Explanation:
Okay, so I need to find the probability that the product of two numbers rolled on two fair 6-sided dice is a multiple of 5. Hmm, let me think about how to approach this.

First, I know that when dealing with probabilities involving dice, it's often helpful to consider the total number of possible outcomes and then figure out how many of those outcomes satisfy the condition. Since each die has 6 sides, the total number of outcomes when throwing two dice is 6 multiplied by 6, which is 36. So, there are 36 possible outcomes in total. Got that down.

Now, the question is about the product being a multiple of 5. Let me recall that a number is a multiple of 5 if it's divisible by 5. So, the product of the two numbers must be divisible by 5. To figure out how many such outcomes there are, maybe I can think about the conditions under which the product is divisible by 5.

I remember that for a product to be divisible by 5, at least one of the numbers in the product must be divisible by 5. Wait, is that right? Let me check. If either of the numbers is a multiple of 5, then their product will be a multiple of 5. Conversely, if neither number is a multiple of 5, then their product won't be a multiple of 5. So, yes, that seems correct. So, the product is a multiple of 5 if and only if at least one of the two numbers is a multiple of 5.

Therefore, instead of calculating the number of favorable outcomes directly, maybe it's easier to calculate the probability of the complementary event (i.e., the product is not a multiple of 5) and then subtract that from 1. That might be simpler because sometimes calculating the complement is easier.

So, let me try that approach. The complementary event is that neither of the two numbers is a multiple of 5. So, first, I need to find the probability that neither die shows a multiple of 5, and then subtract that probability from 1 to get the desired probability.

First, let me figure out how many numbers on a die are multiples of 5. Well, a standard die has numbers 1 through 6. The multiples of 5 in that range are just 5 itself. So, only one number (which is 5) is a multiple of 5. Therefore, the numbers that are not multiples of 5 are 1, 2, 3, 4, 6. That's 5 numbers.

Therefore, the probability that a single die does not show a multiple of 5 is 5/6. Since the two dice are independent, the probability that neither die shows a multiple of 5 is (5/6) * (5/6) = 25/36.

Therefore, the probability that the product is a multiple of 5 is 1 - 25/36 = 11/36. Wait, is that correct? Let me check again.

Alternatively, maybe I can calculate it directly. Let me try that method as well to verify.

So, the number of favorable outcomes where at least one die shows a 5. How do I calculate that? Well, using the principle of inclusion-exclusion. The number of outcomes where the first die is 5 plus the number of outcomes where the second die is 5 minus the number of outcomes where both dice are 5.

So, the number of outcomes where the first die is 5: there are 6 possibilities for the second die, so 6 outcomes. Similarly, the number of outcomes where the second die is 5 is also 6. However, the overlap where both dice are 5 is just 1 outcome (5 and 5). Therefore, the total number of favorable outcomes is 6 + 6 - 1 = 11. Therefore, the probability is 11/36.

So, both methods give me the same answer, 11/36. That seems consistent. Therefore, the probability is 11/36.

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

Another way to approach this is to list all possible outcomes where the product is a multiple of 5. Since the product is a multiple of 5 if at least one die is 5. So, for each die, if one is 5, the product will be a multiple of 5.

So, how many such outcomes are there? Let's think.

First, the number of outcomes where the first die is 5: as before, 6 outcomes (since the second die can be 1-6). Similarly, the number of outcomes where the second die is 5 is 6. However, the outcome where both are 5 is counted twice here, so we subtract 1. So, 6 + 6 - 1 = 11. So, 11 outcomes.

Total outcomes are 36, so 11/36.

Alternatively, if I list them out, maybe? Let's try.

First die is 5: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6). That's 6 outcomes.

Second die is 5: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5). That's another 6 outcomes. But notice that (5,5) is in both lists, so total unique outcomes are 6 + 6 - 1 = 11.

So, the 11 outcomes are:

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5).

Yes, that's 11 outcomes. So, 11/36.

Therefore, the probability is 11/36.

But let me check if there's another way someone might approach this problem and maybe get a different answer. For example, someone might think that the product is a multiple of 5 if both numbers are multiples of 5, but that's not correct. Wait, no. If both numbers are multiples of 5, then the product is a multiple of 25, which is also a multiple of 5. But the condition is just that the product is a multiple of 5, so even if one number is a multiple of 5, regardless of the other number, the product is a multiple of 5.

So, the initial reasoning is correct.

Alternatively, someone might think that the product is a multiple of 5 if the numbers are 5 and something else, but that's not the case. For example, if one die is 5 and the other is 2, the product is 10, which is a multiple of 5. If one die is 5 and the other is 3, product is 15, which is a multiple of 5. So, yes, that's correct.

Alternatively, if someone thought that the product needs to be divisible by 5 and maybe other numbers, but the question is only about being a multiple of 5, so that's not necessary.

Another way to think about it: the product is divisible by 5 if and only if at least one of the numbers is divisible by 5. Since 5 is prime, so the prime factors of the product must include 5. Therefore, at least one of the numbers must have 5 as a factor.

Therefore, the answer is 11/36.

But let me check if there's a mistake in the complementary probability approach.

The complementary event is that neither die is a multiple of 5. So, each die has 5 possibilities (1,2,3,4,6). Therefore, the number of such outcomes is 5*5=25. Therefore, the number of favorable outcomes is 36-25=11. Therefore, probability is 11/36.

Yes, that's the same answer.

Alternatively, if I think about the probability that the first die is not 5 is 5/6, and the second die is not 5 is 5/6, so the probability that both are not 5 is 25/36, so the probability that at least one is 5 is 1 - 25/36 = 11/36.

Therefore, all methods lead to the same answer.

But just to be thorough, let me think if there's any case where even if neither die is a multiple of 5, the product could be a multiple of 5. For example, if one die is 5 and the other is something else, but if neither die is 5, then the product can't be a multiple of 5.

Wait, suppose one die is 5, then the product is a multiple of 5. If neither die is 5, then both numbers are from 1,2,3,4,6. So, their product would be... Let me check. For example, 2 and 5, but 5 isn't allowed here. So, numbers are 1,2,3,4,6. Let me take two numbers from that set. Let me check if their product can be a multiple of 5.

Well, 5 is prime. So, unless one of the numbers is a multiple of 5, the product can't be a multiple of 5. Since neither number is a multiple of 5, their product can't be a multiple of 5. Therefore, the complementary event is indeed correct.

Therefore, the answer is 11/36.

But just to make sure, let me think of another approach. Let me think about the possible products.

But that might be more complicated. Let me try.

The possible products when rolling two dice range from 1*1=1 up to 6*6=36. But enumerating all products and checking which are multiples of 5 would be tedious, but maybe manageable.

But since there are 36 outcomes, maybe I can count how many products are multiples of 5.

But since we already have two methods giving 11/36, maybe this is redundant, but let me try.

First, note that a product is a multiple of 5 if and only if at least one of the numbers is 5. Therefore, the number of such products is 11, as we found before.

Alternatively, if I think of all products that are multiples of 5. Let me list them.

First, when one die is 5:

- 5*1=5
- 5*2=10
- 5*3=15
- 5*4=20
- 5*5=25
- 5*6=30

Similarly, when the other die is 5:

- 1*5=5
- 2*5=10
- 3*5=15
- 4*5=20
- 5*5=25
- 6*5=30

But note that 5*1 and 1*5 both result in 5, which is the same product, but they are different outcomes. However, when counting the number of outcomes, we have to consider that (5,1) and (1,5) are separate.

But in terms of products, the products are 5,10,15,20,25,30. However, each of these products can be achieved in different numbers of ways. But since we are counting the number of outcomes, not the number of distinct products, we need to count all the outcomes where the product is a multiple of 5.

But as we saw earlier, there are 11 such outcomes.

Alternatively, if I think of the products:

- 5: occurs in (1,5) and (5,1)
- 10: occurs in (2,5), (5,2)
- 15: occurs in (3,5), (5,3)
- 20: occurs in (4,5), (5,4)
- 25: occurs in (5,5)
- 30: occurs in (5,6), (6,5)

So, for each product:

- 5: 2 outcomes
- 10: 2 outcomes
- 15: 2 outcomes
- 20: 2 outcomes
- 25: 1 outcome
- 30: 2 outcomes

Adding these up: 2+2+2+2+1+2 = 11.

Yes, that's 11 outcomes. Therefore, confirming again that there are 11 favorable outcomes.

Therefore, the probability is 11/36.

So, after checking multiple approaches—complementary probability, direct counting, enumerating the products—it seems that 11/36 is indeed the correct answer.

I think that's thorough enough. I don't see any mistakes in my reasoning. All methods converge to the same answer, so I feel confident that 11/36 is the correct probability.

**Final Answer**
The probability is \boxed{\dfrac{11}{36}}.

To determine the probability that the product of the numbers rolled on two fair 6-sided dice is a multiple of 5, we begin by analyzing the total number of possible outcomes and the conditions under which the product is divisible by 5.

---

### Step 1: Total Number of Outcomes

Each die has 6 faces, so when rolling two dice, the total number of possible outcomes is:

$$
6 \times 6 = 36
$$

---

### Step 2: Understanding the Condition

A number is a multiple of 5 if it is divisible by 5. For the product of two numbers to be divisible by 5, **at least one of the numbers must be divisible by 5**. This is because 5 is a prime number, and the only way for the product to be divisible by 5 is if one of the factors is divisible by 5.

On a standard die, the only number divisible by 5 is **5**. So, the favorable outcomes are those where **at least one die shows a 5**.

---

### Step 3: Complementary Probability Approach

Instead of directly counting the favorable outcomes, we can use the **complement rule**. The complement of the event "at least one die shows a 5" is "neither die shows a 5".

- The numbers **not divisible by 5** on a die are: 1, 2, 3, 4, 6 → 5 numbers.
- So, the probability that a single die does **not** show a 5 is:

$$
\frac{5}{6}
$$

- The probability that **both** dice do not show a 5 is:

$$
\left(\frac{5}{6}\right)^2 = \frac{25}{36}
$$

- Therefore, the probability that **at least one die shows a 5** (i.e., the product is a multiple of 5) is:

$$
1 - \frac{25}{36} = \frac{11}{36}
$$

---

### Step 4: Direct Counting (Verification)

We can also count the favorable outcomes directly:

- The number of outcomes where the **first die is 5** is 6 (since the second die can be any of 1 through 6).
- The number of outcomes where the **second die is 5** is also 6.
- However, the outcome (5,5) is counted twice, so we subtract 1 to avoid double-counting.

$$
\text{Favorable outcomes} = 6 + 6 - 1 = 11
$$

So, the probability is:

$$
\frac{11}{36}
$$

---

### Final Answer

$$
\boxed{\dfrac{11}{36}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that the product of two numbers rolled on two fair 6-sided dice is a multiple of 5. Hmm, let me think about how to approach this.

First, I know that when dealing with probabilities involving dice, it's often helpful to consider the total number of possible outcomes and then figure out how many of those outcomes satisfy the condition. Since each die has 6 sides, the total number of outcomes when throwing two dice is 6 multiplied by 6, which is 36. So, there are 36 possible outcomes in total. Got that down.

Now, the question is about the product being a multiple of 5. Let me recall that a number is a multiple of 5 if it's divisible by 5. So, the product of the two numbers must be divisible by 5. To figure out how many such outcomes there are, maybe I can think about the conditions under which the product is divisible by 5.

I remember that for a product to be divisible by 5, at least one of the numbers in the product must be divisible by 5. Wait, is that right? Let me check. If either of the numbers is a multiple of 5, then their product will be a multiple of 5. Conversely, if neither number is a multiple of 5, then their product won't be a multiple of 5. So, yes, that seems correct. So, the product is a multiple of 5 if and only if at least one of the two numbers is a multiple of 5.

Therefore, instead of calculating the number of favorable outcomes directly, maybe it's easier to calculate the probability of the complementary event (i.e., the product is not a multiple of 5) and then subtract that from 1. That might be simpler because sometimes calculating the complement is easier.

So, let me try that approach. The complementary event is that neither of the two numbers is a multiple of 5. So, first, I need to find the probability that neither die shows a multiple of 5, and then subtract that probability from 1 to get the desired probability.

First, let me figure out how many numbers on a die are multiples of 5. Well, a standard die has numbers 1 through 6. The multiples of 5 in that range are just 5 itself. So, only one number (which is 5) is a multiple of 5. Therefore, the numbers that are not multiples of 5 are 1, 2, 3, 4, 6. That's 5 numbers.

Therefore, the probability that a single die does not show a multiple of 5 is 5/6. Since the two dice are independent, the probability that neither die shows a multiple of 5 is (5/6) * (5/6) = 25/36.

Therefore, the probability that the product is a multiple of 5 is 1 - 25/36 = 11/36. Wait, is that correct? Let me check again.

Alternatively, maybe I can calculate it directly. Let me try that method as well to verify.

So, the number of favorable outcomes where at least one die shows a 5. How do I calculate that? Well, using the principle of inclusion-exclusion. The number of outcomes where the first die is 5 plus the number of outcomes where the second die is 5 minus the number of outcomes where both dice are 5.

So, the number of outcomes where the first die is 5: there are 6 possibilities for the second die, so 6 outcomes. Similarly, the number of outcomes where the second die is 5 is also 6. However, the overlap where both dice are 5 is just 1 outcome (5 and 5). Therefore, the total number of favorable outcomes is 6 + 6 - 1 = 11. Therefore, the probability is 11/36.

So, both methods give me the same answer, 11/36. That seems consistent. Therefore, the probability is 11/36.

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

Another way to approach this is to list all possible outcomes where the product is a multiple of 5. Since the product is a multiple of 5 if at least one die is 5. So, for each die, if one is 5, the product will be a multiple of 5.

So, how many such outcomes are there? Let's think.

First, the number of outcomes where the first die is 5: as before, 6 outcomes (since the second die can be 1-6). Similarly, the number of outcomes where the second die is 5 is 6. However, the outcome where both are 5 is counted twice here, so we subtract 1. So, 6 + 6 - 1 = 11. So, 11 outcomes.

Total outcomes are 36, so 11/36.

Alternatively, if I list them out, maybe? Let's try.

First die is 5: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6). That's 6 outcomes.

Second die is 5: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5). That's another 6 outcomes. But notice that (5,5) is in both lists, so total unique outcomes are 6 + 6 - 1 = 11.

So, the 11 outcomes are:

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5).

Yes, that's 11 outcomes. So, 11/36.

Therefore, the probability is 11/36.

But let me check if there's another way someone might approach this problem and maybe get a different answer. For example, someone might think that the product is a multiple of 5 if both numbers are multiples of 5, but that's not correct. Wait, no. If both numbers are multiples of 5, then the product is a multiple of 25, which is also a multiple of 5. But the condition is just that the product is a multiple of 5, so even if one number is a multiple of 5, regardless of the other number, the product is a multiple of 5.

So, the initial reasoning is correct.

Alternatively, someone might think that the product is a multiple of 5 if the numbers are 5 and something else, but that's not the case. For example, if one die is 5 and the other is 2, the product is 10, which is a multiple of 5. If one die is 5 and the other is 3, product is 15, which is a multiple of 5. So, yes, that's correct.

Alternatively, if someone thought that the product needs to be divisible by 5 and maybe other numbers, but the question is only about being a multiple of 5, so that's not necessary.

Another way to think about it: the product is divisible by 5 if and only if at least one of the numbers is divisible by 5. Since 5 is prime, so the prime factors of the product must include 5. Therefore, at least one of the numbers must have 5 as a factor.

Therefore, the answer is 11/36.

But let me check if there's a mistake in the complementary probability approach.

The complementary event is that neither die is a multiple of 5. So, each die has 5 possibilities (1,2,3,4,6). Therefore, the number of such outcomes is 5*5=25. Therefore, the number of favorable outcomes is 36-25=11. Therefore, probability is 11/36.

Yes, that's the same answer.

Alternatively, if I think about the probability that the first die is not 5 is 5/6, and the second die is not 5 is 5/6, so the probability that both are not 5 is 25/36, so the probability that at least one is 5 is 1 - 25/36 = 11/36.

Therefore, all methods lead to the same answer.

But just to be thorough, let me think if there's any case where even if neither die is a multiple of 5, the product could be a multiple of 5. For example, if one die is 5 and the other is something else, but if neither die is 5, then the product can't be a multiple of 5.

Wait, suppose one die is 5, then the product is a multiple of 5. If neither die is 5, then both numbers are from 1,2,3,4,6. So, their product would be... Let me check. For example, 2 and 5, but 5 isn't allowed here. So, numbers are 1,2,3,4,6. Let me take two numbers from that set. Let me check if their product can be a multiple of 5.

Well, 5 is prime. So, unless one of the numbers is a multiple of 5, the product can't be a multiple of 5. Since neither number is a multiple of 5, their product can't be a multiple of 5. Therefore, the complementary event is indeed correct.

Therefore, the answer is 11/36.

But just to make sure, let me think of another approach. Let me think about the possible products.

But that might be more complicated. Let me try.

The possible products when rolling two dice range from 1*1=1 up to 6*6=36. But enumerating all products and checking which are multiples of 5 would be tedious, but maybe manageable.

But since there are 36 outcomes, maybe I can count how many products are multiples of 5.

But since we already have two methods giving 11/36, maybe this is redundant, but let me try.

First, note that a product is a multiple of 5 if and only if at least one of the numbers is 5. Therefore, the number of such products is 11, as we found before.

Alternatively, if I think of all products that are multiples of 5. Let me list them.

First, when one die is 5:

- 5*1=5
- 5*2=10
- 5*3=15
- 5*4=20
- 5*5=25
- 5*6=30

Similarly, when the other die is 5:

- 1*5=5
- 2*5=10
- 3*5=15
- 4*5=20
- 5*5=25
- 6*5=30

But note that 5*1 and 1*5 both result in 5, which is the same product, but they are different outcomes. However, when counting the number of outcomes, we have to consider that (5,1) and (1,5) are separate.

But in terms of products, the products are 5,10,15,20,25,30. However, each of these products can be achieved in different numbers of ways. But since we are counting the number of outcomes, not the number of distinct products, we need to count all the outcomes where the product is a multiple of 5.

But as we saw earlier, there are 11 such outcomes.

Alternatively, if I think of the products:

- 5: occurs in (1,5) and (5,1)
- 10: occurs in (2,5), (5,2)
- 15: occurs in (3,5), (5,3)
- 20: occurs in (4,5), (5,4)
- 25: occurs in (5,5)
- 30: occurs in (5,6), (6,5)

So, for each product:

- 5: 2 outcomes
- 10: 2 outcomes
- 15: 2 outcomes
- 20: 2 outcomes
- 25: 1 outcome
- 30: 2 outcomes

Adding these up: 2+2+2+2+1+2 = 11.

Yes, that's 11 outcomes. Therefore, confirming again that there are 11 favorable outcomes.

Therefore, the probability is 11/36.

So, after checking multiple approaches—complementary probability, direct counting, enumerating the products—it seems that 11/36 is indeed the correct answer.

I think that's thorough enough. I don't see any mistakes in my reasoning. All methods converge to the same answer, so I feel confident that 11/36 is the correct probability.

**Final Answer**
The probability is \boxed{\dfrac{11}{36}}.

To determine the probability that the product of the numbers rolled on two fair 6-sided dice is a multiple of 5, we begin by analyzing the total number of possible outcomes and the conditions under which the product is divisible by 5.

---

### Step 1: Total Number of Outcomes

Each die has 6 faces, so when rolling two dice, the total number of possible outcomes is:

$$
6 \times 6 = 36
$$

---

### Step 2: Understanding the Condition

A number is a multiple of 5 if it is divisible by 5. For the product of two numbers to be divisible by 5, **at least one of the numbers must be divisible by 5**. This is because 5 is a prime number, and the only way for the product to be divisible by 5 is if one of the factors is divisible by 5.

On a standard die, the only number divisible by 5 is **5**. So, the favorable outcomes are those where **at least one die shows a 5**.

---

### Step 3: Complementary Probability Approach

Instead of directly counting the favorable outcomes, we can use the **complement rule**. The complement of the event "at least one die shows a 5" is "neither die shows a 5".

- The numbers **not divisible by 5** on a die are: 1, 2, 3, 4, 6 → 5 numbers.
- So, the probability that a single die does **not** show a 5 is:

$$
\frac{5}{6}
$$

- The probability that **both** dice do not show a 5 is:

$$
\left(\frac{5}{6}\right)^2 = \frac{25}{36}
$$

- Therefore, the probability that **at least one die shows a 5** (i.e., the product is a multiple of 5) is:

$$
1 - \frac{25}{36} = \frac{11}{36}
$$

---

### Step 4: Direct Counting (Verification)

We can also count the favorable outcomes directly:

- The number of outcomes where the **first die is 5** is 6 (since the second die can be any of 1 through 6).
- The number of outcomes where the **second die is 5** is also 6.
- However, the outcome (5,5) is counted twice, so we subtract 1 to avoid double-counting.

$$
\text{Favorable outcomes} = 6 + 6 - 1 = 11
$$

So, the probability is:

$$
\frac{11}{36}
$$

---

### Final Answer

$$
\boxed{\dfrac{11}{36}}
$$

Solution: Probability questions are sometimes answered by calculating the ways the event will NOT happen, then subtracting.  In this problem the $1$, $2$, $3$, $4$ and $6$ faces are paired to create $5 \times 5 = 25$ number pairs whose product is NOT multiples of 5.  This leaves $36 - 25 = 11$ ways to get a multiple of $5$, so the probability is $\boxed{\frac{11}{36}}$.


Explanation:
Okay, so I need to find real numbers a and b, where a isn't zero, such that the equation f(x) - f(a) equals (x - a)^2 times (x - b). The function f(x) is given as x³ + 3x² + 1. Alright, let me start by understanding what this equation is saying.

First, f(x) - f(a) is the difference between the function evaluated at x and at a. The right side is a cubic polynomial, specifically (x - a) squared times (x - b). So, expanding the right side should give me a cubic polynomial, and the left side is also a cubic polynomial. Therefore, if I can expand both sides and equate coefficients, maybe I can solve for a and b.

Let me write down both sides.

Left side: f(x) - f(a) = [x³ + 3x² + 1] - [a³ + 3a² + 1] = x³ + 3x² + 1 - a³ - 3a² - 1 = x³ + 3x² - a³ - 3a².

Simplify that: x³ + 3x² - (a³ + 3a²). So that's the left side.

Right side: (x - a)²(x - b). Let me expand that. First, expand (x - a)²: that's x² - 2a x + a². Then multiply by (x - b):

Multiply x² - 2a x + a² by x - b:

First, x²*(x - b) = x³ - b x².

Then, -2a x*(x - b) = -2a x² + 2a b x.

Then, a²*(x - b) = a² x - a² b.

So adding all those terms together:

x³ - b x² - 2a x² + 2a b x + a² x - a² b.

Combine like terms:

x³ + (-b - 2a) x² + (2a b + a²) x - a² b.

So the right side is x³ + (-b - 2a)x² + (2ab + a²)x - a² b.

Now, the left side is x³ + 3x² - (a³ + 3a²). Let me write that as x³ + 3x² + 0x - (a³ + 3a²). So comparing coefficients with the right side.

So equate coefficients for each power of x:

For x³: 1 = 1. That's already equal, so no problem.

For x²: 3 = -b - 2a. So equation (1): -b - 2a = 3.

For x term: 0 = 2ab + a². Equation (2): 2ab + a² = 0.

For the constant term: - (a³ + 3a²) = -a² b. Let me write that as equation (3): -a³ - 3a² = -a² b. Which simplifies to a³ + 3a² = a² b. Assuming a ≠ 0, we can divide both sides by a²: a + 3 = b. So equation (3) gives b = a + 3.

So now, from equation (3), we have b = a + 3. Let me substitute this into equation (1) and equation (2).

First, equation (1): -b - 2a = 3. Substitute b = a + 3:

- (a + 3) - 2a = 3 => -a - 3 - 2a = 3 => -3a - 3 = 3 => -3a = 6 => a = -2.

So a is -2. Then, from equation (3), b = a + 3 = -2 + 3 = 1. So b is 1. Let me check if this satisfies equation (2).

Equation (2): 2ab + a² = 0. Let's plug in a = -2, b = 1:

2*(-2)*(1) + (-2)^2 = -4 + 4 = 0. Perfect, that works. So that's consistent.

Therefore, the ordered pair (a, b) is (-2, 1). Let me just verify this with the original equation to make sure.

Let me compute f(x) - f(a) where a = -2. First, compute f(-2):

f(-2) = (-2)^3 + 3*(-2)^2 + 1 = -8 + 3*(4) + 1 = -8 + 12 + 1 = 5.

So f(x) - f(-2) = x³ + 3x² + 1 - 5 = x³ + 3x² - 4.

Now, compute (x - a)^2 (x - b) with a = -2 and b = 1: (x - (-2))² (x - 1) = (x + 2)^2 (x - 1). Let me expand that:

First, (x + 2)^2 = x² + 4x + 4. Multiply by (x - 1):

(x² + 4x + 4)(x - 1) = x³ - x² + 4x² - 4x + 4x - 4 = x³ + 3x² - 4. Which matches f(x) - f(-2). Perfect, so that works.

Therefore, the answer is (-2, 1). So the ordered pair is (-2, 1). Let me check if there are other possible solutions, but since we had equations that led us uniquely to a = -2 and b = 1, and the problem states that such a and b exist with a ≠ 0, which is satisfied here, so this must be the solution.

**Final Answer**
The ordered pair is \boxed{(-2, 1)}.

To solve the equation

$$
f(x) - f(a) = (x - a)^2 (x - b)
$$

where $ f(x) = x^3 + 3x^2 + 1 $, and given that $ a \neq 0 $, we aim to find the ordered pair $ (a, b) $.

---

### Step 1: Expand the Left-Hand Side

We compute $ f(x) - f(a) $:

$$
f(x) - f(a) = (x^3 + 3x^2 + 1) - (a^3 + 3a^2 + 1) = x^3 + 3x^2 - a^3 - 3a^2
$$

So the left-hand side becomes:

$$
x^3 + 3x^2 - a^3 - 3a^2
$$

---

### Step 2: Expand the Right-Hand Side

We expand $ (x - a)^2 (x - b) $:

First, expand $ (x - a)^2 = x^2 - 2ax + a^2 $

Then multiply by $ (x - b) $:

$$
(x^2 - 2ax + a^2)(x - b) = x^3 - bx^2 - 2ax^2 + 2abx + a^2x - a^2b
$$

Group like terms:

$$
x^3 + (-b - 2a)x^2 + (2ab + a^2)x - a^2b
$$

---

### Step 3: Equate Coefficients

Now equate the left-hand side and the right-hand side:

Left side:
$$
x^3 + 3x^2 + 0x - (a^3 + 3a^2)
$$

Right side:
$$
x^3 + (-b - 2a)x^2 + (2ab + a^2)x - a^2b
$$

Matching coefficients:

1. $ x^3 $: $ 1 = 1 $ → OK
2. $ x^2 $: $ 3 = -b - 2a $ → Equation (1): $ -b - 2a = 3 $
3. $ x $: $ 0 = 2ab + a^2 $ → Equation (2): $ 2ab + a^2 = 0 $
4. Constant term: $ -a^3 - 3a^2 = -a^2b $ → Equation (3): $ a^3 + 3a^2 = a^2b $

---

### Step 4: Solve the System of Equations

From Equation (3):

$$
a^3 + 3a^2 = a^2b \Rightarrow b = a + 3 \quad \text{(since } a \neq 0 \text{)}
$$

Substitute $ b = a + 3 $ into Equation (1):

$$
-(a + 3) - 2a = 3 \Rightarrow -a - 3 - 2a = 3 \Rightarrow -3a - 3 = 3 \Rightarrow -3a = 6 \Rightarrow a = -2
$$

Now find $ b $:

$$
b = a + 3 = -2 + 3 = 1
$$

Check Equation (2):

$$
2ab + a^2 = 2(-2)(1) + (-2)^2 = -4 + 4 = 0 \quad \text{✓}
$$

---

### Step 5: Final Verification

Let’s verify the original equation with $ a = -2 $, $ b = 1 $:

- $ f(x) - f(-2) = x^3 + 3x^2 + 1 - f(-2) $
- $ f(-2) = (-2)^3 + 3(-2)^2 + 1 = -8 + 12 + 1 = 5 $
- So $ f(x) - f(-2) = x^3 + 3x^2 - 4 $

Now compute $ (x + 2)^2(x - 1) $:

- $ (x + 2)^2 = x^2 + 4x + 4 $
- Multiply by $ (x - 1) $:
  $$
  (x^2 + 4x + 4)(x - 1) = x^3 - x^2 + 4x^2 - 4x + 4x - 4 = x^3 + 3x^2 - 4
  $$

This matches the left-hand side, confirming the solution.

---

### Final Answer

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

[Assistant]
Explanation:
Okay, so I need to find real numbers a and b, where a isn't zero, such that the equation f(x) - f(a) equals (x - a)^2 times (x - b). The function f(x) is given as x³ + 3x² + 1. Alright, let me start by understanding what this equation is saying.

First, f(x) - f(a) is the difference between the function evaluated at x and at a. The right side is a cubic polynomial, specifically (x - a) squared times (x - b). So, expanding the right side should give me a cubic polynomial, and the left side is also a cubic polynomial. Therefore, if I can expand both sides and equate coefficients, maybe I can solve for a and b.

Let me write down both sides.

Left side: f(x) - f(a) = [x³ + 3x² + 1] - [a³ + 3a² + 1] = x³ + 3x² + 1 - a³ - 3a² - 1 = x³ + 3x² - a³ - 3a².

Simplify that: x³ + 3x² - (a³ + 3a²). So that's the left side.

Right side: (x - a)²(x - b). Let me expand that. First, expand (x - a)²: that's x² - 2a x + a². Then multiply by (x - b):

Multiply x² - 2a x + a² by x - b:

First, x²*(x - b) = x³ - b x².

Then, -2a x*(x - b) = -2a x² + 2a b x.

Then, a²*(x - b) = a² x - a² b.

So adding all those terms together:

x³ - b x² - 2a x² + 2a b x + a² x - a² b.

Combine like terms:

x³ + (-b - 2a) x² + (2a b + a²) x - a² b.

So the right side is x³ + (-b - 2a)x² + (2ab + a²)x - a² b.

Now, the left side is x³ + 3x² - (a³ + 3a²). Let me write that as x³ + 3x² + 0x - (a³ + 3a²). So comparing coefficients with the right side.

So equate coefficients for each power of x:

For x³: 1 = 1. That's already equal, so no problem.

For x²: 3 = -b - 2a. So equation (1): -b - 2a = 3.

For x term: 0 = 2ab + a². Equation (2): 2ab + a² = 0.

For the constant term: - (a³ + 3a²) = -a² b. Let me write that as equation (3): -a³ - 3a² = -a² b. Which simplifies to a³ + 3a² = a² b. Assuming a ≠ 0, we can divide both sides by a²: a + 3 = b. So equation (3) gives b = a + 3.

So now, from equation (3), we have b = a + 3. Let me substitute this into equation (1) and equation (2).

First, equation (1): -b - 2a = 3. Substitute b = a + 3:

- (a + 3) - 2a = 3 => -a - 3 - 2a = 3 => -3a - 3 = 3 => -3a = 6 => a = -2.

So a is -2. Then, from equation (3), b = a + 3 = -2 + 3 = 1. So b is 1. Let me check if this satisfies equation (2).

Equation (2): 2ab + a² = 0. Let's plug in a = -2, b = 1:

2*(-2)*(1) + (-2)^2 = -4 + 4 = 0. Perfect, that works. So that's consistent.

Therefore, the ordered pair (a, b) is (-2, 1). Let me just verify this with the original equation to make sure.

Let me compute f(x) - f(a) where a = -2. First, compute f(-2):

f(-2) = (-2)^3 + 3*(-2)^2 + 1 = -8 + 3*(4) + 1 = -8 + 12 + 1 = 5.

So f(x) - f(-2) = x³ + 3x² + 1 - 5 = x³ + 3x² - 4.

Now, compute (x - a)^2 (x - b) with a = -2 and b = 1: (x - (-2))² (x - 1) = (x + 2)^2 (x - 1). Let me expand that:

First, (x + 2)^2 = x² + 4x + 4. Multiply by (x - 1):

(x² + 4x + 4)(x - 1) = x³ - x² + 4x² - 4x + 4x - 4 = x³ + 3x² - 4. Which matches f(x) - f(-2). Perfect, so that works.

Therefore, the answer is (-2, 1). So the ordered pair is (-2, 1). Let me check if there are other possible solutions, but since we had equations that led us uniquely to a = -2 and b = 1, and the problem states that such a and b exist with a ≠ 0, which is satisfied here, so this must be the solution.

**Final Answer**
The ordered pair is \boxed{(-2, 1)}.

To solve the equation

$$
f(x) - f(a) = (x - a)^2 (x - b)
$$

where $ f(x) = x^3 + 3x^2 + 1 $, and given that $ a \neq 0 $, we aim to find the ordered pair $ (a, b) $.

---

### Step 1: Expand the Left-Hand Side

We compute $ f(x) - f(a) $:

$$
f(x) - f(a) = (x^3 + 3x^2 + 1) - (a^3 + 3a^2 + 1) = x^3 + 3x^2 - a^3 - 3a^2
$$

So the left-hand side becomes:

$$
x^3 + 3x^2 - a^3 - 3a^2
$$

---

### Step 2: Expand the Right-Hand Side

We expand $ (x - a)^2 (x - b) $:

First, expand $ (x - a)^2 = x^2 - 2ax + a^2 $

Then multiply by $ (x - b) $:

$$
(x^2 - 2ax + a^2)(x - b) = x^3 - bx^2 - 2ax^2 + 2abx + a^2x - a^2b
$$

Group like terms:

$$
x^3 + (-b - 2a)x^2 + (2ab + a^2)x - a^2b
$$

---

### Step 3: Equate Coefficients

Now equate the left-hand side and the right-hand side:

Left side:
$$
x^3 + 3x^2 + 0x - (a^3 + 3a^2)
$$

Right side:
$$
x^3 + (-b - 2a)x^2 + (2ab + a^2)x - a^2b
$$

Matching coefficients:

1. $ x^3 $: $ 1 = 1 $ → OK
2. $ x^2 $: $ 3 = -b - 2a $ → Equation (1): $ -b - 2a = 3 $
3. $ x $: $ 0 = 2ab + a^2 $ → Equation (2): $ 2ab + a^2 = 0 $
4. Constant term: $ -a^3 - 3a^2 = -a^2b $ → Equation (3): $ a^3 + 3a^2 = a^2b $

---

### Step 4: Solve the System of Equations

From Equation (3):

$$
a^3 + 3a^2 = a^2b \Rightarrow b = a + 3 \quad \text{(since } a \neq 0 \text{)}
$$

Substitute $ b = a + 3 $ into Equation (1):

$$
-(a + 3) - 2a = 3 \Rightarrow -a - 3 - 2a = 3 \Rightarrow -3a - 3 = 3 \Rightarrow -3a = 6 \Rightarrow a = -2
$$

Now find $ b $:

$$
b = a + 3 = -2 + 3 = 1
$$

Check Equation (2):

$$
2ab + a^2 = 2(-2)(1) + (-2)^2 = -4 + 4 = 0 \quad \text{✓}
$$

---

### Step 5: Final Verification

Let’s verify the original equation with $ a = -2 $, $ b = 1 $:

- $ f(x) - f(-2) = x^3 + 3x^2 + 1 - f(-2) $
- $ f(-2) = (-2)^3 + 3(-2)^2 + 1 = -8 + 12 + 1 = 5 $
- So $ f(x) - f(-2) = x^3 + 3x^2 - 4 $

Now compute $ (x + 2)^2(x - 1) $:

- $ (x + 2)^2 = x^2 + 4x + 4 $
- Multiply by $ (x - 1) $:
  $$
  (x^2 + 4x + 4)(x - 1) = x^3 - x^2 + 4x^2 - 4x + 4x - 4 = x^3 + 3x^2 - 4
  $$

This matches the left-hand side, confirming the solution.

---

### Final Answer

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

Solution:
By the remainder theorem, $f(x) - f(a)$ is divisible by $x - a,$ so we can take out a factor of $x - a$ accordingly:
\begin{align*}
f(x) - f(a) &= (x^3 + 3x^2 + 1) - (a^3 + 3a^2 + 1) \\
&= (x^3 - a^3) + 3(x^2 - a^2) \\
&= (x - a)(x^2 + ax + a^2) + 3(x - a)(x + a) \\
&= (x - a)(x^2 + ax + a^2 + 3x + 3a) \\
&= (x - a)(x^2 + (a + 3) x + a^2 + 3a).
\end{align*}Thus, we want
\[x^2 + (a + 3) x + a^2 + 3a = (x - a)(x - b) = x^2 - (a + b) x + ab.\]Matching coefficients, we get
\begin{align*}
a + 3 &= -a - b, \\
a^2 + 3a &= ab.
\end{align*}Since $a \neq 0,$ we can divide both sides of the second equation by $a,$ to get $a + 3 = b.$  Then $-a - b = b,$ so $a = -2b.$  Then
\[-2b + 3 = 2b - b,\]which gives us $b = 1$.  Then $a = -2,$ so $(a,b) = \boxed{(-2,1)}.$


Explanation:
Okay, so I need to find the values of 'a' and 'b' such that when I multiply the two quadratic expressions (3t² + 5t + a) and (4t² + bt - 2), the result is the given quartic polynomial 12t⁴ + 26t³ - 8t² - 16t + 6. Then, once I find 'a' and 'b', I have to add them together to get the answer. Alright, let me start by recalling how to multiply two polynomials.

First, I know that when you multiply two polynomials, you have to use the distributive property, which means each term in the first polynomial gets multiplied by each term in the second polynomial. So, let me write down the two polynomials:

First polynomial: 3t² + 5t + a

Second polynomial: 4t² + bt - 2

So, multiplying them together would give me:

(3t²)(4t²) + (3t²)(bt) + (3t²)(-2) + (5t)(4t²) + (5t)(bt) + (5t)(-2) + (a)(4t²) + (a)(bt) + (a)(-2)

Let me compute each term step by step.

Starting with the first term: 3t² multiplied by 4t². Multiplying the coefficients: 3*4 = 12. Then, multiplying the variables: t²*t² = t⁴. So that term is 12t⁴.

Next term: 3t² multiplied by bt. Coefficients: 3*b = 3b. Variables: t²*t = t³. So that term is 3b t³.

Third term: 3t² multiplied by -2. Coefficients: 3*(-2) = -6. Variables: t². So that term is -6t².

Fourth term: 5t multiplied by 4t². Coefficients: 5*4 = 20. Variables: t*t² = t³. So that term is 20t³.

Fifth term: 5t multiplied by bt. Coefficients: 5*b = 5b. Variables: t*t = t². So that term is 5b t².

Sixth term: 5t multiplied by -2. Coefficients: 5*(-2) = -10. Variables: t. So that term is -10t.

Seventh term: a multiplied by 4t². Coefficients: a*4 = 4a. Variables: t². So that term is 4a t².

Eighth term: a multiplied by bt. Coefficients: a*b = ab. Variables: t. So that term is ab t.

Ninth term: a multiplied by -2. Coefficients: a*(-2) = -2a. Variables: none, so just a constant term. So that term is -2a.

Now, let me write all these terms together:

12t⁴ + 3b t³ - 6t² + 20t³ + 5b t² - 10t + 4a t² + ab t - 2a.

Now, I need to combine like terms. Let me group them by their degree:

- For t⁴: only 12t⁴.

- For t³: 3b t³ + 20t³. Let me factor out t³: (3b + 20) t³.

- For t²: -6t² + 5b t² + 4a t². Let me factor out t²: (-6 + 5b + 4a) t².

- For t: -10t + ab t. Factor out t: (-10 + ab) t.

- Constants: -2a.

So putting it all together, the product is:

12t⁴ + (3b + 20) t³ + (-6 + 5b + 4a) t² + (-10 + ab) t - 2a.

Now, according to the problem, this product is equal to 12t⁴ + 26t³ - 8t² - 16t + 6. Therefore, the coefficients of corresponding terms must be equal. So, I can set up equations by equating the coefficients from both polynomials.

Let me list them:

1. Coefficient of t⁴: 12 = 12. Well, that's already matching, so no new information here.

2. Coefficient of t³: 3b + 20 = 26.

3. Coefficient of t²: -6 + 5b + 4a = -8.

4. Coefficient of t: -10 + ab = -16.

5. Constant term: -2a = 6.

So, now I have a system of equations:

Equation (1): 3b + 20 = 26

Equation (2): -6 + 5b + 4a = -8

Equation (3): -10 + ab = -16

Equation (4): -2a = 6

Let me solve these equations step by step. Starting with Equation (4) because it only has 'a' and seems simpler.

Equation (4): -2a = 6. Solving for a:

Divide both sides by -2: a = 6 / (-2) = -3. So, a = -3.

Okay, so a is -3. Let me note that down: a = -3.

Now, let me plug this value of a into Equation (2) to find b.

Equation (2): -6 + 5b + 4a = -8.

Substituting a = -3:

-6 + 5b + 4*(-3) = -8.

Compute 4*(-3) = -12. So:

-6 + 5b -12 = -8.

Combine constants: -6 -12 = -18. So:

-18 + 5b = -8.

Now, solving for b:

Add 18 to both sides: 5b = -8 + 18 = 10.

Therefore, 5b = 10 => b = 10 / 5 = 2. So, b = 2.

Alright, so now I have a = -3 and b = 2. Let me check if these values satisfy Equation (1) and Equation (3) as well. Let me check Equation (1) first.

Equation (1): 3b + 20 = 26.

Substituting b = 2: 3*2 + 20 = 6 + 20 = 26. Which matches. Good.

Now check Equation (3): -10 + ab = -16.

We have a = -3 and b = 2. So ab = (-3)(2) = -6. Therefore:

-10 + (-6) = -16. Which is -16 = -16. Perfect, that works too.

So all equations are satisfied with a = -3 and b = 2. Therefore, the question asks for a + b. So:

a + b = (-3) + 2 = -1.

Wait, but let me double-check everything once more to make sure I didn't make any mistakes. It's easy to make sign errors here.

First, let me verify the multiplication with a = -3 and b = 2.

Original polynomials:

First polynomial: 3t² + 5t + a = 3t² + 5t - 3.

Second polynomial: 4t² + bt - 2 = 4t² + 2t - 2.

Let me multiply them step by step.

Multiply 3t² by each term in the second polynomial:

3t² * 4t² = 12t⁴

3t² * 2t = 6t³

3t² * (-2) = -6t²

Then multiply 5t by each term:

5t * 4t² = 20t³

5t * 2t = 10t²

5t * (-2) = -10t

Then multiply -3 by each term:

-3 * 4t² = -12t²

-3 * 2t = -6t

-3 * (-2) = 6

Now, let me add all these terms together:

12t⁴ + 6t³ -6t² + 20t³ + 10t² -10t -12t² -6t + 6.

Now combine like terms:

t⁴: 12t⁴

t³: 6t³ + 20t³ = 26t³

t²: -6t² + 10t² -12t² = (-6 + 10 -12)t² = (-8)t²

t terms: -10t -6t = -16t

Constants: +6

So altogether, the product is 12t⁴ + 26t³ -8t² -16t +6, which matches exactly the given polynomial. Perfect! So my values for a and b are correct. Therefore, a + b = -3 + 2 = -1.

Wait, but the problem says "What is a + b?" So the answer is -1? Let me check once again if I didn't miscalculate a + b. a is -3, b is 2. So -3 + 2 is indeed -1.

But just to be thorough, let me check if there's any other possible solutions or if I made a mistake in solving the equations. Let me go back through the equations.

From Equation (4): -2a = 6 => a = -3. Correct.

Equation (2): -6 + 5b + 4a = -8. Substituting a = -3:

-6 + 5b + 4*(-3) = -6 + 5b -12 = 5b - 18 = -8 => 5b = 10 => b = 2. Correct.

Equation (1): 3b + 20 = 26 => 3b = 6 => b = 2. Correct.

Equation (3): -10 + ab = -16. With a = -3, b = 2: ab = -6. So -10 + (-6) = -16. Correct.

All equations are satisfied. Therefore, the answer is indeed -1.

But just to make sure, let me think if there's another way to approach this problem, maybe by comparing coefficients differently or using another method. For example, maybe plugging in specific values of t to create equations? Although since we have two variables, maybe that would require two equations, but since we already solved it through coefficient comparison, which worked, maybe that's redundant. But just for thoroughness, let me try plugging in a value of t to see if it holds.

Let me choose t = 1. Let me compute both sides.

Left side: (3(1)^2 + 5(1) + a)(4(1)^2 + b(1) - 2) = (3 + 5 + a)(4 + b - 2) = (8 + a)(2 + b)

Right side: 12(1)^4 + 26(1)^3 -8(1)^2 -16(1) +6 = 12 + 26 -8 -16 +6 = Let's compute:

12 + 26 = 38; 38 -8 = 30; 30 -16 = 14; 14 +6 = 20.

So left side must equal 20. Therefore:

(8 + a)(2 + b) = 20.

We know a = -3 and b = 2. Let's check:

(8 + (-3))(2 + 2) = (5)(4) = 20. Correct. So that works.

Another value, maybe t = 0. Let's check.

Left side: (3(0)^2 + 5(0) + a)(4(0)^2 + b(0) -2) = (0 + 0 + a)(0 + 0 -2) = (a)(-2) = -2a.

Right side: 12(0)^4 + 26(0)^3 -8(0)^2 -16(0) +6 = 0 + 0 -0 -0 +6 = 6.

Therefore, -2a = 6 => a = -3. Which matches our previous result. So that's consistent.

Another check: t = -1.

Left side: (3(-1)^2 + 5(-1) + a)(4(-1)^2 + b(-1) -2) = (3(1) -5 + a)(4(1) -b -2) = (3 -5 + a)(4 - b -2) = (-2 + a)(2 - b)

Right side: 12(-1)^4 + 26(-1)^3 -8(-1)^2 -16(-1) +6 = 12(1) + 26(-1) -8(1) -16(-1) +6 = 12 -26 -8 +16 +6.

Calculating:

12 -26 = -14; -14 -8 = -22; -22 +16 = -6; -6 +6 = 0.

So left side must equal 0. Therefore:

(-2 + a)(2 - b) = 0.

We have a = -3 and b = 2. Let's plug in:

(-2 + (-3))(2 - 2) = (-5)(0) = 0. Correct. So that works as well.

So all these checks confirm that a = -3 and b = 2 are correct. Therefore, a + b = -3 + 2 = -1.

I think that's thorough enough. I don't see any mistakes in my reasoning. So the answer should be -1.

**Final Answer**
\boxed{-1}

To solve the problem, we are given two quadratic polynomials:

$$
(3t^2 + 5t + a)(4t^2 + bt - 2)
$$

and their product is:

$$
12t^4 + 26t^3 - 8t^2 - 16t + 6
$$

Our goal is to find the values of $ a $ and $ b $, and then compute $ a + b $.

---

### Step 1: Multiply the Polynomials

We expand the product:

$$
(3t^2 + 5t + a)(4t^2 + bt - 2)
$$

Using the distributive property, we compute each term:

- $ 3t^2 \cdot 4t^2 = 12t^4 $
- $ 3t^2 \cdot bt = 3bt^3 $
- $ 3t^2 \cdot (-2) = -6t^2 $
- $ 5t \cdot 4t^2 = 20t^3 $
- $ 5t \cdot bt = 5bt^2 $
- $ 5t \cdot (-2) = -10t $
- $ a \cdot 4t^2 = 4at^2 $
- $ a \cdot bt = abt $
- $ a \cdot (-2) = -2a $

Now, combine like terms:

- $ t^4 $: $ 12t^4 $
- $ t^3 $: $ (3b + 20)t^3 $
- $ t^2 $: $ (-6 + 5b + 4a)t^2 $
- $ t $: $ (-10 + ab)t $
- Constant: $ -2a $

So the product becomes:

$$
12t^4 + (3b + 20)t^3 + (-6 + 5b + 4a)t^2 + (-10 + ab)t - 2a
$$

---

### Step 2: Match Coefficients

We are told this equals:

$$
12t^4 + 26t^3 - 8t^2 - 16t + 6
$$

Now, equate the coefficients of corresponding powers of $ t $:

1. $ 12 = 12 $ → No new info
2. $ 3b + 20 = 26 $ → Solve for $ b $
3. $ -6 + 5b + 4a = -8 $ → Equation in $ a $ and $ b $
4. $ -10 + ab = -16 $ → Equation in $ a $ and $ b $
5. $ -2a = 6 $ → Solve for $ a $

---

### Step 3: Solve the System

From equation (5):

$$
-2a = 6 \Rightarrow a = -3
$$

Substitute $ a = -3 $ into equation (2):

$$
3b + 20 = 26 \Rightarrow 3b = 6 \Rightarrow b = 2
$$

Now verify with equation (3):

$$
-6 + 5b + 4a = -6 + 5(2) + 4(-3) = -6 + 10 - 12 = -8 \quad \text{✓}
$$

And equation (4):

$$
-10 + ab = -10 + (-3)(2) = -10 - 6 = -16 \quad \text{✓}
$$

All equations are satisfied.

---

### Step 4: Compute $ a + b $

$$
a + b = -3 + 2 = \boxed{-1}
$$

---

### Final Answer

$$
\boxed{-1}
$$

[Assistant]
Explanation:
Okay, so I need to find the values of 'a' and 'b' such that when I multiply the two quadratic expressions (3t² + 5t + a) and (4t² + bt - 2), the result is the given quartic polynomial 12t⁴ + 26t³ - 8t² - 16t + 6. Then, once I find 'a' and 'b', I have to add them together to get the answer. Alright, let me start by recalling how to multiply two polynomials.

First, I know that when you multiply two polynomials, you have to use the distributive property, which means each term in the first polynomial gets multiplied by each term in the second polynomial. So, let me write down the two polynomials:

First polynomial: 3t² + 5t + a

Second polynomial: 4t² + bt - 2

So, multiplying them together would give me:

(3t²)(4t²) + (3t²)(bt) + (3t²)(-2) + (5t)(4t²) + (5t)(bt) + (5t)(-2) + (a)(4t²) + (a)(bt) + (a)(-2)

Let me compute each term step by step.

Starting with the first term: 3t² multiplied by 4t². Multiplying the coefficients: 3*4 = 12. Then, multiplying the variables: t²*t² = t⁴. So that term is 12t⁴.

Next term: 3t² multiplied by bt. Coefficients: 3*b = 3b. Variables: t²*t = t³. So that term is 3b t³.

Third term: 3t² multiplied by -2. Coefficients: 3*(-2) = -6. Variables: t². So that term is -6t².

Fourth term: 5t multiplied by 4t². Coefficients: 5*4 = 20. Variables: t*t² = t³. So that term is 20t³.

Fifth term: 5t multiplied by bt. Coefficients: 5*b = 5b. Variables: t*t = t². So that term is 5b t².

Sixth term: 5t multiplied by -2. Coefficients: 5*(-2) = -10. Variables: t. So that term is -10t.

Seventh term: a multiplied by 4t². Coefficients: a*4 = 4a. Variables: t². So that term is 4a t².

Eighth term: a multiplied by bt. Coefficients: a*b = ab. Variables: t. So that term is ab t.

Ninth term: a multiplied by -2. Coefficients: a*(-2) = -2a. Variables: none, so just a constant term. So that term is -2a.

Now, let me write all these terms together:

12t⁴ + 3b t³ - 6t² + 20t³ + 5b t² - 10t + 4a t² + ab t - 2a.

Now, I need to combine like terms. Let me group them by their degree:

- For t⁴: only 12t⁴.

- For t³: 3b t³ + 20t³. Let me factor out t³: (3b + 20) t³.

- For t²: -6t² + 5b t² + 4a t². Let me factor out t²: (-6 + 5b + 4a) t².

- For t: -10t + ab t. Factor out t: (-10 + ab) t.

- Constants: -2a.

So putting it all together, the product is:

12t⁴ + (3b + 20) t³ + (-6 + 5b + 4a) t² + (-10 + ab) t - 2a.

Now, according to the problem, this product is equal to 12t⁴ + 26t³ - 8t² - 16t + 6. Therefore, the coefficients of corresponding terms must be equal. So, I can set up equations by equating the coefficients from both polynomials.

Let me list them:

1. Coefficient of t⁴: 12 = 12. Well, that's already matching, so no new information here.

2. Coefficient of t³: 3b + 20 = 26.

3. Coefficient of t²: -6 + 5b + 4a = -8.

4. Coefficient of t: -10 + ab = -16.

5. Constant term: -2a = 6.

So, now I have a system of equations:

Equation (1): 3b + 20 = 26

Equation (2): -6 + 5b + 4a = -8

Equation (3): -10 + ab = -16

Equation (4): -2a = 6

Let me solve these equations step by step. Starting with Equation (4) because it only has 'a' and seems simpler.

Equation (4): -2a = 6. Solving for a:

Divide both sides by -2: a = 6 / (-2) = -3. So, a = -3.

Okay, so a is -3. Let me note that down: a = -3.

Now, let me plug this value of a into Equation (2) to find b.

Equation (2): -6 + 5b + 4a = -8.

Substituting a = -3:

-6 + 5b + 4*(-3) = -8.

Compute 4*(-3) = -12. So:

-6 + 5b -12 = -8.

Combine constants: -6 -12 = -18. So:

-18 + 5b = -8.

Now, solving for b:

Add 18 to both sides: 5b = -8 + 18 = 10.

Therefore, 5b = 10 => b = 10 / 5 = 2. So, b = 2.

Alright, so now I have a = -3 and b = 2. Let me check if these values satisfy Equation (1) and Equation (3) as well. Let me check Equation (1) first.

Equation (1): 3b + 20 = 26.

Substituting b = 2: 3*2 + 20 = 6 + 20 = 26. Which matches. Good.

Now check Equation (3): -10 + ab = -16.

We have a = -3 and b = 2. So ab = (-3)(2) = -6. Therefore:

-10 + (-6) = -16. Which is -16 = -16. Perfect, that works too.

So all equations are satisfied with a = -3 and b = 2. Therefore, the question asks for a + b. So:

a + b = (-3) + 2 = -1.

Wait, but let me double-check everything once more to make sure I didn't make any mistakes. It's easy to make sign errors here.

First, let me verify the multiplication with a = -3 and b = 2.

Original polynomials:

First polynomial: 3t² + 5t + a = 3t² + 5t - 3.

Second polynomial: 4t² + bt - 2 = 4t² + 2t - 2.

Let me multiply them step by step.

Multiply 3t² by each term in the second polynomial:

3t² * 4t² = 12t⁴

3t² * 2t = 6t³

3t² * (-2) = -6t²

Then multiply 5t by each term:

5t * 4t² = 20t³

5t * 2t = 10t²

5t * (-2) = -10t

Then multiply -3 by each term:

-3 * 4t² = -12t²

-3 * 2t = -6t

-3 * (-2) = 6

Now, let me add all these terms together:

12t⁴ + 6t³ -6t² + 20t³ + 10t² -10t -12t² -6t + 6.

Now combine like terms:

t⁴: 12t⁴

t³: 6t³ + 20t³ = 26t³

t²: -6t² + 10t² -12t² = (-6 + 10 -12)t² = (-8)t²

t terms: -10t -6t = -16t

Constants: +6

So altogether, the product is 12t⁴ + 26t³ -8t² -16t +6, which matches exactly the given polynomial. Perfect! So my values for a and b are correct. Therefore, a + b = -3 + 2 = -1.

Wait, but the problem says "What is a + b?" So the answer is -1? Let me check once again if I didn't miscalculate a + b. a is -3, b is 2. So -3 + 2 is indeed -1.

But just to be thorough, let me check if there's any other possible solutions or if I made a mistake in solving the equations. Let me go back through the equations.

From Equation (4): -2a = 6 => a = -3. Correct.

Equation (2): -6 + 5b + 4a = -8. Substituting a = -3:

-6 + 5b + 4*(-3) = -6 + 5b -12 = 5b - 18 = -8 => 5b = 10 => b = 2. Correct.

Equation (1): 3b + 20 = 26 => 3b = 6 => b = 2. Correct.

Equation (3): -10 + ab = -16. With a = -3, b = 2: ab = -6. So -10 + (-6) = -16. Correct.

All equations are satisfied. Therefore, the answer is indeed -1.

But just to make sure, let me think if there's another way to approach this problem, maybe by comparing coefficients differently or using another method. For example, maybe plugging in specific values of t to create equations? Although since we have two variables, maybe that would require two equations, but since we already solved it through coefficient comparison, which worked, maybe that's redundant. But just for thoroughness, let me try plugging in a value of t to see if it holds.

Let me choose t = 1. Let me compute both sides.

Left side: (3(1)^2 + 5(1) + a)(4(1)^2 + b(1) - 2) = (3 + 5 + a)(4 + b - 2) = (8 + a)(2 + b)

Right side: 12(1)^4 + 26(1)^3 -8(1)^2 -16(1) +6 = 12 + 26 -8 -16 +6 = Let's compute:

12 + 26 = 38; 38 -8 = 30; 30 -16 = 14; 14 +6 = 20.

So left side must equal 20. Therefore:

(8 + a)(2 + b) = 20.

We know a = -3 and b = 2. Let's check:

(8 + (-3))(2 + 2) = (5)(4) = 20. Correct. So that works.

Another value, maybe t = 0. Let's check.

Left side: (3(0)^2 + 5(0) + a)(4(0)^2 + b(0) -2) = (0 + 0 + a)(0 + 0 -2) = (a)(-2) = -2a.

Right side: 12(0)^4 + 26(0)^3 -8(0)^2 -16(0) +6 = 0 + 0 -0 -0 +6 = 6.

Therefore, -2a = 6 => a = -3. Which matches our previous result. So that's consistent.

Another check: t = -1.

Left side: (3(-1)^2 + 5(-1) + a)(4(-1)^2 + b(-1) -2) = (3(1) -5 + a)(4(1) -b -2) = (3 -5 + a)(4 - b -2) = (-2 + a)(2 - b)

Right side: 12(-1)^4 + 26(-1)^3 -8(-1)^2 -16(-1) +6 = 12(1) + 26(-1) -8(1) -16(-1) +6 = 12 -26 -8 +16 +6.

Calculating:

12 -26 = -14; -14 -8 = -22; -22 +16 = -6; -6 +6 = 0.

So left side must equal 0. Therefore:

(-2 + a)(2 - b) = 0.

We have a = -3 and b = 2. Let's plug in:

(-2 + (-3))(2 - 2) = (-5)(0) = 0. Correct. So that works as well.

So all these checks confirm that a = -3 and b = 2 are correct. Therefore, a + b = -3 + 2 = -1.

I think that's thorough enough. I don't see any mistakes in my reasoning. So the answer should be -1.

**Final Answer**
\boxed{-1}

To solve the problem, we are given two quadratic polynomials:

$$
(3t^2 + 5t + a)(4t^2 + bt - 2)
$$

and their product is:

$$
12t^4 + 26t^3 - 8t^2 - 16t + 6
$$

Our goal is to find the values of $ a $ and $ b $, and then compute $ a + b $.

---

### Step 1: Multiply the Polynomials

We expand the product:

$$
(3t^2 + 5t + a)(4t^2 + bt - 2)
$$

Using the distributive property, we compute each term:

- $ 3t^2 \cdot 4t^2 = 12t^4 $
- $ 3t^2 \cdot bt = 3bt^3 $
- $ 3t^2 \cdot (-2) = -6t^2 $
- $ 5t \cdot 4t^2 = 20t^3 $
- $ 5t \cdot bt = 5bt^2 $
- $ 5t \cdot (-2) = -10t $
- $ a \cdot 4t^2 = 4at^2 $
- $ a \cdot bt = abt $
- $ a \cdot (-2) = -2a $

Now, combine like terms:

- $ t^4 $: $ 12t^4 $
- $ t^3 $: $ (3b + 20)t^3 $
- $ t^2 $: $ (-6 + 5b + 4a)t^2 $
- $ t $: $ (-10 + ab)t $
- Constant: $ -2a $

So the product becomes:

$$
12t^4 + (3b + 20)t^3 + (-6 + 5b + 4a)t^2 + (-10 + ab)t - 2a
$$

---

### Step 2: Match Coefficients

We are told this equals:

$$
12t^4 + 26t^3 - 8t^2 - 16t + 6
$$

Now, equate the coefficients of corresponding powers of $ t $:

1. $ 12 = 12 $ → No new info
2. $ 3b + 20 = 26 $ → Solve for $ b $
3. $ -6 + 5b + 4a = -8 $ → Equation in $ a $ and $ b $
4. $ -10 + ab = -16 $ → Equation in $ a $ and $ b $
5. $ -2a = 6 $ → Solve for $ a $

---

### Step 3: Solve the System

From equation (5):

$$
-2a = 6 \Rightarrow a = -3
$$

Substitute $ a = -3 $ into equation (2):

$$
3b + 20 = 26 \Rightarrow 3b = 6 \Rightarrow b = 2
$$

Now verify with equation (3):

$$
-6 + 5b + 4a = -6 + 5(2) + 4(-3) = -6 + 10 - 12 = -8 \quad \text{✓}
$$

And equation (4):

$$
-10 + ab = -10 + (-3)(2) = -10 - 6 = -16 \quad \text{✓}
$$

All equations are satisfied.

---

### Step 4: Compute $ a + b $

$$
a + b = -3 + 2 = \boxed{-1}
$$

---

### Final Answer

$$
\boxed{-1}
$$

Solution: The constant term of the product of two polynomials is just the product of the two constant terms. Therefore we know that $6=-2a$, so $a=-3$. We now consider the linear term of the product of our polynomials. It's given by $-16t=(5t\cdot-2)+a\cdot bt\Longrightarrow-16t=-10t+(-3)bt\Longrightarrow b=2$. Therefore our answer is $a+b=\boxed{-1}$.


Explanation:
Okay, so I need to find the radius of a right cylindrical tank. The problem says that the tank is being filled with water at a rate of 20π cubic meters per hour, and as the tank is filled, the water level rises four meters per hour. Hmm, let me think about how to approach this.

First, I remember that the volume of a cylinder is given by the formula V = πr²h, where r is the radius and h is the height (or in this case, the water level). The problem mentions the rate at which the volume is increasing and the rate at which the height is increasing. So, this seems like a related rates problem from calculus. But wait, maybe I don't need calculus here? Let me check.

The volume is increasing at 20π cubic meters per hour, which is dV/dt = 20π. The height is increasing at 4 meters per hour, so dh/dt = 4. I need to find the radius r. Since the volume of the cylinder is V = πr²h, if I differentiate both sides with respect to time t, I can relate dV/dt and dh/dt. Let me try that.

Differentiating V = πr²h with respect to t:

dV/dt = π [ d/dt(r²h) ]

Since r is the radius of the tank, I assume it's constant, right? The tank is a right cylinder with circular bases, so the radius doesn't change as the water level rises. Therefore, r is a constant, and so when differentiating, we can treat r² as a constant. Therefore:

dV/dt = π * r² * dh/dt

Wait, that seems straightforward. So if I plug in the known values:

20π = π * r² * 4

Then, I can divide both sides by π:

20 = 4r²

Then divide both sides by 4:

5 = r²

Therefore, r = √5. So the radius is √5 meters. Is that it? Let me check again.

Wait, let me make sure I didn't skip any steps or make any wrong assumptions. The key assumption here is that the radius is constant. Since the tank is a right cylinder, the radius doesn't change as the water level rises. So, the cross-sectional area of the tank is πr², and the rate at which the volume increases is equal to the cross-sectional area multiplied by the rate at which the height increases. So, dV/dt = A * dh/dt, where A is the cross-sectional area. Therefore, 20π = πr² * 4. Then, solving for r² gives 20π / (4π) = 5, so r² = 5, so r = √5. That seems correct.

But let me think again. Maybe I need to consider something else? For example, is there a possibility that the tank isn't full? But the problem doesn't mention anything about the tank's total height or whether it's being filled to the top. However, since they are giving the rate of change of the water level, it's possible that this relationship holds regardless of the current height, as long as the radius is constant. Therefore, even if the tank is only partially filled, the rate at which the volume increases would still depend on the cross-sectional area. Therefore, my calculation should hold.

Alternatively, if the tank had a varying radius, then we would need more information, but since it's a right cylinder, the radius is constant. Therefore, the answer should be √5 meters. Let me check the units to make sure. The rate of volume is in cubic meters per hour, and the rate of height is in meters per hour. The cross-sectional area would be in square meters, so multiplying by meters per hour gives cubic meters per hour, which matches. So units are consistent.

Therefore, I think my answer is correct. The radius is √5 meters. So, expressed in simplest radical form, that's just √5. Therefore, the answer is √5.

Wait, but let me check once more. Let me write down the equations again.

Given:

dV/dt = 20π m³/h

dh/dt = 4 m/h

Find r.

We know that V = πr²h. Differentiating with respect to t:

dV/dt = πr² dh/dt

So solving for r:

r² = (dV/dt) / (π dh/dt)

Wait, no:

Wait, dV/dt = πr² dh/dt => r² = (dV/dt) / (π dh/dt)

Wait, but substituting numbers:

r² = (20π) / (π * 4) = 20 / 4 = 5. So r² = 5, so r = √5. Yes, same result. So that's correct.

Therefore, I think the answer is √5. So I think that's the radius. I don't see any mistakes here. Maybe I can think of an example. Suppose the radius is √5, then cross-sectional area is π*(√5)^2 = 5π. Then, if the height increases by 4 meters per hour, then the volume increase per hour is 5π * 4 = 20π, which matches the given rate. So that checks out. Therefore, the answer is √5.

**Final Answer**
The radius of the tank is \boxed{\sqrt{5}} meters.

To determine the radius of the cylindrical tank, we begin by analyzing the relationship between the volume of water and the height of the water level as the tank is being filled.

---

### Step 1: Understand the Given Information

- The **rate of change of volume** is given as:
  $$
  \frac{dV}{dt} = 20\pi \text{ cubic meters per hour}
  $$
- The **rate of change of height** is:
  $$
  \frac{dh}{dt} = 4 \text{ meters per hour}
  $$
- The tank is a **right circular cylinder**, so the **radius $ r $** is constant.

---

### Step 2: Use the Volume Formula for a Cylinder

The volume $ V $ of a cylinder is given by:
$$
V = \pi r^2 h
$$
Differentiating both sides with respect to time $ t $, we get:
$$
\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}
$$

This equation relates the rate of change of volume to the rate of change of height, and it allows us to solve for the radius $ r $.

---

### Step 3: Plug in the Known Values

Substitute the given values into the differentiated equation:
$$
20\pi = \pi r^2 \cdot 4
$$

Divide both sides by $ \pi $:
$$
20 = 4r^2
$$

Now solve for $ r^2 $:
$$
r^2 = \frac{20}{4} = 5
$$

Take the square root of both sides:
$$
r = \sqrt{5}
$$

---

### Step 4: Final Answer

The radius of the tank is:
$$
\boxed{\sqrt{5}}
$$

This is the simplest radical form of the radius.

[Assistant]
Explanation:
Okay, so I need to find the radius of a right cylindrical tank. The problem says that the tank is being filled with water at a rate of 20π cubic meters per hour, and as the tank is filled, the water level rises four meters per hour. Hmm, let me think about how to approach this.

First, I remember that the volume of a cylinder is given by the formula V = πr²h, where r is the radius and h is the height (or in this case, the water level). The problem mentions the rate at which the volume is increasing and the rate at which the height is increasing. So, this seems like a related rates problem from calculus. But wait, maybe I don't need calculus here? Let me check.

The volume is increasing at 20π cubic meters per hour, which is dV/dt = 20π. The height is increasing at 4 meters per hour, so dh/dt = 4. I need to find the radius r. Since the volume of the cylinder is V = πr²h, if I differentiate both sides with respect to time t, I can relate dV/dt and dh/dt. Let me try that.

Differentiating V = πr²h with respect to t:

dV/dt = π [ d/dt(r²h) ]

Since r is the radius of the tank, I assume it's constant, right? The tank is a right cylinder with circular bases, so the radius doesn't change as the water level rises. Therefore, r is a constant, and so when differentiating, we can treat r² as a constant. Therefore:

dV/dt = π * r² * dh/dt

Wait, that seems straightforward. So if I plug in the known values:

20π = π * r² * 4

Then, I can divide both sides by π:

20 = 4r²

Then divide both sides by 4:

5 = r²

Therefore, r = √5. So the radius is √5 meters. Is that it? Let me check again.

Wait, let me make sure I didn't skip any steps or make any wrong assumptions. The key assumption here is that the radius is constant. Since the tank is a right cylinder, the radius doesn't change as the water level rises. So, the cross-sectional area of the tank is πr², and the rate at which the volume increases is equal to the cross-sectional area multiplied by the rate at which the height increases. So, dV/dt = A * dh/dt, where A is the cross-sectional area. Therefore, 20π = πr² * 4. Then, solving for r² gives 20π / (4π) = 5, so r² = 5, so r = √5. That seems correct.

But let me think again. Maybe I need to consider something else? For example, is there a possibility that the tank isn't full? But the problem doesn't mention anything about the tank's total height or whether it's being filled to the top. However, since they are giving the rate of change of the water level, it's possible that this relationship holds regardless of the current height, as long as the radius is constant. Therefore, even if the tank is only partially filled, the rate at which the volume increases would still depend on the cross-sectional area. Therefore, my calculation should hold.

Alternatively, if the tank had a varying radius, then we would need more information, but since it's a right cylinder, the radius is constant. Therefore, the answer should be √5 meters. Let me check the units to make sure. The rate of volume is in cubic meters per hour, and the rate of height is in meters per hour. The cross-sectional area would be in square meters, so multiplying by meters per hour gives cubic meters per hour, which matches. So units are consistent.

Therefore, I think my answer is correct. The radius is √5 meters. So, expressed in simplest radical form, that's just √5. Therefore, the answer is √5.

Wait, but let me check once more. Let me write down the equations again.

Given:

dV/dt = 20π m³/h

dh/dt = 4 m/h

Find r.

We know that V = πr²h. Differentiating with respect to t:

dV/dt = πr² dh/dt

So solving for r:

r² = (dV/dt) / (π dh/dt)

Wait, no:

Wait, dV/dt = πr² dh/dt => r² = (dV/dt) / (π dh/dt)

Wait, but substituting numbers:

r² = (20π) / (π * 4) = 20 / 4 = 5. So r² = 5, so r = √5. Yes, same result. So that's correct.

Therefore, I think the answer is √5. So I think that's the radius. I don't see any mistakes here. Maybe I can think of an example. Suppose the radius is √5, then cross-sectional area is π*(√5)^2 = 5π. Then, if the height increases by 4 meters per hour, then the volume increase per hour is 5π * 4 = 20π, which matches the given rate. So that checks out. Therefore, the answer is √5.

**Final Answer**
The radius of the tank is \boxed{\sqrt{5}} meters.

To determine the radius of the cylindrical tank, we begin by analyzing the relationship between the volume of water and the height of the water level as the tank is being filled.

---

### Step 1: Understand the Given Information

- The **rate of change of volume** is given as:
  $$
  \frac{dV}{dt} = 20\pi \text{ cubic meters per hour}
  $$
- The **rate of change of height** is:
  $$
  \frac{dh}{dt} = 4 \text{ meters per hour}
  $$
- The tank is a **right circular cylinder**, so the **radius $ r $** is constant.

---

### Step 2: Use the Volume Formula for a Cylinder

The volume $ V $ of a cylinder is given by:
$$
V = \pi r^2 h
$$
Differentiating both sides with respect to time $ t $, we get:
$$
\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}
$$

This equation relates the rate of change of volume to the rate of change of height, and it allows us to solve for the radius $ r $.

---

### Step 3: Plug in the Known Values

Substitute the given values into the differentiated equation:
$$
20\pi = \pi r^2 \cdot 4
$$

Divide both sides by $ \pi $:
$$
20 = 4r^2
$$

Now solve for $ r^2 $:
$$
r^2 = \frac{20}{4} = 5
$$

Take the square root of both sides:
$$
r = \sqrt{5}
$$

---

### Step 4: Final Answer

The radius of the tank is:
$$
\boxed{\sqrt{5}}
$$

This is the simplest radical form of the radius.

Solution:
The volume of the water increases by $20\pi$ cubic meters each hour, while the height of the water in the tank rises by 4 meters each hour. The volume of a right cylinder is $\pi r^2h$. If we look at the changes in volume and height in just one hour, we can solve for the radius.  \begin{align*}
\pi r^2h_f-\pi r^2h_0&=V_f-V_0\quad\Rightarrow\\
\pi r^2(\Delta h)&=\Delta V\quad\Rightarrow\\
\pi r^2(4)&=20\pi\quad\Rightarrow\\
4r^2&=20\quad\Rightarrow\\
r^2&=5
\end{align*} Since the radius must be positive, $r=\boxed{\sqrt{5}}$ meters.


Explanation:
Okay, so I need to find the greatest common divisor (GCD) of three numbers: 3339, 2961, and 1491. Hmm, let me remember how to do this. I think the GCD of multiple numbers can be found by first finding the GCD of two numbers and then finding the GCD of that result with the third number. So maybe I should start by finding the GCD of 3339 and 2961, and then take that result and find the GCD with 1491. Let me try that approach.

First, let me recall the Euclidean algorithm for finding the GCD of two numbers. The Euclidean algorithm says that GCD(a, b) = GCD(b, a mod b), right? So I can use that to find the GCD of 3339 and 2961. Let me write that down.

Let me compute 3339 divided by 2961. Well, 2961 goes into 3339 once, with a remainder. Let me calculate the remainder: 3339 - 2961 = 378. So, GCD(3339, 2961) = GCD(2961, 378). Now I need to find GCD(2961, 378). Let me do the same step again.

Divide 2961 by 378. Let me see how many times 378 goes into 2961. Let me calculate 378 * 7 = 2646. Then 378 * 8 = 3024, which is too big because 3024 is larger than 2961. So it goes in 7 times with a remainder. Let me subtract 2646 from 2961: 2961 - 2646 = 315. So now, GCD(2961, 378) = GCD(378, 315).

Next step: GCD(378, 315). Let me divide 378 by 315. That goes in once, with a remainder of 378 - 315 = 63. So now it's GCD(315, 63).

Now, GCD(315, 63). Let me divide 315 by 63. 63 * 5 = 315, so the remainder is 0. That means the GCD is 63. So, GCD(3339, 2961) is 63.

Okay, so now I need to find the GCD of 63 and the third number, which is 1491. Let me do that. So GCD(63, 1491). Let me apply the Euclidean algorithm again.

Divide 1491 by 63. Let me see how many times 63 goes into 1491. Let me calculate 63 * 20 = 1260. Then 1491 - 1260 = 231. So the remainder is 231. Therefore, GCD(1491, 63) = GCD(63, 231).

Now, GCD(63, 231). Let me divide 231 by 63. 63 * 3 = 189. Subtract that from 231: 231 - 189 = 42. So now it's GCD(63, 42).

Next step: GCD(63, 42). Divide 63 by 42. That goes in once with a remainder of 63 - 42 = 21. So now it's GCD(42, 21).

Divide 42 by 21. That goes in exactly 2 times with remainder 0. Therefore, the GCD is 21.

So putting it all together, the GCD of 3339, 2961, and 1491 is 21. Wait, let me check if that's correct. Maybe I made a mistake somewhere. Let me verify each step again.

First, GCD(3339, 2961):

3339 divided by 2961 is 1 with remainder 378. Then GCD(2961, 378). 2961 divided by 378: 378*7=2646, 2961-2646=315. Then GCD(378, 315). 378 divided by 315 is 1 with remainder 63. Then GCD(315, 63). 315 divided by 63 is 5 with remainder 0. So yes, GCD is 63. Then GCD(63, 1491). 1491 divided by 63: 63*20=1260, 1491-1260=231. Then GCD(63, 231). 231 divided by 63 is 3 with remainder 42. Then GCD(63, 42). 63 divided by 42 is 1 with remainder 21. Then GCD(42, 21). 42 divided by 21 is 2 with remainder 0. So GCD is 21. That seems correct.

But let me check if 21 divides all three numbers. Let me check 3339 divided by 21. 21*100=2100, 3339-2100=1239. 21*50=1050, 1239-1050=189. 21*9=189. So total is 100+50+9=159. So 21*159=3339. Okay, that works.

Now 2961 divided by 21. Let me do 21*100=2100, 2961-2100=861. 21*40=840, 861-840=21. So that's 100+40+1=141. 21*141=2961. Correct.

Now 1491 divided by 21. Let me see. 21*70=1470. 1491-1470=21. So that's 70+1=71. 21*71=1491. Correct. So 21 divides all three numbers. Is there a larger number that divides all three? Let me check if 63 divides 1491. Wait, earlier steps said that GCD(63, 1491) is 21, so 63 does not divide 1491. Let me check. 1491 divided by 63. 63*20=1260, 1491-1260=231. 231 divided by 63 is 3.75? Wait, 63*3=189, 231-189=42. So 63*3 + 42 = 231. So 1491 divided by 63 is 23. So 63*23=1491? Wait, 63*20=1260, 63*3=189, so 1260+189=1449. But 1449 is less than 1491. 1491-1449=42. So 63*23=1449, remainder 42. Therefore, 63 does not divide 1491. Therefore, 63 is not a common divisor. Therefore, 21 is the GCD.

But just to be thorough, let me check if there is a higher common divisor. Let me check if 21 is the highest. Let me check if 21 is the GCD. Since we used the Euclidean algorithm, which is reliable, and verified that 21 divides all three numbers, and that there is no higher number, since the GCD of 63 and 1491 is 21, which is less than 63. Therefore, 21 is the GCD.

Alternatively, maybe I can factor all three numbers into primes and see. Let me try that approach as a cross-check.

Starting with 3339. Let me factor 3339. Let me check divisibility by small primes. First, check if it's even? No. Sum of digits: 3+3+3+9=18, which is divisible by 9, so 3339 is divisible by 9. Let me divide 3339 by 9. 9*371=3339? Let me check 9*300=2700, 9*70=630, 9*1=9. 2700+630=3330, plus 9 is 3339. Yes. So 3339=9*371. Now factor 371. Let me check divisibility. 371 divided by 7? 7*53=371? 7*50=350, 7*3=21, so 350+21=371. Yes! So 371=7*53. Therefore, 3339=9*7*53. Which is 3^2 * 7 * 53.

Now 2961. Let me factor 2961. Again, check divisibility by 3. Sum of digits: 2+9+6+1=18, which is divisible by 9. So 2961 is divisible by 9. Let me divide 2961 by 9. 9*300=2700, 2961-2700=261. 261 divided by 9 is 29. So total is 300 + 29 = 329. Therefore, 2961=9*329. Now factor 329. Let me check divisibility. 329 divided by 7? 7*47=329? 7*40=280, 7*7=49, so 280+49=329. Yes! So 329=7*47. Therefore, 2961=9*7*47=3^2 *7*47.

Now 1491. Let me factor 1491. Check divisibility by 3: 1+4+9+1=15, which is divisible by 3. So divide by 3: 1491 / 3 = 497. Now factor 497. Let me check divisibility. 497 divided by 7? 7*71=497? 7*70=490, plus 7 is 497. Yes! So 497=7*71. Therefore, 1491=3*7*71.

So now, the prime factors:

- 3339: 3² * 7 * 53
- 2961: 3² * 7 * 47
- 1491: 3 * 7 * 71

So the common prime factors among all three numbers are 3 and 7. The minimum exponents for these primes:

- For 3: the exponents are 2, 2, and 1. So minimum is 1.
- For 7: exponents are 1, 1, 1. So minimum is 1.

Therefore, GCD is 3^1 * 7^1 = 3*7=21. Which matches the previous result. So that confirms that the GCD is indeed 21.

Therefore, I can be confident that the greatest common divisor of 3339, 2961, and 1491 is 21.

**Final Answer**
The greatest common divisor of $3339$, $2961$, and $1491$ is \boxed{21}.

To find the **greatest common divisor (GCD)** of the three numbers $ 3339 $, $ 2961 $, and $ 1491 $, we can use the **Euclidean algorithm**, which is a reliable method for computing the GCD of two numbers. Since we are dealing with three numbers, we can apply the algorithm iteratively: first compute the GCD of the first two numbers, and then compute the GCD of that result with the third number.

---

### Step 1: Compute GCD(3339, 2961)

We apply the Euclidean algorithm:

$$
\begin{align*}
3339 \div 2961 &= 1 \text{ remainder } 378 \\
2961 \div 378 &= 7 \text{ remainder } 315 \\
378 \div 315 &= 1 \text{ remainder } 63 \\
315 \div 63 &= 5 \text{ remainder } 0 \\
\end{align*}
$$

So, $ \gcd(3339, 2961) = 63 $

---

### Step 2: Compute GCD(63, 1491)

Now we compute the GCD of the result from the previous step (63) with the third number (1491):

$$
\begin{align*}
1491 \div 63 &= 23 \text{ remainder } 42 \\
63 \div 42 &= 1 \text{ remainder } 21 \\
42 \div 21 &= 2 \text{ remainder } 0 \\
\end{align*}
$$

So, $ \gcd(63, 1491) = 21 $

---

### Step 3: Final Result

Therefore, the GCD of all three numbers is:

$$
\gcd(3339, 2961, 1491) = \gcd(63, 1491) = 21
$$

---

### Verification via Prime Factorization

To double-check, we can factor each number into its prime components:

- $ 3339 = 3^2 \times 7 \times 53 $
- $ 2961 = 3^2 \times 7 \times 47 $
- $ 1491 = 3 \times 7 \times 71 $

The common prime factors among all three are $ 3 $ and $ 7 $, with the smallest exponents being $ 3^1 $ and $ 7^1 $. Thus:

$$
\gcd = 3^1 \times 7^1 = 21
$$

---

### Final Answer

$$
\boxed{21}
$$

[Assistant]
Explanation:
Okay, so I need to find the greatest common divisor (GCD) of three numbers: 3339, 2961, and 1491. Hmm, let me remember how to do this. I think the GCD of multiple numbers can be found by first finding the GCD of two numbers and then finding the GCD of that result with the third number. So maybe I should start by finding the GCD of 3339 and 2961, and then take that result and find the GCD with 1491. Let me try that approach.

First, let me recall the Euclidean algorithm for finding the GCD of two numbers. The Euclidean algorithm says that GCD(a, b) = GCD(b, a mod b), right? So I can use that to find the GCD of 3339 and 2961. Let me write that down.

Let me compute 3339 divided by 2961. Well, 2961 goes into 3339 once, with a remainder. Let me calculate the remainder: 3339 - 2961 = 378. So, GCD(3339, 2961) = GCD(2961, 378). Now I need to find GCD(2961, 378). Let me do the same step again.

Divide 2961 by 378. Let me see how many times 378 goes into 2961. Let me calculate 378 * 7 = 2646. Then 378 * 8 = 3024, which is too big because 3024 is larger than 2961. So it goes in 7 times with a remainder. Let me subtract 2646 from 2961: 2961 - 2646 = 315. So now, GCD(2961, 378) = GCD(378, 315).

Next step: GCD(378, 315). Let me divide 378 by 315. That goes in once, with a remainder of 378 - 315 = 63. So now it's GCD(315, 63).

Now, GCD(315, 63). Let me divide 315 by 63. 63 * 5 = 315, so the remainder is 0. That means the GCD is 63. So, GCD(3339, 2961) is 63.

Okay, so now I need to find the GCD of 63 and the third number, which is 1491. Let me do that. So GCD(63, 1491). Let me apply the Euclidean algorithm again.

Divide 1491 by 63. Let me see how many times 63 goes into 1491. Let me calculate 63 * 20 = 1260. Then 1491 - 1260 = 231. So the remainder is 231. Therefore, GCD(1491, 63) = GCD(63, 231).

Now, GCD(63, 231). Let me divide 231 by 63. 63 * 3 = 189. Subtract that from 231: 231 - 189 = 42. So now it's GCD(63, 42).

Next step: GCD(63, 42). Divide 63 by 42. That goes in once with a remainder of 63 - 42 = 21. So now it's GCD(42, 21).

Divide 42 by 21. That goes in exactly 2 times with remainder 0. Therefore, the GCD is 21.

So putting it all together, the GCD of 3339, 2961, and 1491 is 21. Wait, let me check if that's correct. Maybe I made a mistake somewhere. Let me verify each step again.

First, GCD(3339, 2961):

3339 divided by 2961 is 1 with remainder 378. Then GCD(2961, 378). 2961 divided by 378: 378*7=2646, 2961-2646=315. Then GCD(378, 315). 378 divided by 315 is 1 with remainder 63. Then GCD(315, 63). 315 divided by 63 is 5 with remainder 0. So yes, GCD is 63. Then GCD(63, 1491). 1491 divided by 63: 63*20=1260, 1491-1260=231. Then GCD(63, 231). 231 divided by 63 is 3 with remainder 42. Then GCD(63, 42). 63 divided by 42 is 1 with remainder 21. Then GCD(42, 21). 42 divided by 21 is 2 with remainder 0. So GCD is 21. That seems correct.

But let me check if 21 divides all three numbers. Let me check 3339 divided by 21. 21*100=2100, 3339-2100=1239. 21*50=1050, 1239-1050=189. 21*9=189. So total is 100+50+9=159. So 21*159=3339. Okay, that works.

Now 2961 divided by 21. Let me do 21*100=2100, 2961-2100=861. 21*40=840, 861-840=21. So that's 100+40+1=141. 21*141=2961. Correct.

Now 1491 divided by 21. Let me see. 21*70=1470. 1491-1470=21. So that's 70+1=71. 21*71=1491. Correct. So 21 divides all three numbers. Is there a larger number that divides all three? Let me check if 63 divides 1491. Wait, earlier steps said that GCD(63, 1491) is 21, so 63 does not divide 1491. Let me check. 1491 divided by 63. 63*20=1260, 1491-1260=231. 231 divided by 63 is 3.75? Wait, 63*3=189, 231-189=42. So 63*3 + 42 = 231. So 1491 divided by 63 is 23. So 63*23=1491? Wait, 63*20=1260, 63*3=189, so 1260+189=1449. But 1449 is less than 1491. 1491-1449=42. So 63*23=1449, remainder 42. Therefore, 63 does not divide 1491. Therefore, 63 is not a common divisor. Therefore, 21 is the GCD.

But just to be thorough, let me check if there is a higher common divisor. Let me check if 21 is the highest. Let me check if 21 is the GCD. Since we used the Euclidean algorithm, which is reliable, and verified that 21 divides all three numbers, and that there is no higher number, since the GCD of 63 and 1491 is 21, which is less than 63. Therefore, 21 is the GCD.

Alternatively, maybe I can factor all three numbers into primes and see. Let me try that approach as a cross-check.

Starting with 3339. Let me factor 3339. Let me check divisibility by small primes. First, check if it's even? No. Sum of digits: 3+3+3+9=18, which is divisible by 9, so 3339 is divisible by 9. Let me divide 3339 by 9. 9*371=3339? Let me check 9*300=2700, 9*70=630, 9*1=9. 2700+630=3330, plus 9 is 3339. Yes. So 3339=9*371. Now factor 371. Let me check divisibility. 371 divided by 7? 7*53=371? 7*50=350, 7*3=21, so 350+21=371. Yes! So 371=7*53. Therefore, 3339=9*7*53. Which is 3^2 * 7 * 53.

Now 2961. Let me factor 2961. Again, check divisibility by 3. Sum of digits: 2+9+6+1=18, which is divisible by 9. So 2961 is divisible by 9. Let me divide 2961 by 9. 9*300=2700, 2961-2700=261. 261 divided by 9 is 29. So total is 300 + 29 = 329. Therefore, 2961=9*329. Now factor 329. Let me check divisibility. 329 divided by 7? 7*47=329? 7*40=280, 7*7=49, so 280+49=329. Yes! So 329=7*47. Therefore, 2961=9*7*47=3^2 *7*47.

Now 1491. Let me factor 1491. Check divisibility by 3: 1+4+9+1=15, which is divisible by 3. So divide by 3: 1491 / 3 = 497. Now factor 497. Let me check divisibility. 497 divided by 7? 7*71=497? 7*70=490, plus 7 is 497. Yes! So 497=7*71. Therefore, 1491=3*7*71.

So now, the prime factors:

- 3339: 3² * 7 * 53
- 2961: 3² * 7 * 47
- 1491: 3 * 7 * 71

So the common prime factors among all three numbers are 3 and 7. The minimum exponents for these primes:

- For 3: the exponents are 2, 2, and 1. So minimum is 1.
- For 7: exponents are 1, 1, 1. So minimum is 1.

Therefore, GCD is 3^1 * 7^1 = 3*7=21. Which matches the previous result. So that confirms that the GCD is indeed 21.

Therefore, I can be confident that the greatest common divisor of 3339, 2961, and 1491 is 21.

**Final Answer**
The greatest common divisor of $3339$, $2961$, and $1491$ is \boxed{21}.

To find the **greatest common divisor (GCD)** of the three numbers $ 3339 $, $ 2961 $, and $ 1491 $, we can use the **Euclidean algorithm**, which is a reliable method for computing the GCD of two numbers. Since we are dealing with three numbers, we can apply the algorithm iteratively: first compute the GCD of the first two numbers, and then compute the GCD of that result with the third number.

---

### Step 1: Compute GCD(3339, 2961)

We apply the Euclidean algorithm:

$$
\begin{align*}
3339 \div 2961 &= 1 \text{ remainder } 378 \\
2961 \div 378 &= 7 \text{ remainder } 315 \\
378 \div 315 &= 1 \text{ remainder } 63 \\
315 \div 63 &= 5 \text{ remainder } 0 \\
\end{align*}
$$

So, $ \gcd(3339, 2961) = 63 $

---

### Step 2: Compute GCD(63, 1491)

Now we compute the GCD of the result from the previous step (63) with the third number (1491):

$$
\begin{align*}
1491 \div 63 &= 23 \text{ remainder } 42 \\
63 \div 42 &= 1 \text{ remainder } 21 \\
42 \div 21 &= 2 \text{ remainder } 0 \\
\end{align*}
$$

So, $ \gcd(63, 1491) = 21 $

---

### Step 3: Final Result

Therefore, the GCD of all three numbers is:

$$
\gcd(3339, 2961, 1491) = \gcd(63, 1491) = 21
$$

---

### Verification via Prime Factorization

To double-check, we can factor each number into its prime components:

- $ 3339 = 3^2 \times 7 \times 53 $
- $ 2961 = 3^2 \times 7 \times 47 $
- $ 1491 = 3 \times 7 \times 71 $

The common prime factors among all three are $ 3 $ and $ 7 $, with the smallest exponents being $ 3^1 $ and $ 7^1 $. Thus:

$$
\gcd = 3^1 \times 7^1 = 21
$$

---

### Final Answer

$$
\boxed{21}
$$

Solution:
We can perform the Euclidean Algorithm two times.

First, we use it for $3339$ and $2961$. \begin{align*}
\text{gcd}\,(3339,2961) &=\text{gcd}\,(3339-2961,2961)\\
&=\text{gcd}\,(378,2961)\\
&=\text{gcd}\,(378,2961-378 \cdot 7)\\
&=\text{gcd}\,(378,315)\\
&=\text{gcd}\,(378-315,315)\\
&=\text{gcd}\,(63,315)\\
\end{align*}Since $63$ is a divisor of $315$, the greatest common divisor of $3339$ and $2961$ is $63$.

Next, we can find the greatest common divisor of $63$ and $1491$, also using the Euclidean Algorithm. \begin{align*}
\text{gcd}\,(63,1491) &=\text{gcd}\,(63,1491-63 \cdot 23)\\
&=\text{gcd}\,(63,42)\\
\end{align*}Since $63=3 \cdot 21$ and $42=2 \cdot 21$, the greatest common divisor is $\boxed{21}$.