Explanation:
Okay, so I need to figure out the value of 9 & 2 using this operation they defined. Let me start by recalling the definition they gave: a & b equals the square root of (ab + a) divided by the square root of (ab - b). So, written out, that's:

a & b = √(ab + a) / √(ab - b)

Alright, so for 9 & 2, that would mean plugging in a = 9 and b = 2 into this formula. Let me write that down step by step.

First, compute the numerator part: √(ab + a). Let me calculate ab first. a is 9 and b is 2, so ab is 9*2 = 18. Then add a to that: 18 + 9 = 27. So the numerator is √27.

Then the denominator part: √(ab - b). Again, ab is 18, and subtract b which is 2: 18 - 2 = 16. So the denominator is √16.

Therefore, putting it all together, 9 & 2 is √27 divided by √16. Let me write that as:

√27 / √16

Hmm, okay. Now, I need to simplify this. Let me think. √16 is easy, that's 4. And √27 can be simplified. Since 27 is 9*3, and √9 is 3, so √27 is 3√3. So then, substituting that in:

(3√3) / 4

Is that the simplest radical form? Let me check if there's any simplification I missed. The numerator is 3√3 and denominator is 4. There's no common factors between 3 and 4, so I think that's as simplified as it gets. So the answer should be 3√3 over 4.

Wait, but just to make sure I didn't make any mistakes in the calculations. Let me go through it again.

First, ab = 9*2 = 18. Then numerator is ab + a = 18 + 9 = 27. Correct. Denominator is ab - b = 18 - 2 = 16. Correct. Then √27 is 3√3, and √16 is 4. So dividing them gives 3√3 / 4. Yeah, that seems right.

Alternatively, maybe I can combine the square roots into one? Like √(27/16). Since √a / √b is √(a/b). So maybe that's another way to look at it. Let me check that:

√(27/16) would be equal to √27 / √16, which is the same as before. Then √27 is 3√3 and √16 is 4, so yeah, same answer. So either way, I get 3√3 / 4. So that seems to check out.

But just to make sure there isn't some trick or something I missed. Let me verify the original problem again. The operation is defined for positive integers a and b. Since 9 and 2 are both positive integers, that's fine. Also, the denominator is √(ab - b). Let me check if that is a positive number. ab - b = b(a - 1). For b = 2, a = 9, so 2*(9 - 1) = 2*8 = 16, which is positive, so the square root is defined. So no issues there. If the denominator had been zero or negative, we'd have a problem, but here it's 16. So all good.

So, yeah, I think I did everything right. The answer is 3√3 over 4. Let me write that as a common fraction in simplest radical form. Since 3 and 4 have no common factors, and √3 can't be simplified further, that's the simplest form. So yeah, 3√3 / 4.

Just to recap:

9 & 2 = √(9*2 + 9) / √(9*2 - 2) = √(18 + 9)/√(18 - 2) = √27 / √16 = 3√3 / 4.

Yep, that all lines up. I think that's the answer.

**Final Answer**
The value of $9 \& 2$ is \boxed{\dfrac{3\sqrt{3}}{4}}.

We are given a custom operation defined for positive integers $ a $ and $ b $ as follows:

$$
a \& b = \frac{\sqrt{ab + a}}{\sqrt{ab - b}}
$$

We are asked to find the value of $ 9 \& 2 $, which means we substitute $ a = 9 $ and $ b = 2 $ into the formula.

---

### Step 1: Calculate the Numerator
First, compute the expression under the square root in the numerator:

$$
ab + a = 9 \cdot 2 + 9 = 18 + 9 = 27
$$

So, the numerator becomes:

$$
\sqrt{27}
$$

---

### Step 2: Simplify the Numerator
We simplify $ \sqrt{27} $ by factoring out perfect squares:

$$
\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}
$$

---

### Step 3: Calculate the Denominator
Now compute the expression under the square root in the denominator:

$$
ab - b = 9 \cdot 2 - 2 = 18 - 2 = 16
$$

So, the denominator becomes:

$$
\sqrt{16} = 4
$$

---

### Step 4: Combine and Simplify
Now plug the simplified values into the original operation definition:

$$
9 \& 2 = \frac{3\sqrt{3}}{4}
$$

This is a common fraction in its simplest radical form, as the denominator is rational, and the radical cannot be simplified further.

---

### Final Answer

$$
\boxed{\dfrac{3\sqrt{3}}{4}}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the value of 9 & 2 using this operation they defined. Let me start by recalling the definition they gave: a & b equals the square root of (ab + a) divided by the square root of (ab - b). So, written out, that's:

a & b = √(ab + a) / √(ab - b)

Alright, so for 9 & 2, that would mean plugging in a = 9 and b = 2 into this formula. Let me write that down step by step.

First, compute the numerator part: √(ab + a). Let me calculate ab first. a is 9 and b is 2, so ab is 9*2 = 18. Then add a to that: 18 + 9 = 27. So the numerator is √27.

Then the denominator part: √(ab - b). Again, ab is 18, and subtract b which is 2: 18 - 2 = 16. So the denominator is √16.

Therefore, putting it all together, 9 & 2 is √27 divided by √16. Let me write that as:

√27 / √16

Hmm, okay. Now, I need to simplify this. Let me think. √16 is easy, that's 4. And √27 can be simplified. Since 27 is 9*3, and √9 is 3, so √27 is 3√3. So then, substituting that in:

(3√3) / 4

Is that the simplest radical form? Let me check if there's any simplification I missed. The numerator is 3√3 and denominator is 4. There's no common factors between 3 and 4, so I think that's as simplified as it gets. So the answer should be 3√3 over 4.

Wait, but just to make sure I didn't make any mistakes in the calculations. Let me go through it again.

First, ab = 9*2 = 18. Then numerator is ab + a = 18 + 9 = 27. Correct. Denominator is ab - b = 18 - 2 = 16. Correct. Then √27 is 3√3, and √16 is 4. So dividing them gives 3√3 / 4. Yeah, that seems right.

Alternatively, maybe I can combine the square roots into one? Like √(27/16). Since √a / √b is √(a/b). So maybe that's another way to look at it. Let me check that:

√(27/16) would be equal to √27 / √16, which is the same as before. Then √27 is 3√3 and √16 is 4, so yeah, same answer. So either way, I get 3√3 / 4. So that seems to check out.

But just to make sure there isn't some trick or something I missed. Let me verify the original problem again. The operation is defined for positive integers a and b. Since 9 and 2 are both positive integers, that's fine. Also, the denominator is √(ab - b). Let me check if that is a positive number. ab - b = b(a - 1). For b = 2, a = 9, so 2*(9 - 1) = 2*8 = 16, which is positive, so the square root is defined. So no issues there. If the denominator had been zero or negative, we'd have a problem, but here it's 16. So all good.

So, yeah, I think I did everything right. The answer is 3√3 over 4. Let me write that as a common fraction in simplest radical form. Since 3 and 4 have no common factors, and √3 can't be simplified further, that's the simplest form. So yeah, 3√3 / 4.

Just to recap:

9 & 2 = √(9*2 + 9) / √(9*2 - 2) = √(18 + 9)/√(18 - 2) = √27 / √16 = 3√3 / 4.

Yep, that all lines up. I think that's the answer.

**Final Answer**
The value of $9 \& 2$ is \boxed{\dfrac{3\sqrt{3}}{4}}.

We are given a custom operation defined for positive integers $ a $ and $ b $ as follows:

$$
a \& b = \frac{\sqrt{ab + a}}{\sqrt{ab - b}}
$$

We are asked to find the value of $ 9 \& 2 $, which means we substitute $ a = 9 $ and $ b = 2 $ into the formula.

---

### Step 1: Calculate the Numerator
First, compute the expression under the square root in the numerator:

$$
ab + a = 9 \cdot 2 + 9 = 18 + 9 = 27
$$

So, the numerator becomes:

$$
\sqrt{27}
$$

---

### Step 2: Simplify the Numerator
We simplify $ \sqrt{27} $ by factoring out perfect squares:

$$
\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}
$$

---

### Step 3: Calculate the Denominator
Now compute the expression under the square root in the denominator:

$$
ab - b = 9 \cdot 2 - 2 = 18 - 2 = 16
$$

So, the denominator becomes:

$$
\sqrt{16} = 4
$$

---

### Step 4: Combine and Simplify
Now plug the simplified values into the original operation definition:

$$
9 \& 2 = \frac{3\sqrt{3}}{4}
$$

This is a common fraction in its simplest radical form, as the denominator is rational, and the radical cannot be simplified further.

---

### Final Answer

$$
\boxed{\dfrac{3\sqrt{3}}{4}}
$$

Solution: We have $9\&2 = \frac{\sqrt{(9)(2)+9}}{\sqrt{(9)(2)-2}} = \frac{\sqrt{27}}{\sqrt{16}} = \boxed{\frac{3\sqrt{3}}{4}}.$


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's see. Let me think about how to approach this.

First, I remember 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 possible outcomes when throwing two dice should be 6 multiplied by 6, which is 36. So, there are 36 possible outcomes in total. That part seems straightforward.

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, which means that the product must have 5 as one of its prime factors. Since we're dealing with the product of two numbers, this would require that at least one of the numbers is a multiple of 5. Wait, is that right? Let me think again.

Actually, no. If the product is a multiple of 5, then at least one of the numbers must be a multiple of 5. Because if neither of the numbers is a multiple of 5, then their product won't have a factor of 5. Conversely, if at least one is a multiple of 5, then the product will be a multiple of 5. So, that seems correct.

Therefore, maybe I can approach this problem by calculating the probability that at least one die shows a 5. But wait, let me confirm this logic. Let me think of some examples. Suppose one die is 5 and the other is 2. Then the product is 10, which is a multiple of 5. If one die is 5 and the other is 5, product is 25, which is also a multiple of 5. If neither die is 5, say 2 and 3, the product is 6, which is not a multiple of 5. If one die is 5 and the other is 1, product is 5, which is a multiple of 5. So yes, it seems that the product is a multiple of 5 if and only if at least one die is a 5.

Wait, but hold on. Suppose one die is 5 and the other is... let me think. If one die is 5, regardless of the other die, the product is going to be a multiple of 5. If neither die is 5, then the product will not be a multiple of 5. Therefore, the number of favorable outcomes is the number of outcomes where at least one die is 5. Therefore, the probability we want is equal to the probability that at least one die shows a 5.

Alternatively, maybe I can use the complementary probability. Instead of calculating the probability of at least one 5, I can calculate 1 minus the probability that neither die shows a 5. That might be easier.

Let me try both approaches and see if they match.

First approach: Directly calculating the number of outcomes with at least one 5.

How many outcomes have at least one 5? Well, for two dice, the number of outcomes where the first die is 5 is 6 (since the second die can be anything from 1 to 6). Similarly, the number of outcomes where the second die is 5 is also 6. However, if I just add them together, I have 6 + 6 = 12. But wait, this counts the outcome (5,5) twice. So, to correct for that, we need to subtract 1. Therefore, total number of favorable outcomes is 12 - 1 = 11? Wait, no. Wait, inclusion-exclusion principle: |A ∪ B| = |A| + |B| - |A ∩ B|. In this case, A is the set where first die is 5, B is the set where the second die is 5. So, |A| is 6, |B| is 6, and |A ∩ B| is 1 (only (5,5)). Therefore, |A ∪ B| = 6 + 6 - 1 = 11. So there are 11 favorable outcomes. Therefore, the probability would be 11/36.

Wait, but let me check if that's correct. Alternatively, maybe I should list all possibilities. For two dice, the number of outcomes where at least one die is 5 is:

First die is 5: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) → 6 outcomes.

Second die is 5: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5) → 6 outcomes.

But (5,5) is counted in both, so total unique outcomes are 6 + 6 - 1 = 11. So that's correct. So 11 outcomes. Therefore, the probability is 11/36. So then, is that the answer?

Wait, but let me think again. Let me check with the complementary approach.

Complementary probability: the probability that neither die is a 5. Then, the probability that the product is not a multiple of 5 is equal to the probability that neither die is a 5. Therefore, the probability we want is 1 minus that probability.

So, what's the probability that neither die is a 5? For each die, the probability of not rolling a 5 is 5/6. Since the dice are independent, the probability that neither is a 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. So same answer. So that seems to confirm that 11/36 is correct.

But wait a second, let me think again. Is there another way to think about this? Let me think if there's a possibility that even if neither die is a 5, the product could still be a multiple of 5. For example, suppose one die is 10 and the other is... but wait, the dice only go up to 6. So the numbers on the dice are 1 through 6. Therefore, the only way to have a product divisible by 5 is if at least one die is 5. Because 5 is the only multiple of 5 on a die. So 5 is the only number on a die that is divisible by 5. Therefore, if neither die is 5, then the product cannot be divisible by 5. Therefore, our previous calculations hold. Therefore, the probability is 11/36.

Wait, but hold on. Let me test with actual examples.

Total outcomes: 36.

Number of outcomes with at least one 5: 11.

Therefore, if I list them:

First die is 5:

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

Second die is 5:

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

But as mentioned, (5,5) is duplicated, so total unique is 6 + 5 = 11. So yes, 11. Then, the probability is 11/36.

But let me verify with another approach. Let me think about how many products are multiples of 5.

Alternatively, perhaps I can think about the possible products. But that might be more complicated.

Alternatively, think about the possible numbers on the dice. Each die has numbers 1 to 6.

So, let me consider all possible pairs. But maybe that's too tedious. Alternatively, think about the numbers.

The product is multiple of 5 iff at least one number is multiple of 5. Since 5 is prime. Since 5 is prime, the product is divisible by 5 if and only if at least one factor is divisible by 5. Since the numbers on the dice are 1,2,3,4,5,6. The only number divisible by 5 is 5 itself. So yes, if at least one die is 5, then product is multiple of 5. Otherwise, not. So that's correct.

Therefore, our answer is 11/36. But wait, let me check again, maybe I made a mistake here. Wait, 5 is on one die. Wait, but suppose that one die is 5 and the other is... for example, suppose one die is 5 and the other is 5. Product is 25, which is multiple of 5. If one die is 5 and the other is 1, product is 5. If one is 5 and other is 2, product is 10. All multiples of 5. If neither is 5, then the numbers are from 1-4 and 1-4, so the product is 1-16, but 5 is not in that range. Wait, but 5 is not in the products? Wait, for example, if you have 2 and 3, product is 6. 1 and 5, but if neither die is 5, then you can't have 5 in the product? Wait, no. Wait, if neither die is 5, then the product can't have a factor of 5. So even if one die is 5, you can have 5 in the product. So that's correct.

Therefore, the logic holds. Therefore, the answer is 11/36. Wait, but let me check again with numbers. Let me count all the outcomes where at least one die is 5.

First die is 5: 6 outcomes.

Second die is 5: 6 outcomes.

But overlapping at (5,5). So total is 6 + 6 - 1 = 11. So 11 outcomes. So 11/36. Therefore, probability is 11/36.

But wait, just to be thorough, maybe I can list all possible pairs where at least one die is 5. Let me try:

First die is 5:

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

Second die is 5:

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

Now, combine them:

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5). That's 11 outcomes. So yes. So 11 outcomes. Therefore, 11/36 is correct.

But let me think again—could there be another case where the product is a multiple of 5 even if neither die is 5? For example, if one die is 10, but we don't have 10 on a die. The maximum is 6. So no, there's no way. If the numbers are 1-6, then the only multiple of 5 is 5 itself. Therefore, if neither die is 5, the product cannot be a multiple of 5. So yes, that's correct.

Therefore, the answer should be 11/36. Wait, but let me check with another method. Let me compute the possible products and count how many are multiples of 5.

But that might take longer, but let me try.

First, the possible numbers on the dice are 1-6. Let me list all possible products. But there are 36 products. It's tedious, but maybe I can categorize them.

Alternatively, think of how many products are multiples of 5. So, for that, as we said, it's when at least one die is 5. So same as before.

But just to make sure, let me think of the products:

If one die is 5, the products are:

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

Similarly, for the other die:

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

So, those are the products. But as we saw earlier, there are 11 distinct outcomes. So the products are 5,10,15,20,25,30 and their counterparts, but since each of these products corresponds to two different outcomes except for (5,5). But the number of outcomes is 11. Therefore, that confirms again that there are 11 outcomes where the product is a multiple of 5.

Therefore, probability is 11/36. So I think that is the answer. But let me check with another approach.

Alternatively, think of the probability that the first die is 5 or the second die is 5. Which is P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Where A is the event that the first die is 5, and B is the event that the second die is 5.

P(A) = 1/6, P(B) = 1/6, P(A ∩ B) = 1/36. Therefore, P(A ∪ B) = 1/6 + 1/6 - 1/36 = 2/6 - 1/36 = 12/36 - 1/36 = 11/36. Same result. Therefore, this confirms again that the probability is 11/36.

Therefore, I think the answer is 11/36.

But wait, let me check once again. Maybe there's an error in assuming that only 5 is the multiple of 5 on the dice. Wait, 5 is on the die, but 10 isn't. Wait, but the product can be 10, which is a multiple of 5, even if one die is 5 and the other is 2, which is allowed. So that's okay.

Alternatively, if you have two dice, and you want the product to be divisible by 5, you need at least one of the numbers to be divisible by 5. Since 5 is prime, and the numbers on the dice are from 1-6, the only number divisible by 5 is 5 itself. Therefore, the logic holds. So no other numbers on the dice can contribute to the product being divisible by 5. Therefore, the answer is indeed 11/36.

But let me check if there's another angle. Suppose someone might think that numbers like 10 could be formed, but since 10 isn't on the die, but the product can be 10. But that's okay because the product being 10 is allowed. So that's acceptable. So even though 10 isn't on the die, the product can be 10. So that's not a problem.

Alternatively, maybe I should think about the probability that the product is not a multiple of 5, which is when both dice are not 5. So that is, as we said, 5/6 * 5/6 = 25/36, so 1 - 25/36 = 11/36. Therefore, same answer.

Therefore, after verifying through multiple approaches—direct counting, complementary probability, inclusion-exclusion principle—I think the answer is 11/36.

But just to be absolutely sure, maybe I can list all the possible products and count how many are multiples of 5. Although that's time-consuming, let me try.

First die: 1, second die: 1-6.

Products:

1*1=1, 1*2=2, 1*3=3, 1*4=4, 1*5=5, 1*6=6.

So, here, only 5 is a multiple of 5. That's 1 case.

First die: 2, second die: 1-6.

Products:

2*1=2, 2*2=4, 2*3=6, 2*4=8, 2*5=10, 2*6=12.

Here, 10 is a multiple of 5. So 1 case.

First die: 3, second die: 1-6.

Products:

3*1=3, 3*2=6, 3*3=9, 3*4=12, 3*5=15, 3*6=18.

Here, 15 is a multiple of 5. 1 case.

First die: 4, second die: 1-6.

Products:

4*1=4, 4*2=8, 4*3=12, 4*4=16, 4*5=20, 4*6=24.

Here, 20 is a multiple of 5. 1 case.

First die: 5, second die: 1-6.

Products:

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

All of these are multiples of 5. That's 6 cases.

First die: 6, second die: 1-6.

Products:

6*1=6, 6*2=12, 6*3=18, 6*4=24, 6*5=30, 6*6=36.

Here, 30 is a multiple of 5. 1 case.

So, let's count how many products are multiples of 5. For each first die:

First die 1: 1 case (product 5)

First die 2: 1 case (product 10)

First die 3: 1 case (product 15)

First die 4: 1 case (product 20)

First die 5: 6 cases (all products multiples of 5)

First die 6: 1 case (product 30)

Total cases: 1+1+1+1+6+1 = 11 cases. There we go. So 11 products are multiples of 5. Therefore, the number of favorable outcomes is 11. Therefore, probability is 11/36. Therefore, this confirms once again that the answer is 11/36.

Therefore, after checking multiple methods—direct counting, complementary counting, inclusion-exclusion, and enumerating all products—I can be confident that the probability is 11/36. So I think that is the correct answer.

**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 approach the problem systematically by identifying favorable outcomes and total possible outcomes.

---

### 🔢 Total Number of Outcomes

Each die has 6 faces, numbered from 1 to 6. Since the dice are independent, the total number of possible outcomes when rolling two dice is:

$$
6 \times 6 = 36
$$

---

### 🟡 Favorable Outcomes

A product is a **multiple of 5** if and only if **at least one of the dice shows a 5**. This is because:

- The number 5 is the only number in 1–6 that includes 5 as a prime factor.
- So, the product will be divisible by 5 **only if** at least one of the dice is **5**.

Let’s calculate the number of such favorable outcomes:

#### Case 1: First die is 5
- Possible second die outcomes: 1, 2, 3, 4, 5, 6 → 6 outcomes

#### Case 2: Second die is 5
- Possible first die outcomes: 1, 2, 3, 4, 5, 6 → 6 outcomes

However, the outcome **(5,5)** is counted **twice** in the above two cases (once in the first, once in the second). So we must **subtract this duplicate**.

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

---

### 📊 Probability Calculation

Now, using basic probability:

$$
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{11}{36}
$$

---

### ✅ Alternate Approach: Complementary Counting

We can double-check using the **complement method**:

- The **complement** of the desired event (product divisible by 5) is that **neither die shows 5**.
- The probability of not rolling a 5 on one die is $ \frac{5}{6} $
- Since the dice are independent, the probability that **both dice are not 5** is:

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

So, the probability that **at least one die is 5** (i.e., the product is divisible by 5) is:

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

---

### ✅ Final Answer

After verifying through both direct counting and the complement method, we find that the number of favorable outcomes is 11 out of 36 total outcomes.

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

[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's see. Let me think about how to approach this.

First, I remember 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 possible outcomes when throwing two dice should be 6 multiplied by 6, which is 36. So, there are 36 possible outcomes in total. That part seems straightforward.

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, which means that the product must have 5 as one of its prime factors. Since we're dealing with the product of two numbers, this would require that at least one of the numbers is a multiple of 5. Wait, is that right? Let me think again.

Actually, no. If the product is a multiple of 5, then at least one of the numbers must be a multiple of 5. Because if neither of the numbers is a multiple of 5, then their product won't have a factor of 5. Conversely, if at least one is a multiple of 5, then the product will be a multiple of 5. So, that seems correct.

Therefore, maybe I can approach this problem by calculating the probability that at least one die shows a 5. But wait, let me confirm this logic. Let me think of some examples. Suppose one die is 5 and the other is 2. Then the product is 10, which is a multiple of 5. If one die is 5 and the other is 5, product is 25, which is also a multiple of 5. If neither die is 5, say 2 and 3, the product is 6, which is not a multiple of 5. If one die is 5 and the other is 1, product is 5, which is a multiple of 5. So yes, it seems that the product is a multiple of 5 if and only if at least one die is a 5.

Wait, but hold on. Suppose one die is 5 and the other is... let me think. If one die is 5, regardless of the other die, the product is going to be a multiple of 5. If neither die is 5, then the product will not be a multiple of 5. Therefore, the number of favorable outcomes is the number of outcomes where at least one die is 5. Therefore, the probability we want is equal to the probability that at least one die shows a 5.

Alternatively, maybe I can use the complementary probability. Instead of calculating the probability of at least one 5, I can calculate 1 minus the probability that neither die shows a 5. That might be easier.

Let me try both approaches and see if they match.

First approach: Directly calculating the number of outcomes with at least one 5.

How many outcomes have at least one 5? Well, for two dice, the number of outcomes where the first die is 5 is 6 (since the second die can be anything from 1 to 6). Similarly, the number of outcomes where the second die is 5 is also 6. However, if I just add them together, I have 6 + 6 = 12. But wait, this counts the outcome (5,5) twice. So, to correct for that, we need to subtract 1. Therefore, total number of favorable outcomes is 12 - 1 = 11? Wait, no. Wait, inclusion-exclusion principle: |A ∪ B| = |A| + |B| - |A ∩ B|. In this case, A is the set where first die is 5, B is the set where the second die is 5. So, |A| is 6, |B| is 6, and |A ∩ B| is 1 (only (5,5)). Therefore, |A ∪ B| = 6 + 6 - 1 = 11. So there are 11 favorable outcomes. Therefore, the probability would be 11/36.

Wait, but let me check if that's correct. Alternatively, maybe I should list all possibilities. For two dice, the number of outcomes where at least one die is 5 is:

First die is 5: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) → 6 outcomes.

Second die is 5: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5) → 6 outcomes.

But (5,5) is counted in both, so total unique outcomes are 6 + 6 - 1 = 11. So that's correct. So 11 outcomes. Therefore, the probability is 11/36. So then, is that the answer?

Wait, but let me think again. Let me check with the complementary approach.

Complementary probability: the probability that neither die is a 5. Then, the probability that the product is not a multiple of 5 is equal to the probability that neither die is a 5. Therefore, the probability we want is 1 minus that probability.

So, what's the probability that neither die is a 5? For each die, the probability of not rolling a 5 is 5/6. Since the dice are independent, the probability that neither is a 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. So same answer. So that seems to confirm that 11/36 is correct.

But wait a second, let me think again. Is there another way to think about this? Let me think if there's a possibility that even if neither die is a 5, the product could still be a multiple of 5. For example, suppose one die is 10 and the other is... but wait, the dice only go up to 6. So the numbers on the dice are 1 through 6. Therefore, the only way to have a product divisible by 5 is if at least one die is 5. Because 5 is the only multiple of 5 on a die. So 5 is the only number on a die that is divisible by 5. Therefore, if neither die is 5, then the product cannot be divisible by 5. Therefore, our previous calculations hold. Therefore, the probability is 11/36.

Wait, but hold on. Let me test with actual examples.

Total outcomes: 36.

Number of outcomes with at least one 5: 11.

Therefore, if I list them:

First die is 5:

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

Second die is 5:

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

But as mentioned, (5,5) is duplicated, so total unique is 6 + 5 = 11. So yes, 11. Then, the probability is 11/36.

But let me verify with another approach. Let me think about how many products are multiples of 5.

Alternatively, perhaps I can think about the possible products. But that might be more complicated.

Alternatively, think about the possible numbers on the dice. Each die has numbers 1 to 6.

So, let me consider all possible pairs. But maybe that's too tedious. Alternatively, think about the numbers.

The product is multiple of 5 iff at least one number is multiple of 5. Since 5 is prime. Since 5 is prime, the product is divisible by 5 if and only if at least one factor is divisible by 5. Since the numbers on the dice are 1,2,3,4,5,6. The only number divisible by 5 is 5 itself. So yes, if at least one die is 5, then product is multiple of 5. Otherwise, not. So that's correct.

Therefore, our answer is 11/36. But wait, let me check again, maybe I made a mistake here. Wait, 5 is on one die. Wait, but suppose that one die is 5 and the other is... for example, suppose one die is 5 and the other is 5. Product is 25, which is multiple of 5. If one die is 5 and the other is 1, product is 5. If one is 5 and other is 2, product is 10. All multiples of 5. If neither is 5, then the numbers are from 1-4 and 1-4, so the product is 1-16, but 5 is not in that range. Wait, but 5 is not in the products? Wait, for example, if you have 2 and 3, product is 6. 1 and 5, but if neither die is 5, then you can't have 5 in the product? Wait, no. Wait, if neither die is 5, then the product can't have a factor of 5. So even if one die is 5, you can have 5 in the product. So that's correct.

Therefore, the logic holds. Therefore, the answer is 11/36. Wait, but let me check again with numbers. Let me count all the outcomes where at least one die is 5.

First die is 5: 6 outcomes.

Second die is 5: 6 outcomes.

But overlapping at (5,5). So total is 6 + 6 - 1 = 11. So 11 outcomes. So 11/36. Therefore, probability is 11/36.

But wait, just to be thorough, maybe I can list all possible pairs where at least one die is 5. Let me try:

First die is 5:

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

Second die is 5:

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

Now, combine them:

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5). That's 11 outcomes. So yes. So 11 outcomes. Therefore, 11/36 is correct.

But let me think again—could there be another case where the product is a multiple of 5 even if neither die is 5? For example, if one die is 10, but we don't have 10 on a die. The maximum is 6. So no, there's no way. If the numbers are 1-6, then the only multiple of 5 is 5 itself. Therefore, if neither die is 5, the product cannot be a multiple of 5. So yes, that's correct.

Therefore, the answer should be 11/36. Wait, but let me check with another method. Let me compute the possible products and count how many are multiples of 5.

But that might take longer, but let me try.

First, the possible numbers on the dice are 1-6. Let me list all possible products. But there are 36 products. It's tedious, but maybe I can categorize them.

Alternatively, think of how many products are multiples of 5. So, for that, as we said, it's when at least one die is 5. So same as before.

But just to make sure, let me think of the products:

If one die is 5, the products are:

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

Similarly, for the other die:

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

So, those are the products. But as we saw earlier, there are 11 distinct outcomes. So the products are 5,10,15,20,25,30 and their counterparts, but since each of these products corresponds to two different outcomes except for (5,5). But the number of outcomes is 11. Therefore, that confirms again that there are 11 outcomes where the product is a multiple of 5.

Therefore, probability is 11/36. So I think that is the answer. But let me check with another approach.

Alternatively, think of the probability that the first die is 5 or the second die is 5. Which is P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Where A is the event that the first die is 5, and B is the event that the second die is 5.

P(A) = 1/6, P(B) = 1/6, P(A ∩ B) = 1/36. Therefore, P(A ∪ B) = 1/6 + 1/6 - 1/36 = 2/6 - 1/36 = 12/36 - 1/36 = 11/36. Same result. Therefore, this confirms again that the probability is 11/36.

Therefore, I think the answer is 11/36.

But wait, let me check once again. Maybe there's an error in assuming that only 5 is the multiple of 5 on the dice. Wait, 5 is on the die, but 10 isn't. Wait, but the product can be 10, which is a multiple of 5, even if one die is 5 and the other is 2, which is allowed. So that's okay.

Alternatively, if you have two dice, and you want the product to be divisible by 5, you need at least one of the numbers to be divisible by 5. Since 5 is prime, and the numbers on the dice are from 1-6, the only number divisible by 5 is 5 itself. Therefore, the logic holds. So no other numbers on the dice can contribute to the product being divisible by 5. Therefore, the answer is indeed 11/36.

But let me check if there's another angle. Suppose someone might think that numbers like 10 could be formed, but since 10 isn't on the die, but the product can be 10. But that's okay because the product being 10 is allowed. So that's acceptable. So even though 10 isn't on the die, the product can be 10. So that's not a problem.

Alternatively, maybe I should think about the probability that the product is not a multiple of 5, which is when both dice are not 5. So that is, as we said, 5/6 * 5/6 = 25/36, so 1 - 25/36 = 11/36. Therefore, same answer.

Therefore, after verifying through multiple approaches—direct counting, complementary probability, inclusion-exclusion principle—I think the answer is 11/36.

But just to be absolutely sure, maybe I can list all the possible products and count how many are multiples of 5. Although that's time-consuming, let me try.

First die: 1, second die: 1-6.

Products:

1*1=1, 1*2=2, 1*3=3, 1*4=4, 1*5=5, 1*6=6.

So, here, only 5 is a multiple of 5. That's 1 case.

First die: 2, second die: 1-6.

Products:

2*1=2, 2*2=4, 2*3=6, 2*4=8, 2*5=10, 2*6=12.

Here, 10 is a multiple of 5. So 1 case.

First die: 3, second die: 1-6.

Products:

3*1=3, 3*2=6, 3*3=9, 3*4=12, 3*5=15, 3*6=18.

Here, 15 is a multiple of 5. 1 case.

First die: 4, second die: 1-6.

Products:

4*1=4, 4*2=8, 4*3=12, 4*4=16, 4*5=20, 4*6=24.

Here, 20 is a multiple of 5. 1 case.

First die: 5, second die: 1-6.

Products:

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

All of these are multiples of 5. That's 6 cases.

First die: 6, second die: 1-6.

Products:

6*1=6, 6*2=12, 6*3=18, 6*4=24, 6*5=30, 6*6=36.

Here, 30 is a multiple of 5. 1 case.

So, let's count how many products are multiples of 5. For each first die:

First die 1: 1 case (product 5)

First die 2: 1 case (product 10)

First die 3: 1 case (product 15)

First die 4: 1 case (product 20)

First die 5: 6 cases (all products multiples of 5)

First die 6: 1 case (product 30)

Total cases: 1+1+1+1+6+1 = 11 cases. There we go. So 11 products are multiples of 5. Therefore, the number of favorable outcomes is 11. Therefore, probability is 11/36. Therefore, this confirms once again that the answer is 11/36.

Therefore, after checking multiple methods—direct counting, complementary counting, inclusion-exclusion, and enumerating all products—I can be confident that the probability is 11/36. So I think that is the correct answer.

**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 approach the problem systematically by identifying favorable outcomes and total possible outcomes.

---

### 🔢 Total Number of Outcomes

Each die has 6 faces, numbered from 1 to 6. Since the dice are independent, the total number of possible outcomes when rolling two dice is:

$$
6 \times 6 = 36
$$

---

### 🟡 Favorable Outcomes

A product is a **multiple of 5** if and only if **at least one of the dice shows a 5**. This is because:

- The number 5 is the only number in 1–6 that includes 5 as a prime factor.
- So, the product will be divisible by 5 **only if** at least one of the dice is **5**.

Let’s calculate the number of such favorable outcomes:

#### Case 1: First die is 5
- Possible second die outcomes: 1, 2, 3, 4, 5, 6 → 6 outcomes

#### Case 2: Second die is 5
- Possible first die outcomes: 1, 2, 3, 4, 5, 6 → 6 outcomes

However, the outcome **(5,5)** is counted **twice** in the above two cases (once in the first, once in the second). So we must **subtract this duplicate**.

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

---

### 📊 Probability Calculation

Now, using basic probability:

$$
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{11}{36}
$$

---

### ✅ Alternate Approach: Complementary Counting

We can double-check using the **complement method**:

- The **complement** of the desired event (product divisible by 5) is that **neither die shows 5**.
- The probability of not rolling a 5 on one die is $ \frac{5}{6} $
- Since the dice are independent, the probability that **both dice are not 5** is:

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

So, the probability that **at least one die is 5** (i.e., the product is divisible by 5) is:

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

---

### ✅ Final Answer

After verifying through both direct counting and the complement method, we find that the number of favorable outcomes is 11 out of 36 total outcomes.

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

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 (which is not zero) and b such that the equation f(x) - f(a) equals (x - a)^2 times (x - b), where f(x) is x cubed plus 3x squared plus 1. Let me start by understanding what this equation is saying.

First, f(x) is a cubic polynomial. When I subtract f(a) from it, that should give me another polynomial, right? And this polynomial is supposed to factor into (x - a)^2 times (x - b). So, essentially, f(x) - f(a) has a double root at x = a and another root at x = b.

Maybe I can start by expanding the right-hand side and then equating coefficients with the left-hand side. Let me try that.

First, let me write down what f(x) - f(a) is.

Given f(x) = x³ + 3x² + 1, then f(a) = a³ + 3a² + 1. Therefore, f(x) - f(a) = (x³ + 3x² + 1) - (a³ + 3a² + 1) = x³ + 3x² - a³ - 3a².

Simplifying that, we get x³ - a³ + 3x² - 3a².

Now, I can factor this expression. Let me recall that x³ - a³ is a difference of cubes, which factors into (x - a)(x² + ax + a²). Similarly, 3x² - 3a² is 3(x² - a²) which factors into 3(x - a)(x + a).

So putting those together, f(x) - f(a) becomes:

(x - a)(x² + ax + a²) + 3(x - a)(x + a)

Factor out the common (x - a) term:

(x - a)[x² + ax + a² + 3(x + a)]

Let me simplify the expression inside the brackets:

x² + ax + a² + 3x + 3a = x² + (a + 3)x + (a² + 3a)

So now, f(x) - f(a) = (x - a)(x² + (a + 3)x + a² + 3a).

But according to the problem statement, this should equal (x - a)^2 (x - b). So, we can write:

(x - a)(x² + (a + 3)x + a² + 3a) = (x - a)^2 (x - b)

Assuming that x ≠ a, we can divide both sides by (x - a), leading to:

x² + (a + 3)x + a² + 3a = (x - a)(x - b)

Now, let me expand the right-hand side:

(x - a)(x - b) = x² - (a + b)x + ab

Therefore, equating the coefficients from both sides:

Left side: x² + (a + 3)x + (a² + 3a)

Right side: x² - (a + b)x + ab

So, setting coefficients equal:

For x² term: 1 = 1, which is already satisfied.

For x term: (a + 3) = - (a + b)

For constant term: (a² + 3a) = ab

So now, we have two equations:

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

2) a² + 3a = ab

Let me write them out:

From equation 1:

a + 3 = -a - b

Let me rearrange this equation:

Bring all terms to one side:

a + 3 + a + b = 0 => 2a + b + 3 = 0 => b = -2a - 3

Okay, so that's an expression for b in terms of a. Let me note that down.

Now equation 2 is:

a² + 3a = ab

But since we know from equation 1 that b = -2a -3, we can substitute that into equation 2.

So replacing b with -2a - 3:

a² + 3a = a*(-2a - 3)

Let me compute the right-hand side:

a*(-2a - 3) = -2a² - 3a

Therefore, equation 2 becomes:

a² + 3a = -2a² - 3a

Let me bring all terms to one side:

a² + 3a + 2a² + 3a = 0 => 3a² + 6a = 0

Factor out 3a:

3a(a + 2) = 0

Therefore, solutions are 3a = 0 => a = 0 or a + 2 = 0 => a = -2

But the problem states that a ≠ 0, so we discard a = 0. Hence, a = -2.

Now, substitute a = -2 into the expression for b:

From equation 1: b = -2a -3 => b = -2*(-2) -3 = 4 - 3 = 1

Therefore, the ordered pair is (a, b) = (-2, 1). Let me check if this works.

Let me verify by plugging back into the original equation.

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

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

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

Now, the right-hand side is (x - (-2))²(x - 1) = (x + 2)^2(x - 1). Let me expand that.

First expand (x + 2)^2: x² + 4x + 4. Then multiply by (x - 1):

(x² + 4x + 4)(x - 1) = x³ - x² + 4x² - 4x + 4x - 4 = x³ + 3x² - 4.

Which is equal to f(x) - f(-2). Therefore, the equation holds.

So the solution is indeed a = -2 and b = 1.

Therefore, the ordered pair is (-2, 1).

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

To solve the given equation:

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

where $ f(x) = x^3 + 3x^2 + 1 $, and $ a \neq 0 $, our goal is to find the real numbers $ a $ and $ b $ that satisfy the identity.

---

### Step 1: Express $ f(x) - f(a) $

Given:
$$
f(x) = x^3 + 3x^2 + 1
$$
Then:
$$
f(a) = a^3 + 3a^2 + 1
$$

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

Group the like terms:
$$
f(x) - f(a) = (x^3 - a^3) + 3(x^2 - a^2)
$$

We can now apply the factoring formulas:

- Difference of cubes: $ x^3 - a^3 = (x - a)(x^2 + ax + a^2) $
- Difference of squares: $ x^2 - a^2 = (x - a)(x + a) $

So:
$$
f(x) - f(a) = (x - a)(x^2 + ax + a^2) + 3(x - a)(x + a)
$$

Factor out $ (x - a) $:
$$
f(x) - f(a) = (x - a)[x^2 + ax + a^2 + 3x + 3a]
$$

Simplify the expression in the brackets:
$$
x^2 + ax + a^2 + 3x + 3a = x^2 + (a + 3)x + (a^2 + 3a)
$$

So we have:
$$
f(x) - f(a) = (x - a)\left(x^2 + (a + 3)x + (a^2 + 3a)\right)
$$

We are told this equals:
$$
(x - a)^2(x - b)
$$

Divide both sides by $ x - a $ (assuming $ x \ne a $), resulting in:
$$
x^2 + (a + 3)x + (a^2 + 3a) = (x - a)(x - b)
$$

Now expand the right-hand side:
$$
(x - a)(x - b) = x^2 - (a + b)x + ab
$$

Set the coefficients of the polynomials on both sides equal:

1. Coefficient of $ x^2 $: $ 1 = 1 $ — matches
2. Coefficient of $ x $: $ a + 3 = -(a + b) $
3. Constant term: $ a^2 + 3a = ab $

---

### Step 2: Solve the System of Equations

From (2):
$$
a + 3 = -a - b \Rightarrow 2a + b = -3 \Rightarrow b = -2a - 3 \quad \text{(Equation A)}
$$

From (3):
$$
a^2 + 3a = ab
$$

Substitute $ b = -2a - 3 $ into the equation:
$$
a^2 + 3a = a(-2a - 3)
$$
$$
a^2 + 3a = -2a^2 - 3a
$$

Bring all terms to one side:
$$
a^2 + 3a + 2a^2 + 3a = 0 \Rightarrow 3a^2 + 6a = 0
$$

Factor:
$$
3a(a + 2) = 0
$$

So:
$$
a = 0 \quad \text{or} \quad a = -2
$$

But $ a \ne 0 $, so:
$$
a = -2
$$

Now substitute into Equation A:
$$
b = -2(-2) - 3 = 4 - 3 = 1
$$

---

### Step 3: Final Check

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

- $ f(x) - f(-2) = x^3 + 3x^2 + 1 - (a^3 + 3a^2 + 1) = x^3 + 3x^2 - 4 $
- $ (x + 2)^2(x - 1) = (x^2 + 4x + 4)(x - 1) = x^3 + 3x^2 - 4 $

Both sides match, so the solution is correct.

---

### Final Answer

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

[Assistant]
Explanation:
Okay, so I need to find real numbers a (which is not zero) and b such that the equation f(x) - f(a) equals (x - a)^2 times (x - b), where f(x) is x cubed plus 3x squared plus 1. Let me start by understanding what this equation is saying.

First, f(x) is a cubic polynomial. When I subtract f(a) from it, that should give me another polynomial, right? And this polynomial is supposed to factor into (x - a)^2 times (x - b). So, essentially, f(x) - f(a) has a double root at x = a and another root at x = b.

Maybe I can start by expanding the right-hand side and then equating coefficients with the left-hand side. Let me try that.

First, let me write down what f(x) - f(a) is.

Given f(x) = x³ + 3x² + 1, then f(a) = a³ + 3a² + 1. Therefore, f(x) - f(a) = (x³ + 3x² + 1) - (a³ + 3a² + 1) = x³ + 3x² - a³ - 3a².

Simplifying that, we get x³ - a³ + 3x² - 3a².

Now, I can factor this expression. Let me recall that x³ - a³ is a difference of cubes, which factors into (x - a)(x² + ax + a²). Similarly, 3x² - 3a² is 3(x² - a²) which factors into 3(x - a)(x + a).

So putting those together, f(x) - f(a) becomes:

(x - a)(x² + ax + a²) + 3(x - a)(x + a)

Factor out the common (x - a) term:

(x - a)[x² + ax + a² + 3(x + a)]

Let me simplify the expression inside the brackets:

x² + ax + a² + 3x + 3a = x² + (a + 3)x + (a² + 3a)

So now, f(x) - f(a) = (x - a)(x² + (a + 3)x + a² + 3a).

But according to the problem statement, this should equal (x - a)^2 (x - b). So, we can write:

(x - a)(x² + (a + 3)x + a² + 3a) = (x - a)^2 (x - b)

Assuming that x ≠ a, we can divide both sides by (x - a), leading to:

x² + (a + 3)x + a² + 3a = (x - a)(x - b)

Now, let me expand the right-hand side:

(x - a)(x - b) = x² - (a + b)x + ab

Therefore, equating the coefficients from both sides:

Left side: x² + (a + 3)x + (a² + 3a)

Right side: x² - (a + b)x + ab

So, setting coefficients equal:

For x² term: 1 = 1, which is already satisfied.

For x term: (a + 3) = - (a + b)

For constant term: (a² + 3a) = ab

So now, we have two equations:

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

2) a² + 3a = ab

Let me write them out:

From equation 1:

a + 3 = -a - b

Let me rearrange this equation:

Bring all terms to one side:

a + 3 + a + b = 0 => 2a + b + 3 = 0 => b = -2a - 3

Okay, so that's an expression for b in terms of a. Let me note that down.

Now equation 2 is:

a² + 3a = ab

But since we know from equation 1 that b = -2a -3, we can substitute that into equation 2.

So replacing b with -2a - 3:

a² + 3a = a*(-2a - 3)

Let me compute the right-hand side:

a*(-2a - 3) = -2a² - 3a

Therefore, equation 2 becomes:

a² + 3a = -2a² - 3a

Let me bring all terms to one side:

a² + 3a + 2a² + 3a = 0 => 3a² + 6a = 0

Factor out 3a:

3a(a + 2) = 0

Therefore, solutions are 3a = 0 => a = 0 or a + 2 = 0 => a = -2

But the problem states that a ≠ 0, so we discard a = 0. Hence, a = -2.

Now, substitute a = -2 into the expression for b:

From equation 1: b = -2a -3 => b = -2*(-2) -3 = 4 - 3 = 1

Therefore, the ordered pair is (a, b) = (-2, 1). Let me check if this works.

Let me verify by plugging back into the original equation.

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

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

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

Now, the right-hand side is (x - (-2))²(x - 1) = (x + 2)^2(x - 1). Let me expand that.

First expand (x + 2)^2: x² + 4x + 4. Then multiply by (x - 1):

(x² + 4x + 4)(x - 1) = x³ - x² + 4x² - 4x + 4x - 4 = x³ + 3x² - 4.

Which is equal to f(x) - f(-2). Therefore, the equation holds.

So the solution is indeed a = -2 and b = 1.

Therefore, the ordered pair is (-2, 1).

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

To solve the given equation:

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

where $ f(x) = x^3 + 3x^2 + 1 $, and $ a \neq 0 $, our goal is to find the real numbers $ a $ and $ b $ that satisfy the identity.

---

### Step 1: Express $ f(x) - f(a) $

Given:
$$
f(x) = x^3 + 3x^2 + 1
$$
Then:
$$
f(a) = a^3 + 3a^2 + 1
$$

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

Group the like terms:
$$
f(x) - f(a) = (x^3 - a^3) + 3(x^2 - a^2)
$$

We can now apply the factoring formulas:

- Difference of cubes: $ x^3 - a^3 = (x - a)(x^2 + ax + a^2) $
- Difference of squares: $ x^2 - a^2 = (x - a)(x + a) $

So:
$$
f(x) - f(a) = (x - a)(x^2 + ax + a^2) + 3(x - a)(x + a)
$$

Factor out $ (x - a) $:
$$
f(x) - f(a) = (x - a)[x^2 + ax + a^2 + 3x + 3a]
$$

Simplify the expression in the brackets:
$$
x^2 + ax + a^2 + 3x + 3a = x^2 + (a + 3)x + (a^2 + 3a)
$$

So we have:
$$
f(x) - f(a) = (x - a)\left(x^2 + (a + 3)x + (a^2 + 3a)\right)
$$

We are told this equals:
$$
(x - a)^2(x - b)
$$

Divide both sides by $ x - a $ (assuming $ x \ne a $), resulting in:
$$
x^2 + (a + 3)x + (a^2 + 3a) = (x - a)(x - b)
$$

Now expand the right-hand side:
$$
(x - a)(x - b) = x^2 - (a + b)x + ab
$$

Set the coefficients of the polynomials on both sides equal:

1. Coefficient of $ x^2 $: $ 1 = 1 $ — matches
2. Coefficient of $ x $: $ a + 3 = -(a + b) $
3. Constant term: $ a^2 + 3a = ab $

---

### Step 2: Solve the System of Equations

From (2):
$$
a + 3 = -a - b \Rightarrow 2a + b = -3 \Rightarrow b = -2a - 3 \quad \text{(Equation A)}
$$

From (3):
$$
a^2 + 3a = ab
$$

Substitute $ b = -2a - 3 $ into the equation:
$$
a^2 + 3a = a(-2a - 3)
$$
$$
a^2 + 3a = -2a^2 - 3a
$$

Bring all terms to one side:
$$
a^2 + 3a + 2a^2 + 3a = 0 \Rightarrow 3a^2 + 6a = 0
$$

Factor:
$$
3a(a + 2) = 0
$$

So:
$$
a = 0 \quad \text{or} \quad a = -2
$$

But $ a \ne 0 $, so:
$$
a = -2
$$

Now substitute into Equation A:
$$
b = -2(-2) - 3 = 4 - 3 = 1
$$

---

### Step 3: Final Check

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

- $ f(x) - f(-2) = x^3 + 3x^2 + 1 - (a^3 + 3a^2 + 1) = x^3 + 3x^2 - 4 $
- $ (x + 2)^2(x - 1) = (x^2 + 4x + 4)(x - 1) = x^3 + 3x^2 - 4 $

Both sides match, so the solution is correct.

---

### 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, let me try to figure out this algebra problem. So, the question is about multiplying two quadratic polynomials and then comparing the result to a given quartic polynomial. The goal is to find the values of 'a' and 'b' and then compute their sum, a + b. Alright, let's start breaking it down.

First, let me write down what I know. The product of (3t² + 5t + a) and (4t² + bt - 2) is given as 12t⁴ + 26t³ - 8t² - 16t + 6. So, I need to multiply these two quadratics and then match the coefficients with the given polynomial to solve for 'a' and 'b'.

Alright, so multiplying two quadratics should give me a quartic polynomial (degree 4), which matches the given result. Let me recall how polynomial multiplication works. Each term in the first polynomial is multiplied by each term in the second polynomial, and then like terms are combined.

Let me write out the multiplication step by step.

First, multiply 3t² by each term in the second polynomial:

3t² * 4t² = 12t⁴

3t² * bt = 3b t³

3t² * (-2) = -6t²

Next, multiply 5t by each term in the second polynomial:

5t * 4t² = 20t³

5t * bt = 5b t²

5t * (-2) = -10t

Then, multiply 'a' by each term in the second polynomial:

a * 4t² = 4a t²

a * bt = ab t

a * (-2) = -2a

Now, let me combine all these terms:

Starting with the highest degree term:

12t⁴

Then, the t³ terms:

3b t³ + 20t³ = (3b + 20) t³

Next, the t² terms:

-6t² + 5b t² + 4a t² = (-6 + 5b + 4a) t²

Then, the t terms:

-10t + ab t = (-10 + ab) t

And the constant term:

-2a

So, putting all together, the product is:

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

Now, this should be equal to the given polynomial: 12t⁴ + 26t³ - 8t² - 16t + 6.

Therefore, we can set up equations by equating the coefficients of corresponding powers of t.

Let me list them:

For t⁴: 12 = 12. That's already matching, so no information here.

For t³: 3b + 20 = 26.

For t²: -6 + 5b + 4a = -8.

For t: -10 + ab = -16.

For the constant term: -2a = 6.

Okay, so now we have a system of equations. Let me write them down:

1. 3b + 20 = 26

2. -6 + 5b + 4a = -8

3. -10 + ab = -16

4. -2a = 6

Let me solve these equations step by step.

Starting with equation 4, since it only has 'a':

-2a = 6 ⇒ a = 6 / (-2) ⇒ a = -3.

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

Now, let's substitute a = -3 into equation 1, which is 3b + 20 = 26. Let's solve for b.

3b + 20 = 26 ⇒ 3b = 26 - 20 ⇒ 3b = 6 ⇒ b = 6 / 3 ⇒ b = 2.

Okay, so b is 2. Let me check if this works with the other equations, particularly equation 2 and equation 3.

First, equation 2: -6 + 5b + 4a = -8. Let's substitute a = -3 and b = 2.

Calculating each term:

-6 + 5*(2) + 4*(-3) = -6 + 10 - 12 = (-6 -12) +10 = -18 +10 = -8. Which matches the right-hand side. Perfect, that works.

Now, equation 3: -10 + ab = -16. Let's substitute a = -3, b = 2.

ab is (-3)*(2) = -6.

So, -10 + (-6) = -16. Which is -16 = -16. Correct, that also works.

So all equations are satisfied with a = -3 and b = 2. Therefore, a + b is (-3) + 2 = -1.

Wait, but let me double-check everything once again to make sure I didn't make any calculation errors.

First, original polynomials:

(3t² + 5t + a)(4t² + bt - 2) with a = -3 and b = 2.

So substituting:

(3t² + 5t - 3)(4t² + 2t - 2)

Let me multiply them again step by step to confirm.

First, 3t² multiplied by 4t² = 12t⁴.

3t² * 2t = 6t³.

3t² * (-2) = -6t².

5t * 4t² = 20t³.

5t * 2t = 10t².

5t * (-2) = -10t.

-3 * 4t² = -12t².

-3 * 2t = -6t.

-3 * (-2) = 6.

Now, combine all terms:

12t⁴

t³ terms: 6t³ + 20t³ = 26t³

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

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

constant term: +6.

So altogether, it's 12t⁴ +26t³ -8t² -16t +6, which matches exactly the given polynomial. Perfect! So my values for a and b are correct.

Therefore, a + b is -3 + 2 = -1.

Wait, but just to make sure, let me check if there's any other way this could be approached or if I made some mistake in interpreting the problem. Let me think again.

The problem states that the product is 12t⁴ +26t³ -8t² -16t +6. So when I multiplied it with a=-3 and b=2, it matched. So that seems correct. Therefore, the answer is -1.

Alternatively, maybe I can check by plugging in specific values of t into the original polynomials and the product?

For example, let me pick t = 0. Let's see:

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

On the other hand, the product of the two polynomials at t = 0 is:

(3(0)^2 +5(0)+a)(4(0)^2 +b(0)-2) = (0 + 0 +a)(0 + 0 -2) = (a)(-2) = -2a. Which should equal 6. Hence, -2a = 6 ⇒ a = -3. Which is consistent with earlier.

Similarly, maybe another value. Let's try t = 1.

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

Calculating that: 12 +26 = 38; 38 -8 = 30; 30 -16 = 14; 14 +6 = 20. So the product at t=1 is 20.

Now, compute the product of the two polynomials at t=1:

(3(1)^2 +5(1) + a)(4(1)^2 +b(1) -2) = (3 +5 +a)(4 +b -2) = (8 + a)(2 + b).

With a = -3 and b = 2, that becomes (8 -3)(2 +2) = (5)(4) = 20. Which matches. Perfect.

Another check: t = -1.

Original polynomial at t=-1:

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

Now, the product of the polynomials at t=-1:

(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).

Substituting a=-3 and b=2: (-2 -3)(2 - 2) = (-5)(0) = 0. Which matches. Perfect again.

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

Wait, but let me check once again for any errors in my initial equations. Let me re-express the multiplication.

Original multiplication steps:

First polynomial: 3t² +5t +a.

Second polynomial: 4t² +bt -2.

Multiplying:

First term 3t² * 4t² = 12t^4.

3t² * bt = 3b t^3.

3t² * (-2) = -6t².

Then 5t * 4t² = 20t³.

5t * bt = 5b t².

5t * (-2) = -10t.

Then a * 4t² = 4a t².

a * bt = ab t.

a * (-2) = -2a.

Adding them all up:

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

Which equals the given polynomial: 12t^4 +26t³ -8t² -16t +6.

So equations:

1. 3b +20 =26 ⇒ 3b=6 ⇒b=2.

2. -6 +5b +4a = -8 ⇒ 5b +4a = -8 +6 ⇒ 5b +4a = -2.

Wait, hold on! Wait, in my earlier step, I said equation 2 was -6 +5b +4a = -8. Which is correct. Then solving with a=-3 and b=2:

-6 +5*2 +4*(-3) = -6 +10 -12 = -8. Which is correct. So yes, equation 2 is satisfied. Then equation 3 is -10 +ab = -16. Which with a=-3, b=2, gives -10 + (-6) = -16. Correct. And equation 4, -2a =6 ⇒a=-3. Correct. So all equations are correct.

Therefore, the answer is a + b = -3 + 2 = -1.

I think that's solid. I don't see any mistakes here. So the answer should be \boxed{-1}.

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

To find the value of $ a + b $, we are given that the product of two quadratic polynomials is:

$$
(3t^2 + 5t + a)(4t^2 + bt - 2) = 12t^4 + 26t^3 - 8t^2 - 16t + 6
$$

---

### Step 1: Multiply the Polynomials

Let’s expand the left-hand side:

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

Start by multiplying each term of the first polynomial by each term of the second, then collect like terms.

- From $3t^2 \cdot (4t^2 + bt - 2)$:
  $$
  12t^4 + 3b t^3 - 6t^2
  $$

- From $5t \cdot (4t^2 + bt - 2)$:
  $$
  20t^3 + 5b t^2 - 10t
  $$

- From $a \cdot (4t^2 + bt - 2)$:
  $$
  4a t^2 + ab t - 2a
  $$

Now combine all terms:

- $t^4$: $12t^4$
- $t^3$: $3b t^3 + 20t^3 = (3b + 20)t^3$
- $t^2$: $-6t^2 + 5b t^2 + 4a t^2 = (-6 + 5b + 4a)t^2$
- $t^1$: $-10t + ab t = (-10 + ab)t$
- Constant: $-2a$

Thus, the expanded product is:

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

---

### Step 2: Match Coefficients with Given Polynomial

We are told that this product is equal to:

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

Now equate the coefficients of like terms:

1. Coefficient of $t^4$: $12 = 12$ → always true
2. $3b + 20 = 26$ → Solve:
   $$
   3b = 6 \Rightarrow b = 2
   $$
3. $-6 + 5b + 4a = -8$
   Substitute $b = 2$:
   $$
   -6 + 10 + 4a = -8 \Rightarrow 4a = -8 + 6 = -2 \Rightarrow a = -\frac{1}{2}
   $$

Wait, but this leads to a problem. Let’s double-check our earlier steps.

---

### Step 3: Re-check the Equation for $a$

Looking back:

From earlier, we had:

$$
-6 + 5b + 4a = -8
$$

Instead of solving this for $a$, let's isolate $-2a$ from the **constant term**:

From the constant term:

$$
-2a = 6 \Rightarrow a = \frac{6}{-2} = -3
$$

Now substitute $a = -3$ into the equation $3b + 20 = 26$:

$$
3b = 6 \Rightarrow b = 2
$$

Now, substitute $a = -3$, $b = 2$ into the other terms to confirm:

- $t^2$ coefficient:
  $$
  -6 + 5(2) + 4(-3) = -6 + 10 - 12 = -8
  $$
- $t$ coefficient:
  $$
  -10 + (-3)(2) = -10 - 6 = -16
  $$
- Constant term:
  $$
  -2a = -2(-3) = 6
  $$

All match the given polynomial exactly.

---

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

Given:

$$
a = -3, \quad b = 2
$$

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

---

### ✅ Final Answer:

$$
\boxed{-1}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this algebra problem. So, the question is about multiplying two quadratic polynomials and then comparing the result to a given quartic polynomial. The goal is to find the values of 'a' and 'b' and then compute their sum, a + b. Alright, let's start breaking it down.

First, let me write down what I know. The product of (3t² + 5t + a) and (4t² + bt - 2) is given as 12t⁴ + 26t³ - 8t² - 16t + 6. So, I need to multiply these two quadratics and then match the coefficients with the given polynomial to solve for 'a' and 'b'.

Alright, so multiplying two quadratics should give me a quartic polynomial (degree 4), which matches the given result. Let me recall how polynomial multiplication works. Each term in the first polynomial is multiplied by each term in the second polynomial, and then like terms are combined.

Let me write out the multiplication step by step.

First, multiply 3t² by each term in the second polynomial:

3t² * 4t² = 12t⁴

3t² * bt = 3b t³

3t² * (-2) = -6t²

Next, multiply 5t by each term in the second polynomial:

5t * 4t² = 20t³

5t * bt = 5b t²

5t * (-2) = -10t

Then, multiply 'a' by each term in the second polynomial:

a * 4t² = 4a t²

a * bt = ab t

a * (-2) = -2a

Now, let me combine all these terms:

Starting with the highest degree term:

12t⁴

Then, the t³ terms:

3b t³ + 20t³ = (3b + 20) t³

Next, the t² terms:

-6t² + 5b t² + 4a t² = (-6 + 5b + 4a) t²

Then, the t terms:

-10t + ab t = (-10 + ab) t

And the constant term:

-2a

So, putting all together, the product is:

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

Now, this should be equal to the given polynomial: 12t⁴ + 26t³ - 8t² - 16t + 6.

Therefore, we can set up equations by equating the coefficients of corresponding powers of t.

Let me list them:

For t⁴: 12 = 12. That's already matching, so no information here.

For t³: 3b + 20 = 26.

For t²: -6 + 5b + 4a = -8.

For t: -10 + ab = -16.

For the constant term: -2a = 6.

Okay, so now we have a system of equations. Let me write them down:

1. 3b + 20 = 26

2. -6 + 5b + 4a = -8

3. -10 + ab = -16

4. -2a = 6

Let me solve these equations step by step.

Starting with equation 4, since it only has 'a':

-2a = 6 ⇒ a = 6 / (-2) ⇒ a = -3.

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

Now, let's substitute a = -3 into equation 1, which is 3b + 20 = 26. Let's solve for b.

3b + 20 = 26 ⇒ 3b = 26 - 20 ⇒ 3b = 6 ⇒ b = 6 / 3 ⇒ b = 2.

Okay, so b is 2. Let me check if this works with the other equations, particularly equation 2 and equation 3.

First, equation 2: -6 + 5b + 4a = -8. Let's substitute a = -3 and b = 2.

Calculating each term:

-6 + 5*(2) + 4*(-3) = -6 + 10 - 12 = (-6 -12) +10 = -18 +10 = -8. Which matches the right-hand side. Perfect, that works.

Now, equation 3: -10 + ab = -16. Let's substitute a = -3, b = 2.

ab is (-3)*(2) = -6.

So, -10 + (-6) = -16. Which is -16 = -16. Correct, that also works.

So all equations are satisfied with a = -3 and b = 2. Therefore, a + b is (-3) + 2 = -1.

Wait, but let me double-check everything once again to make sure I didn't make any calculation errors.

First, original polynomials:

(3t² + 5t + a)(4t² + bt - 2) with a = -3 and b = 2.

So substituting:

(3t² + 5t - 3)(4t² + 2t - 2)

Let me multiply them again step by step to confirm.

First, 3t² multiplied by 4t² = 12t⁴.

3t² * 2t = 6t³.

3t² * (-2) = -6t².

5t * 4t² = 20t³.

5t * 2t = 10t².

5t * (-2) = -10t.

-3 * 4t² = -12t².

-3 * 2t = -6t.

-3 * (-2) = 6.

Now, combine all terms:

12t⁴

t³ terms: 6t³ + 20t³ = 26t³

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

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

constant term: +6.

So altogether, it's 12t⁴ +26t³ -8t² -16t +6, which matches exactly the given polynomial. Perfect! So my values for a and b are correct.

Therefore, a + b is -3 + 2 = -1.

Wait, but just to make sure, let me check if there's any other way this could be approached or if I made some mistake in interpreting the problem. Let me think again.

The problem states that the product is 12t⁴ +26t³ -8t² -16t +6. So when I multiplied it with a=-3 and b=2, it matched. So that seems correct. Therefore, the answer is -1.

Alternatively, maybe I can check by plugging in specific values of t into the original polynomials and the product?

For example, let me pick t = 0. Let's see:

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

On the other hand, the product of the two polynomials at t = 0 is:

(3(0)^2 +5(0)+a)(4(0)^2 +b(0)-2) = (0 + 0 +a)(0 + 0 -2) = (a)(-2) = -2a. Which should equal 6. Hence, -2a = 6 ⇒ a = -3. Which is consistent with earlier.

Similarly, maybe another value. Let's try t = 1.

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

Calculating that: 12 +26 = 38; 38 -8 = 30; 30 -16 = 14; 14 +6 = 20. So the product at t=1 is 20.

Now, compute the product of the two polynomials at t=1:

(3(1)^2 +5(1) + a)(4(1)^2 +b(1) -2) = (3 +5 +a)(4 +b -2) = (8 + a)(2 + b).

With a = -3 and b = 2, that becomes (8 -3)(2 +2) = (5)(4) = 20. Which matches. Perfect.

Another check: t = -1.

Original polynomial at t=-1:

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

Now, the product of the polynomials at t=-1:

(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).

Substituting a=-3 and b=2: (-2 -3)(2 - 2) = (-5)(0) = 0. Which matches. Perfect again.

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

Wait, but let me check once again for any errors in my initial equations. Let me re-express the multiplication.

Original multiplication steps:

First polynomial: 3t² +5t +a.

Second polynomial: 4t² +bt -2.

Multiplying:

First term 3t² * 4t² = 12t^4.

3t² * bt = 3b t^3.

3t² * (-2) = -6t².

Then 5t * 4t² = 20t³.

5t * bt = 5b t².

5t * (-2) = -10t.

Then a * 4t² = 4a t².

a * bt = ab t.

a * (-2) = -2a.

Adding them all up:

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

Which equals the given polynomial: 12t^4 +26t³ -8t² -16t +6.

So equations:

1. 3b +20 =26 ⇒ 3b=6 ⇒b=2.

2. -6 +5b +4a = -8 ⇒ 5b +4a = -8 +6 ⇒ 5b +4a = -2.

Wait, hold on! Wait, in my earlier step, I said equation 2 was -6 +5b +4a = -8. Which is correct. Then solving with a=-3 and b=2:

-6 +5*2 +4*(-3) = -6 +10 -12 = -8. Which is correct. So yes, equation 2 is satisfied. Then equation 3 is -10 +ab = -16. Which with a=-3, b=2, gives -10 + (-6) = -16. Correct. And equation 4, -2a =6 ⇒a=-3. Correct. So all equations are correct.

Therefore, the answer is a + b = -3 + 2 = -1.

I think that's solid. I don't see any mistakes here. So the answer should be \boxed{-1}.

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

To find the value of $ a + b $, we are given that the product of two quadratic polynomials is:

$$
(3t^2 + 5t + a)(4t^2 + bt - 2) = 12t^4 + 26t^3 - 8t^2 - 16t + 6
$$

---

### Step 1: Multiply the Polynomials

Let’s expand the left-hand side:

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

Start by multiplying each term of the first polynomial by each term of the second, then collect like terms.

- From $3t^2 \cdot (4t^2 + bt - 2)$:
  $$
  12t^4 + 3b t^3 - 6t^2
  $$

- From $5t \cdot (4t^2 + bt - 2)$:
  $$
  20t^3 + 5b t^2 - 10t
  $$

- From $a \cdot (4t^2 + bt - 2)$:
  $$
  4a t^2 + ab t - 2a
  $$

Now combine all terms:

- $t^4$: $12t^4$
- $t^3$: $3b t^3 + 20t^3 = (3b + 20)t^3$
- $t^2$: $-6t^2 + 5b t^2 + 4a t^2 = (-6 + 5b + 4a)t^2$
- $t^1$: $-10t + ab t = (-10 + ab)t$
- Constant: $-2a$

Thus, the expanded product is:

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

---

### Step 2: Match Coefficients with Given Polynomial

We are told that this product is equal to:

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

Now equate the coefficients of like terms:

1. Coefficient of $t^4$: $12 = 12$ → always true
2. $3b + 20 = 26$ → Solve:
   $$
   3b = 6 \Rightarrow b = 2
   $$
3. $-6 + 5b + 4a = -8$
   Substitute $b = 2$:
   $$
   -6 + 10 + 4a = -8 \Rightarrow 4a = -8 + 6 = -2 \Rightarrow a = -\frac{1}{2}
   $$

Wait, but this leads to a problem. Let’s double-check our earlier steps.

---

### Step 3: Re-check the Equation for $a$

Looking back:

From earlier, we had:

$$
-6 + 5b + 4a = -8
$$

Instead of solving this for $a$, let's isolate $-2a$ from the **constant term**:

From the constant term:

$$
-2a = 6 \Rightarrow a = \frac{6}{-2} = -3
$$

Now substitute $a = -3$ into the equation $3b + 20 = 26$:

$$
3b = 6 \Rightarrow b = 2
$$

Now, substitute $a = -3$, $b = 2$ into the other terms to confirm:

- $t^2$ coefficient:
  $$
  -6 + 5(2) + 4(-3) = -6 + 10 - 12 = -8
  $$
- $t$ coefficient:
  $$
  -10 + (-3)(2) = -10 - 6 = -16
  $$
- Constant term:
  $$
  -2a = -2(-3) = 6
  $$

All match the given polynomial exactly.

---

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

Given:

$$
a = -3, \quad b = 2
$$

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

---

### ✅ 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 there's this problem about a right cylindrical tank being filled with water. The question is asking for the radius of the tank. Let me try to visualize this. A right cylinder means that the bases are circles and the sides are straight up, right? So it's like a soup can or something. The tank is being filled at a rate of 20π cubic meters per hour. And as it's filled, the water level rises four meters per hour. We need to find the radius.

Let me think about related rates here. Since the tank is cylindrical, the volume of water in it at any time depends on the height of the water. The volume of a cylinder is given by the formula V = πr²h, where r is the radius and h is the height. Here, the radius is constant, so as the water is poured in, the height increases over time.

Given that the rate of filling is 20π cubic meters per hour, that should be the rate of change of volume with respect to time, which is dV/dt. Also, the problem says the water level rises four meters per hour. That should be the rate of change of height with respect to time, dh/dt. So, dh/dt = 4 m/hour. We need to find r.

So, since we know how volume changes with time and how height changes with time, maybe we can relate them using the volume formula. Let me write down what I know:

V = πr²h

Differentiating both sides with respect to time t:

dV/dt = πr² * dh/dt

Wait, is that right? Because when you differentiate V with respect to t, since r is constant, the derivative of πr²h with respect to t is πr² * dh/dt. Yes, that seems correct.

So plugging in the known values:

dV/dt = 20π = πr² * dh/dt = πr² * 4

So, 20π = πr² * 4. Let me write that equation out:

20π = 4πr²

Then, we can divide both sides by π to simplify:

20 = 4r²

Then divide both sides by 4:

5 = r²

Therefore, r = √5. Since radius can't be negative, we take the positive root.

Wait, so is the answer √5? That seems straightforward, but let me check again. Maybe I missed something?

Let me go through the steps again. Volume rate is dV/dt = 20π m³/hour. The height rate dh/dt = 4 m/hour. Volume of cylinder is πr²h. Differentiate both sides with respect to time:

dV/dt = πr² dh/dt. Yes, that's because the radius is constant. So plugging in:

20π = πr² * 4. Divide both sides by π: 20 = 4r² => r² = 5 => r = √5. So that seems correct.

But the problem says "a right cylindrical tank with circular bases". So it's a standard cylinder. Maybe there is something else? Wait, the question says "the water level rises four meters per hour." So that is dh/dt = 4 m/hour. So if the tank is being filled at a rate of 20π m³/hour, which is dV/dt, and we have the relation between dV/dt and dh/dt, so we can solve for r.

Alternatively, maybe they are expecting to use some other formula? But no, the standard formula for the volume of a cylinder is straightforward here. Since the tank is right circular cylinder, the radius is same throughout. So, differentiating V = πr²h with respect to t gives dV/dt = πr² dh/dt. So solving for r gives r = sqrt(dV/dt / (π dh/dt)). Wait, let's plug in numbers:

dV/dt is 20π, dh/dt is 4. So:

r² = (dV/dt) / (π dh/dt) ) = (20π) / (π * 4) ) = 20 / 4 = 5. Therefore, r = sqrt(5).

Yes, that seems correct. So the radius is sqrt(5) meters.

But let me think again if there is any possibility that the problem is not as straightforward. For example, is there a chance that the tank is not full? Like, maybe the tank has a certain height, and as it's being filled, the rate of height increase is given? Wait, but the problem doesn't mention anything about the tank's total height. It just says "as the tank is filled, the water level rises four meters per hour." So regardless of the total height, the rate of change of height is 4 m/hour. Therefore, the calculation should hold true at any time, as long as the tank is being filled. Therefore, we don't need to worry about the total height or anything else. The relationship between dV/dt and dh/dt is always dV/dt = πr² dh/dt, so solving for r gives sqrt(5).

Alternatively, if the tank was not a cylinder, but some other shape, the problem would be different, but since it's a cylinder, we are okay.

Another way to think about it: Suppose we want to find the volume added per hour, which is 20π m³. The volume of a cylinder is πr²h, so if the height increases by 4 meters in an hour, then the volume added is πr²*4. Therefore, πr²*4 = 20π => r²*4 = 20 => r² = 5 => r = √5. Yep, same result.

Therefore, I think that's the answer. The radius is √5 meters.

But wait, the problem says "Express your answer in simplest radical form." Well, √5 is already simplified, so that's fine.

Wait, but maybe I need to check if there's any trick here. For example, maybe the question is referring to the rate of the water level rising as the tank is being filled, but maybe the tank is being filled not from the bottom? But no, the tank is a right cylinder, so it's filled from the bottom. Also, if the tank is being filled at a constant rate, then the rate at which the height increases is constant, which is given as 4 m/hour, so the calculation is straightforward.

Alternatively, maybe there's a misunderstanding about the units? Let me check units. The rate is 20π cubic meters per hour. The dh/dt is 4 meters per hour. So when we do πr² * dh/dt, the units would be πr²*(meters per hour), which is m³ per hour, which matches the units of dV/dt. So the units check out. So if r is in meters, then πr² is in m², multiplied by dh/dt (m/hour) gives m³/hour, which matches. Therefore, the calculation is consistent.

So, I think my answer is correct. The radius is √5 meters.

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

To find the radius of the tank, we are given the following information:

- The tank is a **right circular cylinder**, meaning it has a constant radius and its central axis is perpendicular to its bases.
- Water is being **filled at a rate** of $ \frac{dV}{dt} = 20\pi $ **cubic meters per hour**.
- As it is filled, the **water level rises at a rate** of $ \frac{dh}{dt} = 4 $ **meters per hour**.

---

### Step 1: Use the Volume Formula for a Cylinder

The volume $ V $ of a cylinder is given by:

$$
V = \pi r^2 h
$$

where:
- $ r $ is the radius (what we're trying to find),
- $ h $ is the height (or water level) at time $ t $.

---

### Step 2: Differentiate the Volume with Respect to Time

Differentiate both sides of the volume equation with respect to time $ t $:

$$
\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}
$$

This equation relates the rate at which volume is added to the tank $ \frac{dV}{dt} $, the rate of change of height $ \frac{dh}{dt} $, and the radius $ r $ (which is constant).

---

### Step 3: Plug in the Known Values

We are given:

- $ \frac{dV}{dt} = 20\pi $
- $ \frac{dh}{dt} = 4 $

Substitute these into the equation:

$$
20\pi = \pi r^2 \cdot 4
$$

---

### Step 4: Solve for $ r $

Divide both sides of the equation by $ \pi $:

$$
20 = 4r^2
$$

Divide both sides by 4:

$$
5 = r^2
$$

Take the square root:

$$
r = \sqrt{5}
$$

Since radius must be positive, we take the positive root.

---

### Final Answer

$$
\boxed{\sqrt{5}}
$$

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

[Assistant]
Explanation:
Okay, so there's this problem about a right cylindrical tank being filled with water. The question is asking for the radius of the tank. Let me try to visualize this. A right cylinder means that the bases are circles and the sides are straight up, right? So it's like a soup can or something. The tank is being filled at a rate of 20π cubic meters per hour. And as it's filled, the water level rises four meters per hour. We need to find the radius.

Let me think about related rates here. Since the tank is cylindrical, the volume of water in it at any time depends on the height of the water. The volume of a cylinder is given by the formula V = πr²h, where r is the radius and h is the height. Here, the radius is constant, so as the water is poured in, the height increases over time.

Given that the rate of filling is 20π cubic meters per hour, that should be the rate of change of volume with respect to time, which is dV/dt. Also, the problem says the water level rises four meters per hour. That should be the rate of change of height with respect to time, dh/dt. So, dh/dt = 4 m/hour. We need to find r.

So, since we know how volume changes with time and how height changes with time, maybe we can relate them using the volume formula. Let me write down what I know:

V = πr²h

Differentiating both sides with respect to time t:

dV/dt = πr² * dh/dt

Wait, is that right? Because when you differentiate V with respect to t, since r is constant, the derivative of πr²h with respect to t is πr² * dh/dt. Yes, that seems correct.

So plugging in the known values:

dV/dt = 20π = πr² * dh/dt = πr² * 4

So, 20π = πr² * 4. Let me write that equation out:

20π = 4πr²

Then, we can divide both sides by π to simplify:

20 = 4r²

Then divide both sides by 4:

5 = r²

Therefore, r = √5. Since radius can't be negative, we take the positive root.

Wait, so is the answer √5? That seems straightforward, but let me check again. Maybe I missed something?

Let me go through the steps again. Volume rate is dV/dt = 20π m³/hour. The height rate dh/dt = 4 m/hour. Volume of cylinder is πr²h. Differentiate both sides with respect to time:

dV/dt = πr² dh/dt. Yes, that's because the radius is constant. So plugging in:

20π = πr² * 4. Divide both sides by π: 20 = 4r² => r² = 5 => r = √5. So that seems correct.

But the problem says "a right cylindrical tank with circular bases". So it's a standard cylinder. Maybe there is something else? Wait, the question says "the water level rises four meters per hour." So that is dh/dt = 4 m/hour. So if the tank is being filled at a rate of 20π m³/hour, which is dV/dt, and we have the relation between dV/dt and dh/dt, so we can solve for r.

Alternatively, maybe they are expecting to use some other formula? But no, the standard formula for the volume of a cylinder is straightforward here. Since the tank is right circular cylinder, the radius is same throughout. So, differentiating V = πr²h with respect to t gives dV/dt = πr² dh/dt. So solving for r gives r = sqrt(dV/dt / (π dh/dt)). Wait, let's plug in numbers:

dV/dt is 20π, dh/dt is 4. So:

r² = (dV/dt) / (π dh/dt) ) = (20π) / (π * 4) ) = 20 / 4 = 5. Therefore, r = sqrt(5).

Yes, that seems correct. So the radius is sqrt(5) meters.

But let me think again if there is any possibility that the problem is not as straightforward. For example, is there a chance that the tank is not full? Like, maybe the tank has a certain height, and as it's being filled, the rate of height increase is given? Wait, but the problem doesn't mention anything about the tank's total height. It just says "as the tank is filled, the water level rises four meters per hour." So regardless of the total height, the rate of change of height is 4 m/hour. Therefore, the calculation should hold true at any time, as long as the tank is being filled. Therefore, we don't need to worry about the total height or anything else. The relationship between dV/dt and dh/dt is always dV/dt = πr² dh/dt, so solving for r gives sqrt(5).

Alternatively, if the tank was not a cylinder, but some other shape, the problem would be different, but since it's a cylinder, we are okay.

Another way to think about it: Suppose we want to find the volume added per hour, which is 20π m³. The volume of a cylinder is πr²h, so if the height increases by 4 meters in an hour, then the volume added is πr²*4. Therefore, πr²*4 = 20π => r²*4 = 20 => r² = 5 => r = √5. Yep, same result.

Therefore, I think that's the answer. The radius is √5 meters.

But wait, the problem says "Express your answer in simplest radical form." Well, √5 is already simplified, so that's fine.

Wait, but maybe I need to check if there's any trick here. For example, maybe the question is referring to the rate of the water level rising as the tank is being filled, but maybe the tank is being filled not from the bottom? But no, the tank is a right cylinder, so it's filled from the bottom. Also, if the tank is being filled at a constant rate, then the rate at which the height increases is constant, which is given as 4 m/hour, so the calculation is straightforward.

Alternatively, maybe there's a misunderstanding about the units? Let me check units. The rate is 20π cubic meters per hour. The dh/dt is 4 meters per hour. So when we do πr² * dh/dt, the units would be πr²*(meters per hour), which is m³ per hour, which matches the units of dV/dt. So the units check out. So if r is in meters, then πr² is in m², multiplied by dh/dt (m/hour) gives m³/hour, which matches. Therefore, the calculation is consistent.

So, I think my answer is correct. The radius is √5 meters.

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

To find the radius of the tank, we are given the following information:

- The tank is a **right circular cylinder**, meaning it has a constant radius and its central axis is perpendicular to its bases.
- Water is being **filled at a rate** of $ \frac{dV}{dt} = 20\pi $ **cubic meters per hour**.
- As it is filled, the **water level rises at a rate** of $ \frac{dh}{dt} = 4 $ **meters per hour**.

---

### Step 1: Use the Volume Formula for a Cylinder

The volume $ V $ of a cylinder is given by:

$$
V = \pi r^2 h
$$

where:
- $ r $ is the radius (what we're trying to find),
- $ h $ is the height (or water level) at time $ t $.

---

### Step 2: Differentiate the Volume with Respect to Time

Differentiate both sides of the volume equation with respect to time $ t $:

$$
\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}
$$

This equation relates the rate at which volume is added to the tank $ \frac{dV}{dt} $, the rate of change of height $ \frac{dh}{dt} $, and the radius $ r $ (which is constant).

---

### Step 3: Plug in the Known Values

We are given:

- $ \frac{dV}{dt} = 20\pi $
- $ \frac{dh}{dt} = 4 $

Substitute these into the equation:

$$
20\pi = \pi r^2 \cdot 4
$$

---

### Step 4: Solve for $ r $

Divide both sides of the equation by $ \pi $:

$$
20 = 4r^2
$$

Divide both sides by 4:

$$
5 = r^2
$$

Take the square root:

$$
r = \sqrt{5}
$$

Since radius must be positive, we take the positive root.

---

### Final Answer

$$
\boxed{\sqrt{5}}
$$

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

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.