4. So that's 12.

So all steps check out. Therefore, the simplified value of the expression is 12.

But just to be thorough, let me compute the original expression without factoring, just to see if I get the same result.

Original numerator: √(2.5² - 0.7²). Let me compute 2.5 squared first. 2.5 squared is 6.25. Then 0.7 squared is 0.49. Subtracting those: 6.25 - 0.49. Let me do that subtraction.

6.25 minus 0.49. 6.25 minus 0.4 is 5.85, then minus 0.09 is 5.76. So that's the same as before. Then square root of 5.76 is 2.4.

Denominator: 2.7 - 2.5 = 0.2. So 2.4 divided by 0.2 is 12. Yep, same result.

Therefore, regardless of the method, I get 12.

Wait, but just to make sure there isn't a trick or something I missed. The problem says "simplify," so maybe they want it in fraction form or something? But 12 is already simplified. Let me check if there's another way to represent it.

Alternatively, maybe they want it as a fraction? Let me see. 12 is an integer, so that's as simple as it gets.

Alternatively, maybe I made a mistake in assuming that √(a² - b²) is equal to √[(a - b)(a + b)]? But that's just the difference of squares formula. So that's correct.

Alternatively, maybe there's a different approach? Let me think. For example, if I didn't factor the numerator, but instead calculated each term.

But I think that's what I did. So, I think my answer is correct.

Alternatively, maybe the problem is designed to have some cancellation or something? Let me check.

Wait, 2.5, 0.7, 2.7... Maybe there's a relation between these numbers? Let me see.

2.7 is 2.5 + 0.2. Wait, the denominator is 2.7 - 2.5 = 0.2. So, maybe there's a connection?

But I think the way I did it is correct.

Alternatively, maybe I can write all numbers as fractions to avoid decimals? Let me try that.

Let me convert all decimals to fractions.

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

So, the expression becomes √[( (5/2)^2 - (7/10)^2 ) / (27/10 - 5/2)].

Let me compute numerator and denominator step by step.

First, numerator inside the square root: (5/2)^2 - (7/10)^2.

Calculating (5/2)^2: 25/4.

Calculating (7/10)^2: 49/100.

So, 25/4 - 49/100. To subtract these, 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). Wait, 25/4 is equal to (25*25)/ (4*25)? Wait, no. Wait, to convert 25/4 to denominator 100, multiply numerator and denominator by 25? Wait, 4*25 is 100. So yes.

So 25/4 = (25*25)/100? Wait, no. Wait, 25/4 multiplied by 25/25 is (25*25)/100. Wait, but 25/4 is equal to (25*25)/ (4*25) = 625/100. Wait, but 25/4 is 6.25, and 625/100 is 6.25. Yes, that's correct.

So 25/4 - 49/100 = 625/100 - 49/100 = (625 - 49)/100 = 576/100.

Therefore, the numerator inside the square root is 576/100. Taking the square root of that is √(576/100) = √576 / √100 = 24/10 = 12/5.

Wait, hold on. Wait, earlier I had √5.76 = 2.4, which is 12/5. Because 12 divided by 5 is 2.4. So that's consistent.

Now, the denominator of the entire expression is 27/10 - 5/2. Let me compute that.

Convert 5/2 to tenths: 5/2 = 25/10. Therefore, 27/10 - 25/10 = 2/10 = 1/5.

So the entire expression is (12/5) divided by (1/5). Dividing by 1/5 is the same as multiplying by 5. So (12/5) * 5 = 12.

So that's another way to do it, using fractions, and still get 12.

Therefore, regardless of the method, decimal or fraction, I get 12.

Therefore, I can be confident that the simplified value is 12.

But just to recap all steps once again:

Original expression: √(2.5² - 0.7²) / (2.7 - 2.5)

Compute numerator:

2.5² = 6.25

0.7² = 0.49

Subtract: 6.25 - 0.49 = 5.76

Square root of 5.76 = 2.4

Denominator:

2.7 - 2.5 = 0.2

Divide numerator by denominator: 2.4 / 0.2 = 12

Alternatively, using fractions:

Convert decimals to fractions:

2.5 = 5/2, 0.7 = 7/10, 2.7 = 27/10

Numerator inside sqrt: (5/2)^2 - (7/10)^2 = 25/4 - 49/100 = 625/100 - 49/100 = 576/100

sqrt(576/100) = 24/10 = 12/5

Denominator: 27/10 - 5/2 = 27/10 - 25/10 = 2/10 = 1/5

Divide 12/5 by 1/5 = 12/5 * 5/1 = 12

So both methods confirm the answer is 12.

Therefore, I think that's the correct simplified form.

I don't see any mistakes in my calculations, so I think the answer 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 values inside the parentheses:

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

So the expression 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: Divide the Results**

Now divide the simplified numerator by the denominator:

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

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 to simplify the expression under the square root. Let me try that.

So, 2.5 squared minus 0.7 squared would be (2.5 - 0.7)(2.5 + 0.7). Let me calculate those values.

First, 2.5 minus 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 that's 1.8.

Then, 2.5 plus 0.7. Let me add those. 2.5 + 0.7. 2.5 + 0.5 is 3.0, plus another 0.2 is 3.2. So that's 3.2.

So, the numerator becomes √(1.8 * 3.2). Let me compute 1.8 multiplied by 3.2. Hmm, 1.8 times 3.2. Let me think. Maybe I can do this multiplication step by step.

First, 1 times 3.2 is 3.2. Then 0.8 times 3.2. Let me compute 0.8 * 3.2. Well, 0.8 * 3 is 2.4, and 0.8 * 0.2 is 0.16. Adding those together gives 2.4 + 0.16 = 2.56. Then adding that to the previous 3.2 gives 3.2 + 2.56 = 5.76.

So, the numerator simplifies to √5.76. Now, what's the square root of 5.76? Let me think. I know that 2.4 squared is 5.76 because 2.4 * 2.4. Let me check that. 2 * 2 is 4, 2 * 0.4 is 0.8, 0.4 * 2 is another 0.8, and 0.4 * 0.4 is 0.16. Adding those up: 4 + 0.8 + 0.8 + 0.16 = 4 + 1.6 + 0.16 = 5.76. Yes, that's correct. So √5.76 is 2.4.

Okay, so the numerator simplifies to 2.4. Now, let me look at the denominator: 2.7 minus 2.5. That's straightforward. 2.7 - 2.5. Let me subtract those. 2.7 minus 2.5 is 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, right? Because 1 divided by 0.2 is 5. So 2.4 * 5. Let me calculate that.

2 * 5 is 10, and 0.4 * 5 is 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.

Starting with the numerator: √(2.5² - 0.7²). Factored as (2.5 - 0.7)(2.5 + 0.7) = 1.8 * 3.2 = 5.76. Square root of 5.76 is 2.4. Then denominator is 2.7 - 2.5 = 0.2. Then 2.4 / 0.2 = 12. That seems correct.

But let me verify the multiplication of 1.8 * 3.2 again. Maybe I can do it another way. Let me convert them to fractions to see if that helps.

1.8 is the same as 18/10, and 3.2 is 32/10. Multiplying them together: (18/10)*(32/10) = (18*32)/100. Let me compute 18*32.

Well, 10*32 = 320, 8*32 = 256. So 320 + 256 = 576. Therefore, 576/100 = 5.76. Yep, that's the same as before. So square root of 5.76 is indeed 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 multiply numerator and denominator by 10 to eliminate the decimal, that becomes 24 divided by 2, which is 12. Yep, that's correct.

Alternatively, 0.2 goes into 2.4 how many times? Since 0.2 * 10 = 2.0, so 0.2 * 12 = 2.4. So that's 12.

So all steps check out. Therefore, the simplified value of the expression is 12.

But just to be thorough, let me compute the original expression without factoring, just to see if I get the same result.

Original numerator: √(2.5² - 0.7²). Let me compute 2.5 squared first. 2.5 squared is 6.25. Then 0.7 squared is 0.49. Subtracting those: 6.25 - 0.49. Let me do that subtraction.

6.25 minus 0.49. 6.25 minus 0.4 is 5.85, then minus 0.09 is 5.76. So that's the same as before. Then square root of 5.76 is 2.4.

Denominator: 2.7 - 2.5 = 0.2. So 2.4 divided by 0.2 is 12. Yep, same result.

Therefore, regardless of the method, I get 12.

Wait, but just to make sure there isn't a trick or something I missed. The problem says "simplify," so maybe they want it in fraction form or something? But 12 is already simplified. Let me check if there's another way to represent it.

Alternatively, maybe they want it as a fraction? Let me see. 12 is an integer, so that's as simple as it gets.

Alternatively, maybe I made a mistake in assuming that √(a² - b²) is equal to √[(a - b)(a + b)]? But that's just the difference of squares formula. So that's correct.

Alternatively, maybe there's a different approach? Let me think. For example, if I didn't factor the numerator, but instead calculated each term.

But I think that's what I did. So, I think my answer is correct.

Alternatively, maybe the problem is designed to have some cancellation or something? Let me check.

Wait, 2.5, 0.7, 2.7... Maybe there's a relation between these numbers? Let me see.

2.7 is 2.5 + 0.2. Wait, the denominator is 2.7 - 2.5 = 0.2. So, maybe there's a connection?

But I think the way I did it is correct.

Alternatively, maybe I can write all numbers as fractions to avoid decimals? Let me try that.

Let me convert all decimals to fractions.

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

So, the expression becomes √[( (5/2)^2 - (7/10)^2 ) / (27/10 - 5/2)].

Let me compute numerator and denominator step by step.

First, numerator inside the square root: (5/2)^2 - (7/10)^2.

Calculating (5/2)^2: 25/4.

Calculating (7/10)^2: 49/100.

So, 25/4 - 49/100. To subtract these, 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). Wait, 25/4 is equal to (25*25)/ (4*25)? Wait, no. Wait, to convert 25/4 to denominator 100, multiply numerator and denominator by 25? Wait, 4*25 is 100. So yes.

So 25/4 = (25*25)/100? Wait, no. Wait, 25/4 multiplied by 25/25 is (25*25)/100. Wait, but 25/4 is equal to (25*25)/ (4*25) = 625/100. Wait, but 25/4 is 6.25, and 625/100 is 6.25. Yes, that's correct.

So 25/4 - 49/100 = 625/100 - 49/100 = (625 - 49)/100 = 576/100.

Therefore, the numerator inside the square root is 576/100. Taking the square root of that is √(576/100) = √576 / √100 = 24/10 = 12/5.

Wait, hold on. Wait, earlier I had √5.76 = 2.4, which is 12/5. Because 12 divided by 5 is 2.4. So that's consistent.

Now, the denominator of the entire expression is 27/10 - 5/2. Let me compute that.

Convert 5/2 to tenths: 5/2 = 25/10. Therefore, 27/10 - 25/10 = 2/10 = 1/5.

So the entire expression is (12/5) divided by (1/5). Dividing by 1/5 is the same as multiplying by 5. So (12/5) * 5 = 12.

So that's another way to do it, using fractions, and still get 12.

Therefore, regardless of the method, decimal or fraction, I get 12.

Therefore, I can be confident that the simplified value is 12.

But just to recap all steps once again:

Original expression: √(2.5² - 0.7²) / (2.7 - 2.5)

Compute numerator:

2.5² = 6.25

0.7² = 0.49

Subtract: 6.25 - 0.49 = 5.76

Square root of 5.76 = 2.4

Denominator:

2.7 - 2.5 = 0.2

Divide numerator by denominator: 2.4 / 0.2 = 12

Alternatively, using fractions:

Convert decimals to fractions:

2.5 = 5/2, 0.7 = 7/10, 2.7 = 27/10

Numerator inside sqrt: (5/2)^2 - (7/10)^2 = 25/4 - 49/100 = 625/100 - 49/100 = 576/100

sqrt(576/100) = 24/10 = 12/5

Denominator: 27/10 - 5/2 = 27/10 - 25/10 = 2/10 = 1/5

Divide 12/5 by 1/5 = 12/5 * 5/1 = 12

So both methods confirm the answer is 12.

Therefore, I think that's the correct simplified form.

I don't see any mistakes in my calculations, so I think the answer 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 values inside the parentheses:

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

So the expression 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: Divide the Results**

Now divide the simplified numerator by the denominator:

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

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 $x^3 - 3x^2 + 4x - 1.$