Explanation:
Okay, so I need to find the remainder when this big polynomial f(x) is divided by x² - 1. The problem says not to use long division, which makes sense because that would be really tedious with such a high-degree polynomial. Hmm, I remember that when you divide a polynomial by a quadratic like x² - 1, the remainder should be a linear polynomial, right? So the remainder will be of the form ax + b. My job is to find the coefficients a and b.

Now, how do I find a and b without long division? I think there's something about the Remainder Theorem or maybe using roots of the divisor. Since x² - 1 factors into (x - 1)(x + 1), maybe I can use the fact that if I plug in x = 1 and x = -1 into f(x), those should equal the remainder evaluated at those points. Wait, let me recall: if you divide f(x) by (x - c), the remainder is f(c). But here, the divisor is quadratic, so maybe I need to use both roots?

Yes, I think that's the way. If I have a remainder of ax + b, then when I plug in x = 1, f(1) should equal a(1) + b, and when I plug in x = -1, f(-1) should equal a(-1) + b. Then I can set up two equations and solve for a and b. That seems manageable. Let me try that.

First, let me compute f(1). Let's substitute x = 1 into the polynomial:

f(1) = (1)^10 + 5*(1)^9 - 8*(1)^8 + 7*(1)^7 - (1)^6 - 12*(1)^5 + 4*(1)^4 - 8*(1)^3 + 12*(1)^2 - 5*(1) - 5.

Calculating each term:

1^10 = 1

5*1^9 = 5*1 = 5

-8*1^8 = -8*1 = -8

7*1^7 = 7*1 = 7

-1^6 = -1

-12*1^5 = -12*1 = -12

4*1^4 = 4*1 = 4

-8*1^3 = -8*1 = -8

12*1^2 = 12*1 = 12

-5*1 = -5

-5 remains as is.

Now add all these up:

1 + 5 = 6

6 - 8 = -2

-2 + 7 = 5

5 - 1 = 4

4 - 12 = -8

-8 + 4 = -4

-4 - 8 = -12

-12 + 12 = 0

0 - 5 = -5

-5 - 5 = -10

So f(1) = -10. Therefore, according to the remainder, when x = 1, ax + b = a(1) + b = a + b = -10. So equation one: a + b = -10.

Now let's compute f(-1). Substitute x = -1 into f(x):

f(-1) = (-1)^10 + 5*(-1)^9 - 8*(-1)^8 + 7*(-1)^7 - (-1)^6 - 12*(-1)^5 + 4*(-1)^4 - 8*(-1)^3 + 12*(-1)^2 - 5*(-1) - 5.

Let me compute each term step by step:

(-1)^10 = 1 (since even exponent)

5*(-1)^9 = 5*(-1) = -5

-8*(-1)^8 = -8*(1) = -8

7*(-1)^7 = 7*(-1) = -7

-(-1)^6 = -1 (since (-1)^6 is 1, so negative of that is -1)

-12*(-1)^5 = -12*(-1) = 12

4*(-1)^4 = 4*(1) = 4

-8*(-1)^3 = -8*(-1) = 8

12*(-1)^2 = 12*(1) = 12

-5*(-1) = 5

-5 remains as is.

Now let's add them up:

Start with 1.

1 - 5 = -4

-4 - 8 = -12

-12 - 7 = -19

-19 - 1 = -20

-20 + 12 = -8

-8 + 4 = -4

-4 + 8 = 4

4 + 12 = 16

16 + 5 = 21

21 - 5 = 16

So f(-1) = 16. Therefore, according to the remainder, when x = -1, ax + b = a*(-1) + b = -a + b = 16. So equation two: -a + b = 16.

Now we have two equations:

1) a + b = -10

2) -a + b = 16

Let me solve these equations. Let's subtract equation 1 from equation 2? Wait, maybe add them? Let me see:

If I subtract equation 1 from equation 2:

(-a + b) - (a + b) = 16 - (-10)

Left side: -a + b - a - b = -2a

Right side: 16 + 10 = 26

So -2a = 26 => a = -13

Then plug back into equation 1: -13 + b = -10 => b = -10 + 13 = 3

So the remainder is ax + b = -13x + 3. Let me check if this makes sense.

Wait, let me verify my calculations for f(1) and f(-1) again because it's easy to make a mistake with all those signs.

Starting with f(1):

Original polynomial:

x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

At x=1, all terms become 1, except the coefficients:

So:

1 +5 -8 +7 -1 -12 +4 -8 +12 -5 -5

Let me compute step by step:

Start with 1 +5 = 6

6 -8 = -2

-2 +7 = 5

5 -1 = 4

4 -12 = -8

-8 +4 = -4

-4 -8 = -12

-12 +12 = 0

0 -5 = -5

-5 -5 = -10. Okay, that's correct. So f(1) = -10.

Now f(-1):

Original polynomial:

x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

At x=-1:

(-1)^10 = 1

5*(-1)^9 = 5*(-1) = -5

-8*(-1)^8 = -8*(1) = -8

7*(-1)^7 = 7*(-1) = -7

-(-1)^6 = -1

-12*(-1)^5 = -12*(-1) = 12

4*(-1)^4 = 4*(1) = 4

-8*(-1)^3 = -8*(-1) = 8

12*(-1)^2 = 12*(1) = 12

-5*(-1) = 5

-5 remains.

So adding them up:

1 -5 -8 -7 -1 +12 +4 +8 +12 +5 -5

Let me compute step by step:

Start with 1.

1 -5 = -4

-4 -8 = -12

-12 -7 = -19

-19 -1 = -20

-20 +12 = -8

-8 +4 = -4

-4 +8 = 4

4 +12 = 16

16 +5 = 21

21 -5 = 16. Okay, that's correct. So f(-1) = 16.

So equations:

a + b = -10

- a + b = 16

Subtracting first equation from second:

(-a + b) - (a + b) = 16 - (-10)

Which is -2a = 26 => a = -13. Then b = -10 - a = -10 - (-13) = 3. So remainder is -13x + 3. Let me check if that makes sense.

Alternatively, maybe I can use another method to verify. For example, since x² - 1 divides into f(x) with remainder ax + b, then f(x) = (x² - 1)Q(x) + ax + b. Therefore, if I can express f(x) in terms of x² - 1, but maybe that's similar to long division. Alternatively, maybe I can use substitution.

Alternatively, since x² ≡ 1 mod (x² - 1), so any higher power of x can be reduced by replacing x² with 1. For example, x³ = x*x² ≡ x*1 = x, x^4 = x²*x² ≡ 1*1 = 1, x^5 = x^4*x ≡ 1*x = x, and so on. So in general, even exponents can be reduced to 1, odd exponents to x. Therefore, maybe I can rewrite f(x) by replacing x² with 1, x³ with x, x^4 with 1, etc., and then compute the result. That would give me the remainder.

Let me try that approach. Let me take each term of f(x) and reduce it modulo x² - 1.

Original polynomial:

f(x) = x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

Let me handle each term:

x^10: Since x² ≡ 1, so x^10 = (x²)^5 ≡ 1^5 = 1

5x^9: x^9 = x^(2*4 + 1) = (x²)^4 * x ≡ 1^4 * x = x, so 5x^9 ≡ 5x

-8x^8: x^8 = (x²)^4 ≡ 1^4 = 1, so -8x^8 ≡ -8*1 = -8

7x^7: x^7 = x^(2*3 + 1) = (x²)^3 * x ≡ 1^3 * x = x, so 7x^7 ≡ 7x

- x^6: x^6 = (x²)^3 ≡ 1^3 = 1, so -x^6 ≡ -1

-12x^5: x^5 = x^(2*2 + 1) = (x²)^2 * x ≡ 1^2 * x = x, so -12x^5 ≡ -12x

4x^4: x^4 = (x²)^2 ≡ 1^2 = 1, so 4x^4 ≡ 4*1 = 4

-8x^3: x^3 = x^(2*1 + 1) = (x²)^1 * x ≡ 1 * x = x, so -8x^3 ≡ -8x

12x^2: x^2 ≡ 1, so 12x^2 ≡ 12*1 = 12

-5x: remains as -5x

-5: remains as -5

Now, let me write all these reduced terms:

1 + 5x -8 +7x -1 -12x +4 -8x +12 -5x -5

Now combine like terms:

First, constants:

1 -8 -1 +4 +12 -5 = Let's compute step by step:

Start with 1 -8 = -7

-7 -1 = -8

-8 +4 = -4

-4 +12 = 8

8 -5 = 3

Now the x terms:

5x +7x -12x -8x -5x

Compute coefficients:

5 +7 = 12; 12 -12 = 0; 0 -8 = -8; -8 -5 = -13. So total x term is -13x.

Therefore, combining constants and x terms: -13x + 3. Which matches the previous result! So that confirms that the remainder is indeed -13x + 3. Therefore, the answer should be -13x + 3. But let me check once more to be sure.

Alternatively, maybe I can plug in x = 1 and x = -1 into the remainder and see if they match f(1) and f(-1). Let me check:

For x = 1: remainder is -13(1) + 3 = -13 + 3 = -10, which matches f(1) = -10. Good.

For x = -1: remainder is -13(-1) + 3 = 13 + 3 = 16, which matches f(-1) = 16. Perfect. So that's correct.

Therefore, the remainder is -13x + 3. But let me check if the problem expects the answer in a specific format. The question says "find the remainder", so it's acceptable to write it as -13x + 3, but sometimes people prefer writing it with positive leading coefficient. However, since the question doesn't specify, both forms are correct. But maybe they want it in the form ax + b, so -13x + 3 is fine.

Alternatively, maybe I can write it as 3 -13x, but both are equivalent. Let me check if there's a standard way. Typically, the remainder is written as ax + b where a and b are constants, so -13x + 3 is acceptable.

Therefore, I think that's the answer. Let me recap:

By using the fact that the remainder when dividing by x² - 1 is linear, and evaluating f(1) and f(-1), we set up equations to solve for the coefficients. Alternatively, by reducing each term modulo x² - 1, we also arrived at the same result. Both methods confirm that the remainder is -13x + 3.

**Final Answer**
The remainder is \boxed{-13x + 3}.

To find the remainder when the polynomial
$$
f(x) = x^{10} + 5x^9 - 8x^8 + 7x^7 - x^6 - 12x^5 + 4x^4 - 8x^3 + 12x^2 - 5x - 5
$$
is divided by $ x^2 - 1 $, we can use the **Remainder Theorem** in a clever way.

---

### Step 1: Understand the Structure of the Divisor

The divisor is $ x^2 - 1 $, which factors as $ (x - 1)(x + 1) $. When dividing a polynomial $ f(x) $ by a quadratic, the **remainder** will be a **linear polynomial** of the form:
$$
R(x) = ax + b
$$

This is because the degree of the remainder must be less than the degree of the divisor (which is 2), so the remainder is of degree at most 1.

---

### Step 2: Use the Roots of the Divisor

Since $ x^2 - 1 = 0 $ has roots $ x = 1 $ and $ x = -1 $, we can use the **Remainder Theorem** for polynomials:

- $ f(1) = R(1) = a(1) + b = a + b $
- $ f(-1) = R(-1) = a(-1) + b = -a + b $

So, we can compute $ f(1) $ and $ f(-1) $, and solve the system of equations to find $ a $ and $ b $.

---

### Step 3: Compute $ f(1) $ and $ f(-1) $

#### Compute $ f(1) $:

$$
f(1) = 1^{10} + 5 \cdot 1^9 - 8 \cdot 1^8 + 7 \cdot 1^7 - 1^6 - 12 \cdot 1^5 + 4 \cdot 1^4 - 8 \cdot 1^3 + 12 \cdot 1^2 - 5 \cdot 1 - 5
$$
$$
= 1 + 5 - 8 + 7 - 1 - 12 + 4 - 8 + 12 - 5 - 5 = -10
$$

#### Compute $ f(-1) $:

$$
f(-1) = (-1)^{10} + 5 \cdot (-1)^9 - 8 \cdot (-1)^8 + 7 \cdot (-1)^7 - (-1)^6 - 12 \cdot (-1)^5 + 4 \cdot (-1)^4 - 8 \cdot (-1)^3 + 12 \cdot (-1)^2 - 5 \cdot (-1) - 5
$$
$$
= 1 - 5 - 8 - 7 - 1 + 12 + 4 + 8 + 12 + 5 - 5 = 16
$$

---

### Step 4: Solve the System of Equations

We now have:
$$
\begin{cases}
a + b = -10 \\
-a + b = 16
\end{cases}
$$

Subtract the first equation from the second:
$$
(-a + b) - (a + b) = 16 - (-10) \Rightarrow -2a = 26 \Rightarrow a = -13
$$

Substitute $ a = -13 $ into the first equation:
$$
-13 + b = -10 \Rightarrow b = 3
$$

---

### Step 5: Write the Remainder

The remainder is:
$$
R(x) = ax + b = -13x + 3
$$

---

### Final Answer

$$
\boxed{-13x + 3}
$$

[Assistant]
Explanation:
Okay, so I need to find the remainder when this big polynomial f(x) is divided by x² - 1. The problem says not to use long division, which makes sense because that would be really tedious with such a high-degree polynomial. Hmm, I remember that when you divide a polynomial by a quadratic like x² - 1, the remainder should be a linear polynomial, right? So the remainder will be of the form ax + b. My job is to find the coefficients a and b.

Now, how do I find a and b without long division? I think there's something about the Remainder Theorem or maybe using roots of the divisor. Since x² - 1 factors into (x - 1)(x + 1), maybe I can use the fact that if I plug in x = 1 and x = -1 into f(x), those should equal the remainder evaluated at those points. Wait, let me recall: if you divide f(x) by (x - c), the remainder is f(c). But here, the divisor is quadratic, so maybe I need to use both roots?

Yes, I think that's the way. If I have a remainder of ax + b, then when I plug in x = 1, f(1) should equal a(1) + b, and when I plug in x = -1, f(-1) should equal a(-1) + b. Then I can set up two equations and solve for a and b. That seems manageable. Let me try that.

First, let me compute f(1). Let's substitute x = 1 into the polynomial:

f(1) = (1)^10 + 5*(1)^9 - 8*(1)^8 + 7*(1)^7 - (1)^6 - 12*(1)^5 + 4*(1)^4 - 8*(1)^3 + 12*(1)^2 - 5*(1) - 5.

Calculating each term:

1^10 = 1

5*1^9 = 5*1 = 5

-8*1^8 = -8*1 = -8

7*1^7 = 7*1 = 7

-1^6 = -1

-12*1^5 = -12*1 = -12

4*1^4 = 4*1 = 4

-8*1^3 = -8*1 = -8

12*1^2 = 12*1 = 12

-5*1 = -5

-5 remains as is.

Now add all these up:

1 + 5 = 6

6 - 8 = -2

-2 + 7 = 5

5 - 1 = 4

4 - 12 = -8

-8 + 4 = -4

-4 - 8 = -12

-12 + 12 = 0

0 - 5 = -5

-5 - 5 = -10

So f(1) = -10. Therefore, according to the remainder, when x = 1, ax + b = a(1) + b = a + b = -10. So equation one: a + b = -10.

Now let's compute f(-1). Substitute x = -1 into f(x):

f(-1) = (-1)^10 + 5*(-1)^9 - 8*(-1)^8 + 7*(-1)^7 - (-1)^6 - 12*(-1)^5 + 4*(-1)^4 - 8*(-1)^3 + 12*(-1)^2 - 5*(-1) - 5.

Let me compute each term step by step:

(-1)^10 = 1 (since even exponent)

5*(-1)^9 = 5*(-1) = -5

-8*(-1)^8 = -8*(1) = -8

7*(-1)^7 = 7*(-1) = -7

-(-1)^6 = -1 (since (-1)^6 is 1, so negative of that is -1)

-12*(-1)^5 = -12*(-1) = 12

4*(-1)^4 = 4*(1) = 4

-8*(-1)^3 = -8*(-1) = 8

12*(-1)^2 = 12*(1) = 12

-5*(-1) = 5

-5 remains as is.

Now let's add them up:

Start with 1.

1 - 5 = -4

-4 - 8 = -12

-12 - 7 = -19

-19 - 1 = -20

-20 + 12 = -8

-8 + 4 = -4

-4 + 8 = 4

4 + 12 = 16

16 + 5 = 21

21 - 5 = 16

So f(-1) = 16. Therefore, according to the remainder, when x = -1, ax + b = a*(-1) + b = -a + b = 16. So equation two: -a + b = 16.

Now we have two equations:

1) a + b = -10

2) -a + b = 16

Let me solve these equations. Let's subtract equation 1 from equation 2? Wait, maybe add them? Let me see:

If I subtract equation 1 from equation 2:

(-a + b) - (a + b) = 16 - (-10)

Left side: -a + b - a - b = -2a

Right side: 16 + 10 = 26

So -2a = 26 => a = -13

Then plug back into equation 1: -13 + b = -10 => b = -10 + 13 = 3

So the remainder is ax + b = -13x + 3. Let me check if this makes sense.

Wait, let me verify my calculations for f(1) and f(-1) again because it's easy to make a mistake with all those signs.

Starting with f(1):

Original polynomial:

x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

At x=1, all terms become 1, except the coefficients:

So:

1 +5 -8 +7 -1 -12 +4 -8 +12 -5 -5

Let me compute step by step:

Start with 1 +5 = 6

6 -8 = -2

-2 +7 = 5

5 -1 = 4

4 -12 = -8

-8 +4 = -4

-4 -8 = -12

-12 +12 = 0

0 -5 = -5

-5 -5 = -10. Okay, that's correct. So f(1) = -10.

Now f(-1):

Original polynomial:

x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

At x=-1:

(-1)^10 = 1

5*(-1)^9 = 5*(-1) = -5

-8*(-1)^8 = -8*(1) = -8

7*(-1)^7 = 7*(-1) = -7

-(-1)^6 = -1

-12*(-1)^5 = -12*(-1) = 12

4*(-1)^4 = 4*(1) = 4

-8*(-1)^3 = -8*(-1) = 8

12*(-1)^2 = 12*(1) = 12

-5*(-1) = 5

-5 remains.

So adding them up:

1 -5 -8 -7 -1 +12 +4 +8 +12 +5 -5

Let me compute step by step:

Start with 1.

1 -5 = -4

-4 -8 = -12

-12 -7 = -19

-19 -1 = -20

-20 +12 = -8

-8 +4 = -4

-4 +8 = 4

4 +12 = 16

16 +5 = 21

21 -5 = 16. Okay, that's correct. So f(-1) = 16.

So equations:

a + b = -10

- a + b = 16

Subtracting first equation from second:

(-a + b) - (a + b) = 16 - (-10)

Which is -2a = 26 => a = -13. Then b = -10 - a = -10 - (-13) = 3. So remainder is -13x + 3. Let me check if that makes sense.

Alternatively, maybe I can use another method to verify. For example, since x² - 1 divides into f(x) with remainder ax + b, then f(x) = (x² - 1)Q(x) + ax + b. Therefore, if I can express f(x) in terms of x² - 1, but maybe that's similar to long division. Alternatively, maybe I can use substitution.

Alternatively, since x² ≡ 1 mod (x² - 1), so any higher power of x can be reduced by replacing x² with 1. For example, x³ = x*x² ≡ x*1 = x, x^4 = x²*x² ≡ 1*1 = 1, x^5 = x^4*x ≡ 1*x = x, and so on. So in general, even exponents can be reduced to 1, odd exponents to x. Therefore, maybe I can rewrite f(x) by replacing x² with 1, x³ with x, x^4 with 1, etc., and then compute the result. That would give me the remainder.

Let me try that approach. Let me take each term of f(x) and reduce it modulo x² - 1.

Original polynomial:

f(x) = x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

Let me handle each term:

x^10: Since x² ≡ 1, so x^10 = (x²)^5 ≡ 1^5 = 1

5x^9: x^9 = x^(2*4 + 1) = (x²)^4 * x ≡ 1^4 * x = x, so 5x^9 ≡ 5x

-8x^8: x^8 = (x²)^4 ≡ 1^4 = 1, so -8x^8 ≡ -8*1 = -8

7x^7: x^7 = x^(2*3 + 1) = (x²)^3 * x ≡ 1^3 * x = x, so 7x^7 ≡ 7x

- x^6: x^6 = (x²)^3 ≡ 1^3 = 1, so -x^6 ≡ -1

-12x^5: x^5 = x^(2*2 + 1) = (x²)^2 * x ≡ 1^2 * x = x, so -12x^5 ≡ -12x

4x^4: x^4 = (x²)^2 ≡ 1^2 = 1, so 4x^4 ≡ 4*1 = 4

-8x^3: x^3 = x^(2*1 + 1) = (x²)^1 * x ≡ 1 * x = x, so -8x^3 ≡ -8x

12x^2: x^2 ≡ 1, so 12x^2 ≡ 12*1 = 12

-5x: remains as -5x

-5: remains as -5

Now, let me write all these reduced terms:

1 + 5x -8 +7x -1 -12x +4 -8x +12 -5x -5

Now combine like terms:

First, constants:

1 -8 -1 +4 +12 -5 = Let's compute step by step:

Start with 1 -8 = -7

-7 -1 = -8

-8 +4 = -4

-4 +12 = 8

8 -5 = 3

Now the x terms:

5x +7x -12x -8x -5x

Compute coefficients:

5 +7 = 12; 12 -12 = 0; 0 -8 = -8; -8 -5 = -13. So total x term is -13x.

Therefore, combining constants and x terms: -13x + 3. Which matches the previous result! So that confirms that the remainder is indeed -13x + 3. Therefore, the answer should be -13x + 3. But let me check once more to be sure.

Alternatively, maybe I can plug in x = 1 and x = -1 into the remainder and see if they match f(1) and f(-1). Let me check:

For x = 1: remainder is -13(1) + 3 = -13 + 3 = -10, which matches f(1) = -10. Good.

For x = -1: remainder is -13(-1) + 3 = 13 + 3 = 16, which matches f(-1) = 16. Perfect. So that's correct.

Therefore, the remainder is -13x + 3. But let me check if the problem expects the answer in a specific format. The question says "find the remainder", so it's acceptable to write it as -13x + 3, but sometimes people prefer writing it with positive leading coefficient. However, since the question doesn't specify, both forms are correct. But maybe they want it in the form ax + b, so -13x + 3 is fine.

Alternatively, maybe I can write it as 3 -13x, but both are equivalent. Let me check if there's a standard way. Typically, the remainder is written as ax + b where a and b are constants, so -13x + 3 is acceptable.

Therefore, I think that's the answer. Let me recap:

By using the fact that the remainder when dividing by x² - 1 is linear, and evaluating f(1) and f(-1), we set up equations to solve for the coefficients. Alternatively, by reducing each term modulo x² - 1, we also arrived at the same result. Both methods confirm that the remainder is -13x + 3.

**Final Answer**
The remainder is \boxed{-13x + 3}.

To find the remainder when the polynomial
$$
f(x) = x^{10} + 5x^9 - 8x^8 + 7x^7 - x^6 - 12x^5 + 4x^4 - 8x^3 + 12x^2 - 5x - 5
$$
is divided by $ x^2 - 1 $, we can use the **Remainder Theorem** in a clever way.

---

### Step 1: Understand the Structure of the Divisor

The divisor is $ x^2 - 1 $, which factors as $ (x - 1)(x + 1) $. When dividing a polynomial $ f(x) $ by a quadratic, the **remainder** will be a **linear polynomial** of the form:
$$
R(x) = ax + b
$$

This is because the degree of the remainder must be less than the degree of the divisor (which is 2), so the remainder is of degree at most 1.

---

### Step 2: Use the Roots of the Divisor

Since $ x^2 - 1 = 0 $ has roots $ x = 1 $ and $ x = -1 $, we can use the **Remainder Theorem** for polynomials:

- $ f(1) = R(1) = a(1) + b = a + b $
- $ f(-1) = R(-1) = a(-1) + b = -a + b $

So, we can compute $ f(1) $ and $ f(-1) $, and solve the system of equations to find $ a $ and $ b $.

---

### Step 3: Compute $ f(1) $ and $ f(-1) $

#### Compute $ f(1) $:

$$
f(1) = 1^{10} + 5 \cdot 1^9 - 8 \cdot 1^8 + 7 \cdot 1^7 - 1^6 - 12 \cdot 1^5 + 4 \cdot 1^4 - 8 \cdot 1^3 + 12 \cdot 1^2 - 5 \cdot 1 - 5
$$
$$
= 1 + 5 - 8 + 7 - 1 - 12 + 4 - 8 + 12 - 5 - 5 = -10
$$

#### Compute $ f(-1) $:

$$
f(-1) = (-1)^{10} + 5 \cdot (-1)^9 - 8 \cdot (-1)^8 + 7 \cdot (-1)^7 - (-1)^6 - 12 \cdot (-1)^5 + 4 \cdot (-1)^4 - 8 \cdot (-1)^3 + 12 \cdot (-1)^2 - 5 \cdot (-1) - 5
$$
$$
= 1 - 5 - 8 - 7 - 1 + 12 + 4 + 8 + 12 + 5 - 5 = 16
$$

---

### Step 4: Solve the System of Equations

We now have:
$$
\begin{cases}
a + b = -10 \\
-a + b = 16
\end{cases}
$$

Subtract the first equation from the second:
$$
(-a + b) - (a + b) = 16 - (-10) \Rightarrow -2a = 26 \Rightarrow a = -13
$$

Substitute $ a = -13 $ into the first equation:
$$
-13 + b = -10 \Rightarrow b = 3
$$

---

### Step 5: Write the Remainder

The remainder is:
$$
R(x) = ax + b = -13x + 3
$$

---

### Final Answer

$$
\boxed{-13x + 3}
$$

Solution:
We have
$$f(x) = q(x)(x^2-1) + r(x),$$where $q(x)$ is the quotient and $r(x)$ is the remainder. Since $x^2-1$ is quadratic, the remainder is at most linear; let us write $r(x) = ax+b$.

Observe that $x=-1$ and $x=1$ are both zeroes of $x^2-1$. Thus $f(1)=r(1)$ and $f(-1)=r(-1)$.

We can use the given formula for $f(x)$ to compute $f(1)=-10$ and $f(-1)=16$. Thus we have the system of equations
$$\begin{cases}
-10 = a\cdot (1) + b,\\
\phantom{-}16 = a\cdot (-1) + b.
\end{cases}$$Adding these equations yields $6=2b$ and hence $b=3$. Substituting into either equation then yields $a=-13$.

Therefore, $r(x) = ax+b = \boxed{-13x+3}$.


Explanation:
Okay, so I need to compute A^27 + A^31 + A^40 where A is the matrix [[3, 7], [-1, -2]]. Hmm, matrix exponentiation can be tricky, especially with such high exponents. I remember that for diagonal matrices, you can just raise the diagonal entries to the power, but this matrix isn't diagonal. Maybe I can diagonalize it? If I can find eigenvalues and eigenvectors, then maybe I can express A as PDP^{-1} and then A^n would be PD^nP^{-1}. That sounds like a plan. Let me try that.

First, let me find the eigenvalues of A. The eigenvalues are the roots of the characteristic equation det(A - λI) = 0. So let me compute that determinant.

The matrix A - λI is [[3 - λ, 7], [-1, -2 - λ]]. The determinant is (3 - λ)(-2 - λ) - (7)(-1). Let me compute that:

(3 - λ)(-2 - λ) = 3*(-2) + 3*(-λ) - λ*(-2) - λ*λ = -6 - 3λ + 2λ + λ² = λ² - λ - 6.

Then subtract (7)(-1) which is -7*(-1) = +7? Wait, no. Wait, determinant is (top left * bottom right) - (top right * bottom left). So it's (3 - λ)(-2 - λ) - (7)(-1). So that's (λ² - λ - 6) + 7. Wait, let me recalculate:

First, expand (3 - λ)(-2 - λ):

Multiply 3 by -2: -6

Multiply 3 by -λ: -3λ

Multiply -λ by -2: +2λ

Multiply -λ by -λ: +λ²

So altogether: -6 -3λ + 2λ + λ² = λ² - λ -6. Then subtract (7)(-1) which is -7*(-1) = +7? Wait, no. Wait, determinant is (top left * bottom right) - (top right * bottom left). So the first term is (3 - λ)(-2 - λ) and the second term is 7*(-1). So determinant is (3 - λ)(-2 - λ) - (7)(-1). So that is (λ² - λ -6) - (-7) = λ² - λ -6 +7 = λ² - λ +1. Wait, is that right?

Wait, let me check again:

Original determinant:

(3 - λ)(-2 - λ) - (7)(-1) = ?

First term: (3 - λ)(-2 - λ) = let me compute it again:

Multiply (3 - λ)(-2 - λ) = 3*(-2) + 3*(-λ) - λ*(-2) - λ*λ = -6 -3λ + 2λ + λ² = λ² - λ -6. Then subtract (7)(-1) which is -7*(-1) = +7? Wait, no. Wait, the determinant is (top left * bottom right) - (top right * bottom left). So the second term is 7*(-1). So the determinant is (λ² - λ -6) - [7*(-1)] = λ² - λ -6 +7 = λ² - λ +1. Yes, that's correct. So the characteristic equation is λ² - λ +1 = 0. Let me solve that.

Using quadratic formula: λ = [1 ± sqrt(1 - 4*1*1)] / 2 = [1 ± sqrt(-3)] / 2. So the eigenvalues are complex: (1 ± i√3)/2. Hmm, complex eigenvalues. That complicates things a bit. But maybe I can still diagonalize it using complex numbers? Or perhaps there's another approach?

Alternatively, maybe I can find a recurrence relation or use the fact that the matrix satisfies its own characteristic equation? Since the characteristic equation is λ² - λ +1 = 0, so A² - A + I = 0. Therefore, A² = A - I. That seems useful. Let me check that.

If A² - A + I = 0, then A² = A - I. So that's a relation that allows me to express higher powers of A in terms of lower powers. That might be helpful for computing A^n. Let me verify this with the matrix.

Compute A²:

A is [[3, 7], [-1, -2]]. Let me compute A²:

First row first column: 3*3 + 7*(-1) = 9 -7 = 2

First row second column: 3*7 + 7*(-2) = 21 -14 = 7

Second row first column: -1*3 + (-2)*(-1) = -3 + 2 = -1

Second row second column: -1*7 + (-2)*(-2) = -7 + 4 = -3

So A² is [[2, 7], [-1, -3]]

Now compute A - I: A - I is [[3 -1, 7], [-1, -2 -1]] = [[2, 7], [-1, -3]]. Which matches A². Great! So indeed, A² = A - I. So that's the relation. Therefore, we can use this recurrence to find higher powers of A.

So, since A² = A - I, maybe we can find a pattern or a recurrence relation for A^n. Let me think. Let me denote that A² = A - I. Then perhaps we can find a linear recurrence for A^n?

Let me try to find a recurrence. Let me suppose that A^n = p_n A + q_n I, where p_n and q_n are coefficients to be determined. Let me check if this is possible.

Assume A^n = p_n A + q_n I. Then, for n=1, A^1 = 1*A + 0*I, so p_1=1, q_1=0.

For n=2, A² = A - I, so p_2=1, q_2=-1.

Now, let's try to find a recurrence relation for p_n and q_n. Let me compute A^{n+1} = A^n * A. If A^n = p_n A + q_n I, then A^{n+1} = (p_n A + q_n I) * A = p_n A² + q_n A. But since A² = A - I, substitute that in:

A^{n+1} = p_n (A - I) + q_n A = p_n A - p_n I + q_n A = (p_n + q_n) A - p_n I.

Therefore, A^{n+1} = (p_n + q_n) A - p_n I. Therefore, comparing with the assumed form A^{n+1} = p_{n+1} A + q_{n+1} I, we get:

p_{n+1} = p_n + q_n

q_{n+1} = -p_n

So we have the recurrence relations:

p_{n+1} = p_n + q_n

q_{n+1} = -p_n

With initial conditions p_1 = 1, q_1 = 0.

So we can compute p_n and q_n for any n. Let me try to compute them for some n to see if there's a pattern.

Let me compute p_1 = 1, q_1 = 0.

Then p_2 = p_1 + q_1 = 1 + 0 = 1

q_2 = -p_1 = -1

Then p_3 = p_2 + q_2 = 1 + (-1) = 0

q_3 = -p_2 = -1

p_4 = p_3 + q_3 = 0 + (-1) = -1

q_4 = -p_3 = -0 = 0

p_5 = p_4 + q_4 = -1 + 0 = -1

q_5 = -p_4 = -(-1) = 1

p_6 = p_5 + q_5 = -1 + 1 = 0

q_6 = -p_5 = -(-1) = 1

p_7 = p_6 + q_6 = 0 + 1 = 1

q_7 = -p_6 = -0 = 0

p_8 = p_7 + q_7 = 1 + 0 = 1

q_8 = -p_7 = -1

Wait, this seems to be repeating the earlier values. Let me check:

p_1=1, q_1=0

p_2=1, q_2=-1

p_3=0, q_3=-1

p_4=-1, q_4=0

p_5=-1, q_5=1

p_6=0, q_6=1

p_7=1, q_7=0

p_8=1, q_8=-1

Yes, so the sequence for p_n and q_n is periodic with period 6? Let me check:

From p_1 to p_7: 1,1,0,-1,-1,0,1,...

Similarly for q_n: 0,-1,-1,0,1,1,0,...

So the cycle is 6. Therefore, the sequence of p_n and q_n is periodic with period 6. Therefore, if we can find n modulo 6, we can find p_n and q_n.

Therefore, since we need to compute A^27, A^31, A^40, let me compute p_n and q_n for those exponents.

But first, let me confirm the periodicity. Let me check p_7=1, which is same as p_1=1. Similarly, p_8=1=p_2. So yes, period 6. Therefore, for any n, p_n = p_{n mod 6}, but need to adjust for the offset. Let me check:

Let me note that the sequence for p_n is:

n: 1 2 3 4 5 6 7 8 9 10 11 12...

p_n:1,1,0,-1,-1,0,1,1,0,-1,-1,0...

Similarly for q_n:

n:1 2 3 4 5 6 7 8 9 10 11 12...

q_n:0,-1,-1,0,1,1,0,-1,-1,0,1,1...

So the period is 6. Therefore, for any n, p_n = p_{n mod 6}, but if n mod 6 is 0, then it's p_6. Similarly for q_n.

So let me compute for each exponent:

First, compute A^27. Let me find n=27. Let me compute 27 mod 6. 6*4=24, 27-24=3. So 27 mod 6 is 3. Therefore, p_27 = p_3 = 0, q_27 = q_3 = -1. Therefore, A^27 = p_27 A + q_27 I = 0*A + (-1)*I = -I.

Wait, that's interesting. Let me check with n=3. For n=3, p_3=0, q_3=-1. So A^3 = 0*A -1*I = -I. Let me verify that by computing A^3 directly. Since A^2 = A - I, then A^3 = A^2 * A = (A - I) * A = A^2 - A = (A - I) - A = -I. Yes! So A^3 = -I. Therefore, A^3 = -I. Therefore, A^3 = -I, so A^6 = (A^3)^2 = (-I)^2 = I. Therefore, A^6 = I. Therefore, A^6 is identity matrix. Therefore, A^6 = I. Therefore, A^n has period 6. Therefore, A^{k} = A^{k mod 6} if we adjust for the modulus. Wait, but since A^6 = I, then A^{6m + r} = (A^6)^m * A^r = I^m * A^r = A^r. Therefore, A^n = A^{n mod 6} if n mod 6 ≠ 0, else A^6 = I. Therefore, for exponents, we can reduce them modulo 6. Therefore, for example:

A^27: 27 divided by 6 is 4*6=24, remainder 3. So A^27 = A^3 = -I.

Similarly, A^31: 31 divided by 6 is 5*6=30, remainder 1. So A^31 = A^1 = A.

A^40: 40 divided by 6 is 6*6=36, remainder 4. So A^40 = A^4.

Therefore, the original expression is A^27 + A^31 + A^40 = (-I) + A + A^4.

So now, I need to compute (-I) + A + A^4. Let me compute each term.

First, let me compute A^4. Since A^3 = -I, then A^4 = A^3 * A = (-I)*A = -A. Therefore, A^4 = -A. Therefore, substituting back:

(-I) + A + (-A) = -I + A - A = -I. Therefore, the entire expression simplifies to -I. Wait, that seems too simple. Let me check again.

Wait, A^4 = -A? Let me verify that. Since A^3 = -I, then A^4 = A^3 * A = (-I) * A = -A. Yes, that's correct. Therefore, A^4 = -A. Therefore, substituting into the expression:

A^27 + A^31 + A^40 = (-I) + A + (-A) = -I + A - A = -I. Therefore, the answer is -I, which is [[-1, 0], [0, -1]].

Wait, but let me check again step by step to make sure I didn't make a mistake.

First, A^3 = -I. Then A^4 = A^3 * A = (-I) * A = -A. Then A^4 = -A. Then A^27 = A^{6*4 + 3} = (A^6)^4 * A^3 = I^4 * A^3 = A^3 = -I. Similarly, A^31 = A^{6*5 + 1} = I^5 * A^1 = A. A^40 = A^{6*6 + 4} = I^6 * A^4 = A^4 = -A. Therefore, adding them up: -I + A - A = -I. So yes, that seems correct.

But let me check with another approach. Let me compute A^4 directly. Since A^2 = A - I, then A^3 = A^2 * A = (A - I) * A = A^2 - A = (A - I) - A = -I. Then A^4 = A^3 * A = (-I) * A = -A. So that's correct. Therefore, the conclusion is that the sum is -I. Therefore, the answer is -I, which is the matrix [[-1, 0], [0, -1]].

But let me check with another exponent. For example, A^6 = (A^3)^2 = (-I)^2 = I. So A^6 = I. Therefore, A^6 is identity. Therefore, A^27 = A^{6*4 + 3} = (A^6)^4 * A^3 = I^4 * A^3 = A^3 = -I. Similarly, A^31 = A^{6*5 + 1} = A^1 = A. A^40 = A^{6*6 + 4} = A^4 = -A. Therefore, adding them up: -I + A - A = -I. So that's correct.

But let me check if there's a possibility that I made a mistake in the periodicity. Let me compute A^4 manually. Earlier, I computed A^2 as [[2,7],[-1,-3]]. Then A^3 = A^2 * A. Let me compute that:

A^2 is [[2,7],[-1,-3]], A is [[3,7],[-1,-2]]. Multiply them:

First row first column: 2*3 + 7*(-1) = 6 -7 = -1

First row second column: 2*7 + 7*(-2) = 14 -14 = 0

Second row first column: -1*3 + (-3)*(-1) = -3 + 3 = 0

Second row second column: -1*7 + (-3)*(-2) = -7 + 6 = -1

So A^3 is [[-1, 0], [0, -1]] which is -I. Correct. Then A^4 = A^3 * A = (-I)*A = -A. So A^4 is [[-3, -7], [1, 2]]. Let me check that:

Original A is [[3,7],[-1,-2]]. Multiply by -1: [[-3, -7], [1, 2]]. Yes, that's correct. So A^4 = -A. Therefore, adding A^27 + A^31 + A^40 = -I + A + (-A) = -I. Therefore, the answer is -I. Therefore, the final answer is the matrix with -1 on the diagonal and 0 elsewhere.

But let me check once more with another exponent. Let me compute A^6. Since A^3 = -I, then A^6 = (A^3)^2 = (-I)^2 = I. So A^6 is identity. Therefore, A^6 = I. Therefore, A^6 = I. Therefore, A^27 = A^{6*4 + 3} = (A^6)^4 * A^3 = I^4 * A^3 = A^3 = -I. Correct. Similarly, A^31 = A^{6*5 + 1} = A^1 = A. A^40 = A^{6*6 + 4} = A^4 = -A. Therefore, adding them up: -I + A - A = -I. So yes, that's correct.

Therefore, the answer is -I, which is [[-1, 0], [0, -1]]. Therefore, boxed, it would be \boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}.

But let me check once more if there's a mistake in the logic. Suppose I made a mistake in the recurrence relations. Let me check for n=4. According to the earlier recurrence, p_4 = -1, q_4 = 0. Therefore, A^4 = p_4 A + q_4 I = -1*A + 0*I = -A. Which matches our manual calculation. Similarly, for n=5, p_5=-1, q_5=1. Therefore, A^5 = -A + I. Let me check that. Since A^4 = -A, then A^5 = A^4 * A = (-A)*A = -A^2. But A^2 = A - I, so -A^2 = -A + I. Which matches. Therefore, yes, A^5 = -A + I. Therefore, the recurrence is correct.

Therefore, the conclusion is correct. Therefore, the answer is -I. Therefore, the final answer is the matrix with -1s on the diagonal and 0s elsewhere.

**Final Answer**
\boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}

To compute the expression
$$
\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40}
$$
where
$$
\mathbf{A} = \begin{pmatrix} 3 & 7 \\ -1 & -2 \end{pmatrix},
$$
we begin by analyzing the structure of the matrix and its powers.

---

### Step 1: Characteristic Equation and Matrix Identity

We compute the characteristic polynomial of $\mathbf{A}$:

$$
\det(\mathbf{A} - \lambda \mathbf{I}) = \begin{vmatrix} 3 - \lambda & 7 \\ -1 & -2 - \lambda \end{vmatrix}
= (3 - \lambda)(-2 - \lambda) - (7)(-1)
= \lambda^2 - \lambda + 1.
$$

This gives the characteristic equation:
$$
\lambda^2 - \lambda + 1 = 0.
$$

The roots are complex:
$$
\lambda = \frac{1 \pm i\sqrt{3}}{2}.
$$

However, instead of working with complex eigenvalues, we can use the **Cayley-Hamilton Theorem**, which tells us that a matrix satisfies its own characteristic equation. Thus:
$$
\mathbf{A}^2 - \mathbf{A} + \mathbf{I} = \mathbf{0} \quad \Rightarrow \quad \mathbf{A}^2 = \mathbf{A} - \mathbf{I}.
$$

This is a key identity that allows us to express higher powers of $\mathbf{A}$ in terms of lower powers.

---

### Step 2: Finding a Pattern in Powers of $\mathbf{A}$

Using the identity $\mathbf{A}^2 = \mathbf{A} - \mathbf{I}$, we can compute:

- $\mathbf{A}^3 = \mathbf{A}^2 \cdot \mathbf{A} = (\mathbf{A} - \mathbf{I}) \cdot \mathbf{A} = \mathbf{A}^2 - \mathbf{A} = (\mathbf{A} - \mathbf{I}) - \mathbf{A} = -\mathbf{I}$,
- $\mathbf{A}^4 = \mathbf{A}^3 \cdot \mathbf{A} = (-\mathbf{I}) \cdot \mathbf{A} = -\mathbf{A}$,
- $\mathbf{A}^5 = \mathbf{A}^4 \cdot \mathbf{A} = (-\mathbf{A}) \cdot \mathbf{A} = -\mathbf{A}^2 = -(\mathbf{A} - \mathbf{I}) = -\mathbf{A} + \mathbf{I}$,
- $\mathbf{A}^6 = \mathbf{A}^5 \cdot \mathbf{A} = (-\mathbf{A} + \mathbf{I}) \cdot \mathbf{A} = -\mathbf{A}^2 + \mathbf{A} = -(\mathbf{A} - \mathbf{I}) + \mathbf{A} = \mathbf{I}$.

So we find that:
$$
\mathbf{A}^6 = \mathbf{I} \quad \Rightarrow \quad \text{The powers of } \mathbf{A} \text{ are periodic with period 6}.
$$

---

### Step 3: Reduce Exponents Modulo 6

We now reduce the exponents modulo 6:

- $27 \mod 6 = 3 \Rightarrow \mathbf{A}^{27} = \mathbf{A}^3 = -\mathbf{I}$,
- $31 \mod 6 = 1 \Rightarrow \mathbf{A}^{31} = \mathbf{A}^1 = \mathbf{A}$,
- $40 \mod 6 = 4 \Rightarrow \mathbf{A}^{40} = \mathbf{A}^4 = -\mathbf{A}$.

---

### Step 4: Compute the Final Expression

Now compute:
$$
\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40} = (-\mathbf{I}) + \mathbf{A} + (-\mathbf{A}) = -\mathbf{I}.
$$

---

### Final Answer

$$
\boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}
$$

[Assistant]
Explanation:
Okay, so I need to compute A^27 + A^31 + A^40 where A is the matrix [[3, 7], [-1, -2]]. Hmm, matrix exponentiation can be tricky, especially with such high exponents. I remember that for diagonal matrices, you can just raise the diagonal entries to the power, but this matrix isn't diagonal. Maybe I can diagonalize it? If I can find eigenvalues and eigenvectors, then maybe I can express A as PDP^{-1} and then A^n would be PD^nP^{-1}. That sounds like a plan. Let me try that.

First, let me find the eigenvalues of A. The eigenvalues are the roots of the characteristic equation det(A - λI) = 0. So let me compute that determinant.

The matrix A - λI is [[3 - λ, 7], [-1, -2 - λ]]. The determinant is (3 - λ)(-2 - λ) - (7)(-1). Let me compute that:

(3 - λ)(-2 - λ) = 3*(-2) + 3*(-λ) - λ*(-2) - λ*λ = -6 - 3λ + 2λ + λ² = λ² - λ - 6.

Then subtract (7)(-1) which is -7*(-1) = +7? Wait, no. Wait, determinant is (top left * bottom right) - (top right * bottom left). So it's (3 - λ)(-2 - λ) - (7)(-1). So that's (λ² - λ - 6) + 7. Wait, let me recalculate:

First, expand (3 - λ)(-2 - λ):

Multiply 3 by -2: -6

Multiply 3 by -λ: -3λ

Multiply -λ by -2: +2λ

Multiply -λ by -λ: +λ²

So altogether: -6 -3λ + 2λ + λ² = λ² - λ -6. Then subtract (7)(-1) which is -7*(-1) = +7? Wait, no. Wait, determinant is (top left * bottom right) - (top right * bottom left). So the first term is (3 - λ)(-2 - λ) and the second term is 7*(-1). So determinant is (3 - λ)(-2 - λ) - (7)(-1). So that is (λ² - λ -6) - (-7) = λ² - λ -6 +7 = λ² - λ +1. Wait, is that right?

Wait, let me check again:

Original determinant:

(3 - λ)(-2 - λ) - (7)(-1) = ?

First term: (3 - λ)(-2 - λ) = let me compute it again:

Multiply (3 - λ)(-2 - λ) = 3*(-2) + 3*(-λ) - λ*(-2) - λ*λ = -6 -3λ + 2λ + λ² = λ² - λ -6. Then subtract (7)(-1) which is -7*(-1) = +7? Wait, no. Wait, the determinant is (top left * bottom right) - (top right * bottom left). So the second term is 7*(-1). So the determinant is (λ² - λ -6) - [7*(-1)] = λ² - λ -6 +7 = λ² - λ +1. Yes, that's correct. So the characteristic equation is λ² - λ +1 = 0. Let me solve that.

Using quadratic formula: λ = [1 ± sqrt(1 - 4*1*1)] / 2 = [1 ± sqrt(-3)] / 2. So the eigenvalues are complex: (1 ± i√3)/2. Hmm, complex eigenvalues. That complicates things a bit. But maybe I can still diagonalize it using complex numbers? Or perhaps there's another approach?

Alternatively, maybe I can find a recurrence relation or use the fact that the matrix satisfies its own characteristic equation? Since the characteristic equation is λ² - λ +1 = 0, so A² - A + I = 0. Therefore, A² = A - I. That seems useful. Let me check that.

If A² - A + I = 0, then A² = A - I. So that's a relation that allows me to express higher powers of A in terms of lower powers. That might be helpful for computing A^n. Let me verify this with the matrix.

Compute A²:

A is [[3, 7], [-1, -2]]. Let me compute A²:

First row first column: 3*3 + 7*(-1) = 9 -7 = 2

First row second column: 3*7 + 7*(-2) = 21 -14 = 7

Second row first column: -1*3 + (-2)*(-1) = -3 + 2 = -1

Second row second column: -1*7 + (-2)*(-2) = -7 + 4 = -3

So A² is [[2, 7], [-1, -3]]

Now compute A - I: A - I is [[3 -1, 7], [-1, -2 -1]] = [[2, 7], [-1, -3]]. Which matches A². Great! So indeed, A² = A - I. So that's the relation. Therefore, we can use this recurrence to find higher powers of A.

So, since A² = A - I, maybe we can find a pattern or a recurrence relation for A^n. Let me think. Let me denote that A² = A - I. Then perhaps we can find a linear recurrence for A^n?

Let me try to find a recurrence. Let me suppose that A^n = p_n A + q_n I, where p_n and q_n are coefficients to be determined. Let me check if this is possible.

Assume A^n = p_n A + q_n I. Then, for n=1, A^1 = 1*A + 0*I, so p_1=1, q_1=0.

For n=2, A² = A - I, so p_2=1, q_2=-1.

Now, let's try to find a recurrence relation for p_n and q_n. Let me compute A^{n+1} = A^n * A. If A^n = p_n A + q_n I, then A^{n+1} = (p_n A + q_n I) * A = p_n A² + q_n A. But since A² = A - I, substitute that in:

A^{n+1} = p_n (A - I) + q_n A = p_n A - p_n I + q_n A = (p_n + q_n) A - p_n I.

Therefore, A^{n+1} = (p_n + q_n) A - p_n I. Therefore, comparing with the assumed form A^{n+1} = p_{n+1} A + q_{n+1} I, we get:

p_{n+1} = p_n + q_n

q_{n+1} = -p_n

So we have the recurrence relations:

p_{n+1} = p_n + q_n

q_{n+1} = -p_n

With initial conditions p_1 = 1, q_1 = 0.

So we can compute p_n and q_n for any n. Let me try to compute them for some n to see if there's a pattern.

Let me compute p_1 = 1, q_1 = 0.

Then p_2 = p_1 + q_1 = 1 + 0 = 1

q_2 = -p_1 = -1

Then p_3 = p_2 + q_2 = 1 + (-1) = 0

q_3 = -p_2 = -1

p_4 = p_3 + q_3 = 0 + (-1) = -1

q_4 = -p_3 = -0 = 0

p_5 = p_4 + q_4 = -1 + 0 = -1

q_5 = -p_4 = -(-1) = 1

p_6 = p_5 + q_5 = -1 + 1 = 0

q_6 = -p_5 = -(-1) = 1

p_7 = p_6 + q_6 = 0 + 1 = 1

q_7 = -p_6 = -0 = 0

p_8 = p_7 + q_7 = 1 + 0 = 1

q_8 = -p_7 = -1

Wait, this seems to be repeating the earlier values. Let me check:

p_1=1, q_1=0

p_2=1, q_2=-1

p_3=0, q_3=-1

p_4=-1, q_4=0

p_5=-1, q_5=1

p_6=0, q_6=1

p_7=1, q_7=0

p_8=1, q_8=-1

Yes, so the sequence for p_n and q_n is periodic with period 6? Let me check:

From p_1 to p_7: 1,1,0,-1,-1,0,1,...

Similarly for q_n: 0,-1,-1,0,1,1,0,...

So the cycle is 6. Therefore, the sequence of p_n and q_n is periodic with period 6. Therefore, if we can find n modulo 6, we can find p_n and q_n.

Therefore, since we need to compute A^27, A^31, A^40, let me compute p_n and q_n for those exponents.

But first, let me confirm the periodicity. Let me check p_7=1, which is same as p_1=1. Similarly, p_8=1=p_2. So yes, period 6. Therefore, for any n, p_n = p_{n mod 6}, but need to adjust for the offset. Let me check:

Let me note that the sequence for p_n is:

n: 1 2 3 4 5 6 7 8 9 10 11 12...

p_n:1,1,0,-1,-1,0,1,1,0,-1,-1,0...

Similarly for q_n:

n:1 2 3 4 5 6 7 8 9 10 11 12...

q_n:0,-1,-1,0,1,1,0,-1,-1,0,1,1...

So the period is 6. Therefore, for any n, p_n = p_{n mod 6}, but if n mod 6 is 0, then it's p_6. Similarly for q_n.

So let me compute for each exponent:

First, compute A^27. Let me find n=27. Let me compute 27 mod 6. 6*4=24, 27-24=3. So 27 mod 6 is 3. Therefore, p_27 = p_3 = 0, q_27 = q_3 = -1. Therefore, A^27 = p_27 A + q_27 I = 0*A + (-1)*I = -I.

Wait, that's interesting. Let me check with n=3. For n=3, p_3=0, q_3=-1. So A^3 = 0*A -1*I = -I. Let me verify that by computing A^3 directly. Since A^2 = A - I, then A^3 = A^2 * A = (A - I) * A = A^2 - A = (A - I) - A = -I. Yes! So A^3 = -I. Therefore, A^3 = -I. Therefore, A^3 = -I, so A^6 = (A^3)^2 = (-I)^2 = I. Therefore, A^6 = I. Therefore, A^6 is identity matrix. Therefore, A^6 = I. Therefore, A^n has period 6. Therefore, A^{k} = A^{k mod 6} if we adjust for the modulus. Wait, but since A^6 = I, then A^{6m + r} = (A^6)^m * A^r = I^m * A^r = A^r. Therefore, A^n = A^{n mod 6} if n mod 6 ≠ 0, else A^6 = I. Therefore, for exponents, we can reduce them modulo 6. Therefore, for example:

A^27: 27 divided by 6 is 4*6=24, remainder 3. So A^27 = A^3 = -I.

Similarly, A^31: 31 divided by 6 is 5*6=30, remainder 1. So A^31 = A^1 = A.

A^40: 40 divided by 6 is 6*6=36, remainder 4. So A^40 = A^4.

Therefore, the original expression is A^27 + A^31 + A^40 = (-I) + A + A^4.

So now, I need to compute (-I) + A + A^4. Let me compute each term.

First, let me compute A^4. Since A^3 = -I, then A^4 = A^3 * A = (-I)*A = -A. Therefore, A^4 = -A. Therefore, substituting back:

(-I) + A + (-A) = -I + A - A = -I. Therefore, the entire expression simplifies to -I. Wait, that seems too simple. Let me check again.

Wait, A^4 = -A? Let me verify that. Since A^3 = -I, then A^4 = A^3 * A = (-I) * A = -A. Yes, that's correct. Therefore, A^4 = -A. Therefore, substituting into the expression:

A^27 + A^31 + A^40 = (-I) + A + (-A) = -I + A - A = -I. Therefore, the answer is -I, which is [[-1, 0], [0, -1]].

Wait, but let me check again step by step to make sure I didn't make a mistake.

First, A^3 = -I. Then A^4 = A^3 * A = (-I) * A = -A. Then A^4 = -A. Then A^27 = A^{6*4 + 3} = (A^6)^4 * A^3 = I^4 * A^3 = A^3 = -I. Similarly, A^31 = A^{6*5 + 1} = I^5 * A^1 = A. A^40 = A^{6*6 + 4} = I^6 * A^4 = A^4 = -A. Therefore, adding them up: -I + A - A = -I. So yes, that seems correct.

But let me check with another approach. Let me compute A^4 directly. Since A^2 = A - I, then A^3 = A^2 * A = (A - I) * A = A^2 - A = (A - I) - A = -I. Then A^4 = A^3 * A = (-I) * A = -A. So that's correct. Therefore, the conclusion is that the sum is -I. Therefore, the answer is -I, which is the matrix [[-1, 0], [0, -1]].

But let me check with another exponent. For example, A^6 = (A^3)^2 = (-I)^2 = I. So A^6 = I. Therefore, A^6 is identity. Therefore, A^27 = A^{6*4 + 3} = (A^6)^4 * A^3 = I^4 * A^3 = A^3 = -I. Similarly, A^31 = A^{6*5 + 1} = A^1 = A. A^40 = A^{6*6 + 4} = A^4 = -A. Therefore, adding them up: -I + A - A = -I. So that's correct.

But let me check if there's a possibility that I made a mistake in the periodicity. Let me compute A^4 manually. Earlier, I computed A^2 as [[2,7],[-1,-3]]. Then A^3 = A^2 * A. Let me compute that:

A^2 is [[2,7],[-1,-3]], A is [[3,7],[-1,-2]]. Multiply them:

First row first column: 2*3 + 7*(-1) = 6 -7 = -1

First row second column: 2*7 + 7*(-2) = 14 -14 = 0

Second row first column: -1*3 + (-3)*(-1) = -3 + 3 = 0

Second row second column: -1*7 + (-3)*(-2) = -7 + 6 = -1

So A^3 is [[-1, 0], [0, -1]] which is -I. Correct. Then A^4 = A^3 * A = (-I)*A = -A. So A^4 is [[-3, -7], [1, 2]]. Let me check that:

Original A is [[3,7],[-1,-2]]. Multiply by -1: [[-3, -7], [1, 2]]. Yes, that's correct. So A^4 = -A. Therefore, adding A^27 + A^31 + A^40 = -I + A + (-A) = -I. Therefore, the answer is -I. Therefore, the final answer is the matrix with -1 on the diagonal and 0 elsewhere.

But let me check once more with another exponent. Let me compute A^6. Since A^3 = -I, then A^6 = (A^3)^2 = (-I)^2 = I. So A^6 is identity. Therefore, A^6 = I. Therefore, A^6 = I. Therefore, A^27 = A^{6*4 + 3} = (A^6)^4 * A^3 = I^4 * A^3 = A^3 = -I. Correct. Similarly, A^31 = A^{6*5 + 1} = A^1 = A. A^40 = A^{6*6 + 4} = A^4 = -A. Therefore, adding them up: -I + A - A = -I. So yes, that's correct.

Therefore, the answer is -I, which is [[-1, 0], [0, -1]]. Therefore, boxed, it would be \boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}.

But let me check once more if there's a mistake in the logic. Suppose I made a mistake in the recurrence relations. Let me check for n=4. According to the earlier recurrence, p_4 = -1, q_4 = 0. Therefore, A^4 = p_4 A + q_4 I = -1*A + 0*I = -A. Which matches our manual calculation. Similarly, for n=5, p_5=-1, q_5=1. Therefore, A^5 = -A + I. Let me check that. Since A^4 = -A, then A^5 = A^4 * A = (-A)*A = -A^2. But A^2 = A - I, so -A^2 = -A + I. Which matches. Therefore, yes, A^5 = -A + I. Therefore, the recurrence is correct.

Therefore, the conclusion is correct. Therefore, the answer is -I. Therefore, the final answer is the matrix with -1s on the diagonal and 0s elsewhere.

**Final Answer**
\boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}

To compute the expression
$$
\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40}
$$
where
$$
\mathbf{A} = \begin{pmatrix} 3 & 7 \\ -1 & -2 \end{pmatrix},
$$
we begin by analyzing the structure of the matrix and its powers.

---

### Step 1: Characteristic Equation and Matrix Identity

We compute the characteristic polynomial of $\mathbf{A}$:

$$
\det(\mathbf{A} - \lambda \mathbf{I}) = \begin{vmatrix} 3 - \lambda & 7 \\ -1 & -2 - \lambda \end{vmatrix}
= (3 - \lambda)(-2 - \lambda) - (7)(-1)
= \lambda^2 - \lambda + 1.
$$

This gives the characteristic equation:
$$
\lambda^2 - \lambda + 1 = 0.
$$

The roots are complex:
$$
\lambda = \frac{1 \pm i\sqrt{3}}{2}.
$$

However, instead of working with complex eigenvalues, we can use the **Cayley-Hamilton Theorem**, which tells us that a matrix satisfies its own characteristic equation. Thus:
$$
\mathbf{A}^2 - \mathbf{A} + \mathbf{I} = \mathbf{0} \quad \Rightarrow \quad \mathbf{A}^2 = \mathbf{A} - \mathbf{I}.
$$

This is a key identity that allows us to express higher powers of $\mathbf{A}$ in terms of lower powers.

---

### Step 2: Finding a Pattern in Powers of $\mathbf{A}$

Using the identity $\mathbf{A}^2 = \mathbf{A} - \mathbf{I}$, we can compute:

- $\mathbf{A}^3 = \mathbf{A}^2 \cdot \mathbf{A} = (\mathbf{A} - \mathbf{I}) \cdot \mathbf{A} = \mathbf{A}^2 - \mathbf{A} = (\mathbf{A} - \mathbf{I}) - \mathbf{A} = -\mathbf{I}$,
- $\mathbf{A}^4 = \mathbf{A}^3 \cdot \mathbf{A} = (-\mathbf{I}) \cdot \mathbf{A} = -\mathbf{A}$,
- $\mathbf{A}^5 = \mathbf{A}^4 \cdot \mathbf{A} = (-\mathbf{A}) \cdot \mathbf{A} = -\mathbf{A}^2 = -(\mathbf{A} - \mathbf{I}) = -\mathbf{A} + \mathbf{I}$,
- $\mathbf{A}^6 = \mathbf{A}^5 \cdot \mathbf{A} = (-\mathbf{A} + \mathbf{I}) \cdot \mathbf{A} = -\mathbf{A}^2 + \mathbf{A} = -(\mathbf{A} - \mathbf{I}) + \mathbf{A} = \mathbf{I}$.

So we find that:
$$
\mathbf{A}^6 = \mathbf{I} \quad \Rightarrow \quad \text{The powers of } \mathbf{A} \text{ are periodic with period 6}.
$$

---

### Step 3: Reduce Exponents Modulo 6

We now reduce the exponents modulo 6:

- $27 \mod 6 = 3 \Rightarrow \mathbf{A}^{27} = \mathbf{A}^3 = -\mathbf{I}$,
- $31 \mod 6 = 1 \Rightarrow \mathbf{A}^{31} = \mathbf{A}^1 = \mathbf{A}$,
- $40 \mod 6 = 4 \Rightarrow \mathbf{A}^{40} = \mathbf{A}^4 = -\mathbf{A}$.

---

### Step 4: Compute the Final Expression

Now compute:
$$
\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40} = (-\mathbf{I}) + \mathbf{A} + (-\mathbf{A}) = -\mathbf{I}.
$$

---

### Final Answer

$$
\boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}
$$

Solution:
Note that
\[\mathbf{A}^2 = \begin{pmatrix} 2 & 7 \\ -1 & -3 \end{pmatrix}\]and
\[\mathbf{A}^3 = \mathbf{A} \mathbf{A}^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -\mathbf{I}.\]Then
\begin{align*}
\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40} &= (\mathbf{A}^3)^9 + (\mathbf{A}^3)^{10} \mathbf{A} + (\mathbf{A}^3)^{13} \mathbf{A} \\
&= (-\mathbf{I})^9 + (-\mathbf{I})^{10} \mathbf{A} + (-\mathbf{I})^{13} \mathbf{A} \\
&= -\mathbf{I} + \mathbf{A} - \mathbf{A} \\
&= -\mathbf{I} = \boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}.
\end{align*}


Explanation:
Okay, so I need to find an odd integer t between 0 and 23 (so t is from 1 to 22, but since it's odd, t is 1, 3, 5,...,21) such that t + 2 is the inverse of t modulo 23. Hmm, let me recall what an inverse modulo means.

The inverse of a number t modulo 23 is another number x such that t * x ≡ 1 mod 23. So in this case, the problem says that t + 2 is the inverse of t. That means t * (t + 2) ≡ 1 mod 23. So I need to solve the congruence equation t(t + 2) ≡ 1 mod 23.

Let me write that out: t² + 2t ≡ 1 mod 23. If I rearrange that, it becomes t² + 2t - 1 ≡ 0 mod 23. So this is a quadratic congruence equation. Maybe I can solve it by completing the square or using the quadratic formula? Let me try completing the square first.

Starting with t² + 2t - 1 ≡ 0 mod 23. Let me add 1 to both sides to get t² + 2t ≡ 1 mod 23. Then, completing the square: take half of the coefficient of t, which is 1, square it, so 1² = 1. So add 1 to both sides: t² + 2t + 1 ≡ 1 + 1 mod 23. That gives (t + 1)² ≡ 2 mod 23.

So now, the equation is (t + 1)² ≡ 2 mod 23. Therefore, t + 1 is a square root of 2 modulo 23. So I need to find integers x such that x² ≡ 2 mod 23, then t = x - 1. Then check which of those t's are odd and between 0 and 23.

But how do I find square roots modulo 23? Let me recall that 23 is a prime number, so maybe I can use some methods for finding square roots modulo primes. One method is to check all numbers from 1 to 22 and see which ones squared give 2 mod 23. Since 23 is not too big, this might be feasible.

Alternatively, maybe there's a formula or something. Let me try the brute force approach first. Let me compute squares modulo 23:

Let me list squares mod 23:

0² = 0

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25 ≡ 25 - 23 = 2

Oh! Wait, 5² is 25, which is 2 mod 23. So 5 is a square root of 2 mod 23. Therefore, x ≡ 5 or x ≡ -5 mod 23. Since -5 mod 23 is 23 - 5 = 18. So the square roots of 2 mod 23 are 5 and 18.

Therefore, t + 1 ≡ 5 mod 23 or t + 1 ≡ 18 mod 23. Therefore, solving for t:

Case 1: t + 1 ≡ 5 mod 23 ⇒ t ≡ 5 - 1 = 4 mod 23. So t = 4 or 4 + 23k. But since t is between 0 and 23, t = 4.

Case 2: t + 1 ≡ 18 mod 23 ⇒ t ≡ 18 - 1 = 17 mod 23. So t = 17 or 17 + 23k. Again, between 0 and 23, t = 17.

So the possible solutions are t = 4 and t = 17. But the problem says that t must be odd. Let me check: 4 is even, 17 is odd. Therefore, the unique odd integer t is 17.

Wait, but let me verify this. Let me check if t = 17 satisfies the original condition.

First, check if t + 2 is the inverse of t modulo 23. So t + 2 = 17 + 2 = 19. So we need to check if 17 * 19 ≡ 1 mod 23. Let me compute 17 * 19.

17 * 19: Let me compute 17*20 = 340, subtract 17: 340 - 17 = 323. Now divide 323 by 23 and find the remainder.

Let me compute 23*14 = 322. Therefore, 323 - 322 = 1. Therefore, 323 ≡ 1 mod 23. So yes, 17 * 19 ≡ 1 mod 23. Therefore, 19 is indeed the inverse of 17 mod 23.

So t = 17 is a solution. Also, t = 4: check if t + 2 = 6 is the inverse of 4 mod 23. Let me check 4 * 6 = 24 ≡ 1 mod 23? 24 - 23 = 1, yes. So 4 * 6 ≡ 1 mod 23. Therefore, t = 4 is also a solution. But since the problem says "unique odd integer t", and 4 is even, so 17 is the answer.

But let me make sure there are no other solutions. Since we had the equation (t + 1)^2 ≡ 2 mod 23, which had two solutions, 5 and 18 mod 23, leading to t = 4 and t = 17. Since 23 is prime, the equation x² ≡ a mod p has either 0 or 2 solutions. Since 2 is a quadratic residue mod 23 (as we found 5² ≡ 2), there are exactly two solutions. Therefore, only t = 4 and t = 17. Since the problem says "unique odd integer", so 17 is the answer.

But just to be thorough, let me check if there are other square roots of 2 mod 23. For example, maybe I missed some? Let me check squares again:

Compute squares mod 23:

0² = 0

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25 ≡ 2

6² = 36 ≡ 36 - 23 = 13

7² = 49 ≡ 49 - 2*23 = 49 - 46 = 3

8² = 64 ≡ 64 - 2*23 = 64 - 46 = 18

9² = 81 ≡ 81 - 3*23 = 81 - 69 = 12

10² = 100 ≡ 100 - 4*23 = 100 - 92 = 8

11² = 121 ≡ 121 - 5*23 = 121 - 115 = 6

12² = 144 ≡ 144 - 6*23 = 144 - 138 = 6

Wait, 12² is 144. 23*6 = 138, so 144 - 138 = 6. So 12² ≡ 6 mod 23.

13²: 13² = 169. 23*7 = 161, so 169 - 161 = 8. So 13² ≡ 8 mod 23.

14²: 14² = 196. 23*8 = 184, 196 - 184 = 12. So 14² ≡ 12 mod 23.

15²: 225. 23*9 = 207, 225 - 207 = 18. So 15² ≡ 18 mod 23.

16²: 256. 23*11 = 253, 256 - 253 = 3. So 16² ≡ 3 mod 23.

17²: 289. 23*12 = 276, 289 - 276 = 13. So 17² ≡ 13 mod 23.

18²: 324. 23*14 = 322, 324 - 322 = 2. So 18² ≡ 2 mod 23.

19²: 361. 23*15 = 345, 361 - 345 = 16. So 19² ≡ 16 mod 23.

20²: 400. 23*17 = 391, 400 - 391 = 9. So 20² ≡ 9 mod 23.

21²: 441. 23*19 = 437, 441 - 437 = 4. So 21² ≡ 4 mod 23.

22²: 484. 23*21 = 483, 484 - 483 = 1. So 22² ≡ 1 mod 23.

So indeed, only 5² and 18² ≡ 2 mod 23. Therefore, only two solutions. Therefore, t = 4 and t = 17. Since the problem asks for the unique odd integer t, which is 17.

But just to make sure, let me check if there's another way to approach this problem. Maybe using the inverse formula?

Alternatively, since t and t + 2 are inverses modulo 23, then t*(t + 2) ≡ 1 mod 23. Which is the same equation as before. So same result.

Alternatively, maybe using the fact that if t is inverse of t + 2, then t + 2 ≡ t^{-1} mod 23. Therefore, multiplying both sides by t: t*(t + 2) ≡ 1 mod 23. Which again gives the same equation.

Alternatively, maybe using the property that if t + 2 ≡ t^{-1}, then t + 2 ≡ t^{-1} ⇒ t + 2 ≡ t^{-1} ⇒ multiplying both sides by t: t^2 + 2t ≡ 1 mod 23. Which again is the same equation.

So seems like the same path.

Alternatively, maybe trying to solve t^2 + 2t - 1 ≡ 0 mod 23. Using quadratic formula.

In modular arithmetic, the quadratic formula works if the modulus is prime and the discriminant is a quadratic residue.

So for equation ax² + bx + c ≡ 0 mod p, solutions are x ≡ [-b ± sqrt(b² - 4ac)] / (2a) mod p.

In our case, equation is t² + 2t - 1 ≡ 0 mod 23. So a = 1, b = 2, c = -1.

Discriminant D = b² - 4ac = 4 - 4*1*(-1) = 4 + 4 = 8.

So sqrt(D) = sqrt(8) mod 23. Wait, but earlier we found that sqrt(2) is 5 and 18. So sqrt(8) would be sqrt(2*4) = 2*sqrt(2). So sqrt(8) mod 23 is 2*5 = 10 or 2*18 = 36 ≡ 36 - 23 = 13 mod 23.

Therefore, solutions would be t ≡ [-2 ± sqrt(8)] / 2 mod 23.

So sqrt(8) is 10 or 13. Therefore:

First solution: [-2 + 10]/2 = 8/2 = 4 mod 23.

Second solution: [-2 + 13]/2 = 11/2 mod 23. Wait, 11 divided by 2 mod 23. Since 2*12 = 24 ≡ 1 mod 23, so inverse of 2 is 12. Therefore, 11/2 ≡ 11*12 mod 23. Let me compute 11*12 = 132. 132 divided by 23: 23*5 = 115, 132 - 115 = 17. So 11*12 ≡ 17 mod 23. Therefore, the second solution is 17.

Therefore, same solutions: t ≡ 4 and t ≡ 17 mod 23. So again, same result.

Therefore, confirming that t = 4 and t = 17. Since the problem asks for the unique odd integer t between 0 and 23, which is 17.

Therefore, the answer is 17.

But just to make sure, let me check if there's any other possible way that maybe I missed something. For example, maybe negative numbers? But since t is between 0 and 23, we don't have to consider negative numbers. Also, the problem says t is an integer, so positive.

Alternatively, maybe there's a mistake in assuming that both solutions are valid? Let me check t = 4 again.

t = 4, t + 2 = 6. Check if 4 * 6 ≡ 1 mod 23. 4*6 = 24 ≡ 1 mod 23. Yes, that's correct. So t = 4 is a solution, but it's even.

t = 17, t + 2 = 19. 17*19 = 323. 323 divided by 23: 23*14 = 322, so 323 - 322 = 1. Therefore, 17*19 ≡ 1 mod 23. Correct.

So both are solutions, but only 17 is odd.

Therefore, the unique odd integer t is 17.

I think that's thorough. I considered multiple approaches: completing the square, quadratic formula, brute force checking squares, verifying the solutions. All lead to the same conclusion. So I think 17 is the correct answer.

**Final Answer**
The unique odd integer $ t $ is \boxed{17}.

To solve the problem, we are given that $ t $ is an **odd integer** such that $ 0 < t < 23 $, and that $ t + 2 $ is the **inverse** of $ t $ modulo 23. That is, we are looking for an integer $ t $ satisfying:

$$
t \cdot (t + 2) \equiv 1 \pmod{23}
$$

---

### Step 1: Translate the condition into a congruence

We expand the left-hand side:

$$
t(t + 2) = t^2 + 2t
$$

So the congruence becomes:

$$
t^2 + 2t \equiv 1 \pmod{23}
$$

Rewriting:

$$
t^2 + 2t - 1 \equiv 0 \pmod{23}
$$

This is a quadratic congruence in $ t $.

---

### Step 2: Solve the quadratic congruence

We can solve this using the **quadratic formula** in modular arithmetic. For a general quadratic congruence:

$$
at^2 + bt + c \equiv 0 \pmod{p}
$$

the solutions are:

$$
t \equiv \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \pmod{p}
$$

In our case, $ a = 1 $, $ b = 2 $, $ c = -1 $, and $ p = 23 $. So:

$$
t \equiv \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2} \pmod{23}
$$

$$
t \equiv \frac{-2 \pm \sqrt{4 + 4}}{2} \pmod{23}
$$

$$
t \equiv \frac{-2 \pm \sqrt{8}}{2} \pmod{23}
$$

Now, we need to compute $ \sqrt{8} \mod 23 $. Since $ 8 = 2 \cdot 4 $, and we already know that $ 5^2 \equiv 2 \pmod{23} $, then:

$$
\sqrt{8} \equiv \sqrt{2 \cdot 4} \equiv 2\sqrt{2} \equiv 2 \cdot 5 = 10 \pmod{23}
$$

So the two square roots of 8 modulo 23 are $ 10 $ and $ -10 \equiv 13 \pmod{23} $.

Now plug these into the quadratic formula:

- First solution:
  $$
  t \equiv \frac{-2 + 10}{2} = \frac{8}{2} = 4 \pmod{23}
  $$

- Second solution:
  $$
  t \equiv \frac{-2 + 13}{2} = \frac{11}{2} \pmod{23}
  $$

To compute $ \frac{11}{2} \mod 23 $, we find the modular inverse of 2 modulo 23. Since $ 2 \cdot 12 = 24 \equiv 1 \pmod{23} $, the inverse of 2 is 12. So:

$$
\frac{11}{2} \equiv 11 \cdot 12 = 132 \equiv 17 \pmod{23}
$$

Thus, the two solutions are:

$$
t \equiv 4 \pmod{23} \quad \text{and} \quad t \equiv 17 \pmod{23}
$$

---

### Step 3: Apply the condition that $ t $ is **odd**

From the two solutions:

- $ t = 4 $ is **even**
- $ t = 17 $ is **odd**

Since the problem asks for the **unique odd** integer $ t $ such that $ 0 < t < 23 $, the only valid solution is:

$$
\boxed{17}
$$

---

### Final Verification

Let’s verify that $ t = 17 $ satisfies the original condition:

- $ t + 2 = 19 $
- $ 17 \cdot 19 = 323 $
- $ 323 \mod 23 = 1 $ (since $ 23 \cdot 14 = 322 $, and $ 323 - 322 = 1 $)

So $ 17 \cdot 19 \equiv 1 \pmod{23} $, confirming that $ t = 17 $ is indeed the inverse of $ t + 2 $ modulo 23.

---

### Final Answer

$$
\boxed{17}
$$

[Assistant]
Explanation:
Okay, so I need to find an odd integer t between 0 and 23 (so t is from 1 to 22, but since it's odd, t is 1, 3, 5,...,21) such that t + 2 is the inverse of t modulo 23. Hmm, let me recall what an inverse modulo means.

The inverse of a number t modulo 23 is another number x such that t * x ≡ 1 mod 23. So in this case, the problem says that t + 2 is the inverse of t. That means t * (t + 2) ≡ 1 mod 23. So I need to solve the congruence equation t(t + 2) ≡ 1 mod 23.

Let me write that out: t² + 2t ≡ 1 mod 23. If I rearrange that, it becomes t² + 2t - 1 ≡ 0 mod 23. So this is a quadratic congruence equation. Maybe I can solve it by completing the square or using the quadratic formula? Let me try completing the square first.

Starting with t² + 2t - 1 ≡ 0 mod 23. Let me add 1 to both sides to get t² + 2t ≡ 1 mod 23. Then, completing the square: take half of the coefficient of t, which is 1, square it, so 1² = 1. So add 1 to both sides: t² + 2t + 1 ≡ 1 + 1 mod 23. That gives (t + 1)² ≡ 2 mod 23.

So now, the equation is (t + 1)² ≡ 2 mod 23. Therefore, t + 1 is a square root of 2 modulo 23. So I need to find integers x such that x² ≡ 2 mod 23, then t = x - 1. Then check which of those t's are odd and between 0 and 23.

But how do I find square roots modulo 23? Let me recall that 23 is a prime number, so maybe I can use some methods for finding square roots modulo primes. One method is to check all numbers from 1 to 22 and see which ones squared give 2 mod 23. Since 23 is not too big, this might be feasible.

Alternatively, maybe there's a formula or something. Let me try the brute force approach first. Let me compute squares modulo 23:

Let me list squares mod 23:

0² = 0

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25 ≡ 25 - 23 = 2

Oh! Wait, 5² is 25, which is 2 mod 23. So 5 is a square root of 2 mod 23. Therefore, x ≡ 5 or x ≡ -5 mod 23. Since -5 mod 23 is 23 - 5 = 18. So the square roots of 2 mod 23 are 5 and 18.

Therefore, t + 1 ≡ 5 mod 23 or t + 1 ≡ 18 mod 23. Therefore, solving for t:

Case 1: t + 1 ≡ 5 mod 23 ⇒ t ≡ 5 - 1 = 4 mod 23. So t = 4 or 4 + 23k. But since t is between 0 and 23, t = 4.

Case 2: t + 1 ≡ 18 mod 23 ⇒ t ≡ 18 - 1 = 17 mod 23. So t = 17 or 17 + 23k. Again, between 0 and 23, t = 17.

So the possible solutions are t = 4 and t = 17. But the problem says that t must be odd. Let me check: 4 is even, 17 is odd. Therefore, the unique odd integer t is 17.

Wait, but let me verify this. Let me check if t = 17 satisfies the original condition.

First, check if t + 2 is the inverse of t modulo 23. So t + 2 = 17 + 2 = 19. So we need to check if 17 * 19 ≡ 1 mod 23. Let me compute 17 * 19.

17 * 19: Let me compute 17*20 = 340, subtract 17: 340 - 17 = 323. Now divide 323 by 23 and find the remainder.

Let me compute 23*14 = 322. Therefore, 323 - 322 = 1. Therefore, 323 ≡ 1 mod 23. So yes, 17 * 19 ≡ 1 mod 23. Therefore, 19 is indeed the inverse of 17 mod 23.

So t = 17 is a solution. Also, t = 4: check if t + 2 = 6 is the inverse of 4 mod 23. Let me check 4 * 6 = 24 ≡ 1 mod 23? 24 - 23 = 1, yes. So 4 * 6 ≡ 1 mod 23. Therefore, t = 4 is also a solution. But since the problem says "unique odd integer t", and 4 is even, so 17 is the answer.

But let me make sure there are no other solutions. Since we had the equation (t + 1)^2 ≡ 2 mod 23, which had two solutions, 5 and 18 mod 23, leading to t = 4 and t = 17. Since 23 is prime, the equation x² ≡ a mod p has either 0 or 2 solutions. Since 2 is a quadratic residue mod 23 (as we found 5² ≡ 2), there are exactly two solutions. Therefore, only t = 4 and t = 17. Since the problem says "unique odd integer", so 17 is the answer.

But just to be thorough, let me check if there are other square roots of 2 mod 23. For example, maybe I missed some? Let me check squares again:

Compute squares mod 23:

0² = 0

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25 ≡ 2

6² = 36 ≡ 36 - 23 = 13

7² = 49 ≡ 49 - 2*23 = 49 - 46 = 3

8² = 64 ≡ 64 - 2*23 = 64 - 46 = 18

9² = 81 ≡ 81 - 3*23 = 81 - 69 = 12

10² = 100 ≡ 100 - 4*23 = 100 - 92 = 8

11² = 121 ≡ 121 - 5*23 = 121 - 115 = 6

12² = 144 ≡ 144 - 6*23 = 144 - 138 = 6

Wait, 12² is 144. 23*6 = 138, so 144 - 138 = 6. So 12² ≡ 6 mod 23.

13²: 13² = 169. 23*7 = 161, so 169 - 161 = 8. So 13² ≡ 8 mod 23.

14²: 14² = 196. 23*8 = 184, 196 - 184 = 12. So 14² ≡ 12 mod 23.

15²: 225. 23*9 = 207, 225 - 207 = 18. So 15² ≡ 18 mod 23.

16²: 256. 23*11 = 253, 256 - 253 = 3. So 16² ≡ 3 mod 23.

17²: 289. 23*12 = 276, 289 - 276 = 13. So 17² ≡ 13 mod 23.

18²: 324. 23*14 = 322, 324 - 322 = 2. So 18² ≡ 2 mod 23.

19²: 361. 23*15 = 345, 361 - 345 = 16. So 19² ≡ 16 mod 23.

20²: 400. 23*17 = 391, 400 - 391 = 9. So 20² ≡ 9 mod 23.

21²: 441. 23*19 = 437, 441 - 437 = 4. So 21² ≡ 4 mod 23.

22²: 484. 23*21 = 483, 484 - 483 = 1. So 22² ≡ 1 mod 23.

So indeed, only 5² and 18² ≡ 2 mod 23. Therefore, only two solutions. Therefore, t = 4 and t = 17. Since the problem asks for the unique odd integer t, which is 17.

But just to make sure, let me check if there's another way to approach this problem. Maybe using the inverse formula?

Alternatively, since t and t + 2 are inverses modulo 23, then t*(t + 2) ≡ 1 mod 23. Which is the same equation as before. So same result.

Alternatively, maybe using the fact that if t is inverse of t + 2, then t + 2 ≡ t^{-1} mod 23. Therefore, multiplying both sides by t: t*(t + 2) ≡ 1 mod 23. Which again gives the same equation.

Alternatively, maybe using the property that if t + 2 ≡ t^{-1}, then t + 2 ≡ t^{-1} ⇒ t + 2 ≡ t^{-1} ⇒ multiplying both sides by t: t^2 + 2t ≡ 1 mod 23. Which again is the same equation.

So seems like the same path.

Alternatively, maybe trying to solve t^2 + 2t - 1 ≡ 0 mod 23. Using quadratic formula.

In modular arithmetic, the quadratic formula works if the modulus is prime and the discriminant is a quadratic residue.

So for equation ax² + bx + c ≡ 0 mod p, solutions are x ≡ [-b ± sqrt(b² - 4ac)] / (2a) mod p.

In our case, equation is t² + 2t - 1 ≡ 0 mod 23. So a = 1, b = 2, c = -1.

Discriminant D = b² - 4ac = 4 - 4*1*(-1) = 4 + 4 = 8.

So sqrt(D) = sqrt(8) mod 23. Wait, but earlier we found that sqrt(2) is 5 and 18. So sqrt(8) would be sqrt(2*4) = 2*sqrt(2). So sqrt(8) mod 23 is 2*5 = 10 or 2*18 = 36 ≡ 36 - 23 = 13 mod 23.

Therefore, solutions would be t ≡ [-2 ± sqrt(8)] / 2 mod 23.

So sqrt(8) is 10 or 13. Therefore:

First solution: [-2 + 10]/2 = 8/2 = 4 mod 23.

Second solution: [-2 + 13]/2 = 11/2 mod 23. Wait, 11 divided by 2 mod 23. Since 2*12 = 24 ≡ 1 mod 23, so inverse of 2 is 12. Therefore, 11/2 ≡ 11*12 mod 23. Let me compute 11*12 = 132. 132 divided by 23: 23*5 = 115, 132 - 115 = 17. So 11*12 ≡ 17 mod 23. Therefore, the second solution is 17.

Therefore, same solutions: t ≡ 4 and t ≡ 17 mod 23. So again, same result.

Therefore, confirming that t = 4 and t = 17. Since the problem asks for the unique odd integer t between 0 and 23, which is 17.

Therefore, the answer is 17.

But just to make sure, let me check if there's any other possible way that maybe I missed something. For example, maybe negative numbers? But since t is between 0 and 23, we don't have to consider negative numbers. Also, the problem says t is an integer, so positive.

Alternatively, maybe there's a mistake in assuming that both solutions are valid? Let me check t = 4 again.

t = 4, t + 2 = 6. Check if 4 * 6 ≡ 1 mod 23. 4*6 = 24 ≡ 1 mod 23. Yes, that's correct. So t = 4 is a solution, but it's even.

t = 17, t + 2 = 19. 17*19 = 323. 323 divided by 23: 23*14 = 322, so 323 - 322 = 1. Therefore, 17*19 ≡ 1 mod 23. Correct.

So both are solutions, but only 17 is odd.

Therefore, the unique odd integer t is 17.

I think that's thorough. I considered multiple approaches: completing the square, quadratic formula, brute force checking squares, verifying the solutions. All lead to the same conclusion. So I think 17 is the correct answer.

**Final Answer**
The unique odd integer $ t $ is \boxed{17}.

To solve the problem, we are given that $ t $ is an **odd integer** such that $ 0 < t < 23 $, and that $ t + 2 $ is the **inverse** of $ t $ modulo 23. That is, we are looking for an integer $ t $ satisfying:

$$
t \cdot (t + 2) \equiv 1 \pmod{23}
$$

---

### Step 1: Translate the condition into a congruence

We expand the left-hand side:

$$
t(t + 2) = t^2 + 2t
$$

So the congruence becomes:

$$
t^2 + 2t \equiv 1 \pmod{23}
$$

Rewriting:

$$
t^2 + 2t - 1 \equiv 0 \pmod{23}
$$

This is a quadratic congruence in $ t $.

---

### Step 2: Solve the quadratic congruence

We can solve this using the **quadratic formula** in modular arithmetic. For a general quadratic congruence:

$$
at^2 + bt + c \equiv 0 \pmod{p}
$$

the solutions are:

$$
t \equiv \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \pmod{p}
$$

In our case, $ a = 1 $, $ b = 2 $, $ c = -1 $, and $ p = 23 $. So:

$$
t \equiv \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2} \pmod{23}
$$

$$
t \equiv \frac{-2 \pm \sqrt{4 + 4}}{2} \pmod{23}
$$

$$
t \equiv \frac{-2 \pm \sqrt{8}}{2} \pmod{23}
$$

Now, we need to compute $ \sqrt{8} \mod 23 $. Since $ 8 = 2 \cdot 4 $, and we already know that $ 5^2 \equiv 2 \pmod{23} $, then:

$$
\sqrt{8} \equiv \sqrt{2 \cdot 4} \equiv 2\sqrt{2} \equiv 2 \cdot 5 = 10 \pmod{23}
$$

So the two square roots of 8 modulo 23 are $ 10 $ and $ -10 \equiv 13 \pmod{23} $.

Now plug these into the quadratic formula:

- First solution:
  $$
  t \equiv \frac{-2 + 10}{2} = \frac{8}{2} = 4 \pmod{23}
  $$

- Second solution:
  $$
  t \equiv \frac{-2 + 13}{2} = \frac{11}{2} \pmod{23}
  $$

To compute $ \frac{11}{2} \mod 23 $, we find the modular inverse of 2 modulo 23. Since $ 2 \cdot 12 = 24 \equiv 1 \pmod{23} $, the inverse of 2 is 12. So:

$$
\frac{11}{2} \equiv 11 \cdot 12 = 132 \equiv 17 \pmod{23}
$$

Thus, the two solutions are:

$$
t \equiv 4 \pmod{23} \quad \text{and} \quad t \equiv 17 \pmod{23}
$$

---

### Step 3: Apply the condition that $ t $ is **odd**

From the two solutions:

- $ t = 4 $ is **even**
- $ t = 17 $ is **odd**

Since the problem asks for the **unique odd** integer $ t $ such that $ 0 < t < 23 $, the only valid solution is:

$$
\boxed{17}
$$

---

### Final Verification

Let’s verify that $ t = 17 $ satisfies the original condition:

- $ t + 2 = 19 $
- $ 17 \cdot 19 = 323 $
- $ 323 \mod 23 = 1 $ (since $ 23 \cdot 14 = 322 $, and $ 323 - 322 = 1 $)

So $ 17 \cdot 19 \equiv 1 \pmod{23} $, confirming that $ t = 17 $ is indeed the inverse of $ t + 2 $ modulo 23.

---

### Final Answer

$$
\boxed{17}
$$

Solution:
We could find the answer by trial and error -- testing each candidate $t$ to see if $t\cdot (t+2)\equiv 1\pmod{23}$. However, here is another way:

We can easily see that $4\cdot 6=24\equiv 1\pmod{23}$, so $4$ fulfills the main requirement that its inverse is $2$ more than it. Unfortunately, $4$ isn't odd. But we also have \begin{align*}
(-4)\cdot (-6) &= 4\cdot 6 \\
&\equiv 1\pmod{23},
\end{align*} so $-4$ and $-6$ are each other's inverses $\pmod{23}$. Since $-4\equiv 19\pmod{23}$ and $-6\equiv 17\pmod{23}$, the answer $t=\boxed{17}$ satisfies the requirements of the problem.

(We can even check that $17\cdot 19 = 323 = 14\cdot 23 + 1$.)


Explanation:
Okay, so I need to find the x-intercept (a), the y-intercept (b), and the slope (m) of the line given by the equation x/4 + y/12 = 1. Then add them all together for the final answer. Let me start by recalling what each of these terms means.

First, the x-intercept is the point where the line crosses the x-axis. At that point, the y-coordinate is 0. Similarly, the y-intercept is where the line crosses the y-axis, so the x-coordinate there is 0. The slope, m, is the steepness of the line, calculated as rise over run, or change in y over change in x.

But let me make sure I remember the intercepts correctly. For the equation of a line in standard form, Ax + By = C, the x-intercept is found by setting y = 0 and solving for x, and the y-intercept is found by setting x = 0 and solving for y. Alternatively, if the equation is in intercept form, which I think is x/a + y/b = 1, then a is the x-intercept and b is the y-intercept. Wait, the given equation is x/4 + y/12 = 1. So that seems to be in intercept form. So that would mean that a is 4 and b is 12? Is that right?

Let me check. If I set y = 0, then x/4 + 0 = 1, so x = 4. So the x-intercept is (4, 0), so a = 4. Similarly, if I set x = 0, then 0 + y/12 = 1, so y = 12. Therefore, the y-intercept is (0, 12), so b = 12. So that seems straightforward. Therefore, a = 4 and b = 12. So then, a + b is 4 + 12 = 16. But then I need to find m, the slope.

So how do I find the slope? Well, the equation is given in standard form, but maybe I should convert it to slope-intercept form (y = mx + c) to easily identify the slope. Let me try that.

Starting with the original equation: x/4 + y/12 = 1. Let me solve for y.

First, subtract x/4 from both sides: y/12 = 1 - x/4.

Then multiply both sides by 12: y = 12*(1 - x/4) = 12*1 - 12*(x/4) = 12 - 3x.

So that simplifies to y = -3x + 12. Therefore, the slope m is -3, and the y-intercept is 12, which matches what I found earlier. So m = -3.

Therefore, a = 4, b = 12, m = -3. Then a + b + m = 4 + 12 + (-3) = 4 + 12 - 3 = 13. Wait, 4 + 12 is 16, minus 3 is 13. So the answer is 13?

Wait, let me check again. Maybe I made a mistake in calculating the slope? Let me verify.

Original equation: x/4 + y/12 = 1. Let me rearrange it again. Multiply both sides by 12 to eliminate denominators. 12*(x/4) + 12*(y/12) = 12*1. That simplifies to 3x + y = 12. Then, solving for y: y = -3x + 12. Yes, same result. So slope is -3. So that's correct.

Alternatively, if I use the standard form Ax + By = C, then the slope is -A/B. Let me recall that formula. If the equation is Ax + By = C, then solving for y gives y = (-A/B)x + C/B. Therefore, slope m = -A/B. In our case, the original equation after multiplying by 12 is 3x + y = 12. So A is 3, B is 1. Therefore, slope m = -3/1 = -3. Yep, same answer. So that's correct.

Therefore, a = 4, b = 12, m = -3. Therefore, adding them together: 4 + 12 + (-3) = 13. So the answer is 13. Wait, but let me check once more to make sure I didn't mix up any numbers.

Original equation: x/4 + y/12 = 1. So intercept form is x/a + y/b = 1, so a = 4, b = 12. Correct. Then slope is -A/B when in standard form. If I write the original equation as (1/4)x + (1/12)y = 1, so A is 1/4, B is 1/12. Then slope is -A/B = -(1/4)/(1/12) = -(1/4)*(12/1) = -3. Correct. So that's consistent. So slope is indeed -3.

Therefore, adding a + b + m = 4 + 12 + (-3) = 13. So I think that's the answer. Wait, but let me check once more if I didn't miscalculate 4 + 12 - 3. 4 + 12 is 16, 16 - 3 is 13. Yes. So 13 is the answer.

Alternatively, maybe the problem expects something else? Let me check if there's another way to interpret the question. The problem says "the line with equation x/4 + y/12 = 1". So that's the same as 3x + y = 12, as we saw. So intercepts are 4 and 12, slope is -3. So adding them gives 13. I think that's correct.

Alternatively, maybe I need to check if the question is referring to something else? For example, sometimes people might confuse intercepts with something else, but no, the x-intercept is when y=0, so x=4. The y-intercept is when x=0, so y=12. So that's correct. The slope is -3. So adding them gives 4 + 12 + (-3) = 13. So I think that's correct.

Alternatively, maybe the problem is expecting me to write the answer as a fraction or something? But 13 is an integer. Let me check once again.

Original equation: x/4 + y/12 = 1. Let me plug in the intercepts to check. For x-intercept, y=0: x/4 + 0 = 1 => x=4. Correct. For y-intercept, x=0: 0 + y/12 = 1 => y=12. Correct. Then slope: between the two intercepts, from (4, 0) to (0, 12). The slope is (12 - 0)/(0 - 4) = 12 / (-4) = -3. Correct. So slope is -3. Therefore, a + b + m = 4 + 12 + (-3) = 13. So yes, that's correct.

Therefore, I think the answer is 13. I don't see any mistakes in my reasoning. So I think that's the correct answer.

**Final Answer**
The value of $a + b + m$ is \boxed{13}.

To solve the problem, we are given the equation of a line:

$$
\frac{x}{4} + \frac{y}{12} = 1
$$

We are asked to find the **x-intercept** $ a $, the **y-intercept** $ b $, and the **slope** $ m $, and then compute the sum $ a + b + m $.

---

### Step 1: Identify the x-intercept $ a $

The **x-intercept** occurs when $ y = 0 $. Substituting into the equation:

$$
\frac{x}{4} + \frac{0}{12} = 1 \Rightarrow \frac{x}{4} = 1 \Rightarrow x = 4
$$

So, the x-intercept is $ a = 4 $.

---

### Step 2: Identify the y-intercept $ b $

The **y-intercept** occurs when $ x = 0 $. Substituting into the equation:

$$
\frac{0}{4} + \frac{y}{12} = 1 \Rightarrow \frac{y}{12} = 1 \Rightarrow y = 12
$$

So, the y-intercept is $ b = 12 $.

---

### Step 3: Find the slope $ m $

We can rewrite the equation in **slope-intercept form** $ y = mx + c $ to identify the slope.

Start with the original equation:

$$
\frac{x}{4} + \frac{y}{12} = 1
$$

Multiply both sides by 12 to eliminate denominators:

$$
3x + y = 12
$$

Now solve for $ y $:

$$
y = -3x + 12
$$

From this, we see that the **slope** $ m = -3 $.

---

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

Now that we have:

- $ a = 4 $
- $ b = 12 $
- $ m = -3 $

We compute:

$$
a + b + m = 4 + 12 + (-3) = 13
$$

---

### Final Answer

$$
\boxed{13}
$$

[Assistant]
Explanation:
Okay, so I need to find the x-intercept (a), the y-intercept (b), and the slope (m) of the line given by the equation x/4 + y/12 = 1. Then add them all together for the final answer. Let me start by recalling what each of these terms means.

First, the x-intercept is the point where the line crosses the x-axis. At that point, the y-coordinate is 0. Similarly, the y-intercept is where the line crosses the y-axis, so the x-coordinate there is 0. The slope, m, is the steepness of the line, calculated as rise over run, or change in y over change in x.

But let me make sure I remember the intercepts correctly. For the equation of a line in standard form, Ax + By = C, the x-intercept is found by setting y = 0 and solving for x, and the y-intercept is found by setting x = 0 and solving for y. Alternatively, if the equation is in intercept form, which I think is x/a + y/b = 1, then a is the x-intercept and b is the y-intercept. Wait, the given equation is x/4 + y/12 = 1. So that seems to be in intercept form. So that would mean that a is 4 and b is 12? Is that right?

Let me check. If I set y = 0, then x/4 + 0 = 1, so x = 4. So the x-intercept is (4, 0), so a = 4. Similarly, if I set x = 0, then 0 + y/12 = 1, so y = 12. Therefore, the y-intercept is (0, 12), so b = 12. So that seems straightforward. Therefore, a = 4 and b = 12. So then, a + b is 4 + 12 = 16. But then I need to find m, the slope.

So how do I find the slope? Well, the equation is given in standard form, but maybe I should convert it to slope-intercept form (y = mx + c) to easily identify the slope. Let me try that.

Starting with the original equation: x/4 + y/12 = 1. Let me solve for y.

First, subtract x/4 from both sides: y/12 = 1 - x/4.

Then multiply both sides by 12: y = 12*(1 - x/4) = 12*1 - 12*(x/4) = 12 - 3x.

So that simplifies to y = -3x + 12. Therefore, the slope m is -3, and the y-intercept is 12, which matches what I found earlier. So m = -3.

Therefore, a = 4, b = 12, m = -3. Then a + b + m = 4 + 12 + (-3) = 4 + 12 - 3 = 13. Wait, 4 + 12 is 16, minus 3 is 13. So the answer is 13?

Wait, let me check again. Maybe I made a mistake in calculating the slope? Let me verify.

Original equation: x/4 + y/12 = 1. Let me rearrange it again. Multiply both sides by 12 to eliminate denominators. 12*(x/4) + 12*(y/12) = 12*1. That simplifies to 3x + y = 12. Then, solving for y: y = -3x + 12. Yes, same result. So slope is -3. So that's correct.

Alternatively, if I use the standard form Ax + By = C, then the slope is -A/B. Let me recall that formula. If the equation is Ax + By = C, then solving for y gives y = (-A/B)x + C/B. Therefore, slope m = -A/B. In our case, the original equation after multiplying by 12 is 3x + y = 12. So A is 3, B is 1. Therefore, slope m = -3/1 = -3. Yep, same answer. So that's correct.

Therefore, a = 4, b = 12, m = -3. Therefore, adding them together: 4 + 12 + (-3) = 13. So the answer is 13. Wait, but let me check once more to make sure I didn't mix up any numbers.

Original equation: x/4 + y/12 = 1. So intercept form is x/a + y/b = 1, so a = 4, b = 12. Correct. Then slope is -A/B when in standard form. If I write the original equation as (1/4)x + (1/12)y = 1, so A is 1/4, B is 1/12. Then slope is -A/B = -(1/4)/(1/12) = -(1/4)*(12/1) = -3. Correct. So that's consistent. So slope is indeed -3.

Therefore, adding a + b + m = 4 + 12 + (-3) = 13. So I think that's the answer. Wait, but let me check once more if I didn't miscalculate 4 + 12 - 3. 4 + 12 is 16, 16 - 3 is 13. Yes. So 13 is the answer.

Alternatively, maybe the problem expects something else? Let me check if there's another way to interpret the question. The problem says "the line with equation x/4 + y/12 = 1". So that's the same as 3x + y = 12, as we saw. So intercepts are 4 and 12, slope is -3. So adding them gives 13. I think that's correct.

Alternatively, maybe I need to check if the question is referring to something else? For example, sometimes people might confuse intercepts with something else, but no, the x-intercept is when y=0, so x=4. The y-intercept is when x=0, so y=12. So that's correct. The slope is -3. So adding them gives 4 + 12 + (-3) = 13. So I think that's correct.

Alternatively, maybe the problem is expecting me to write the answer as a fraction or something? But 13 is an integer. Let me check once again.

Original equation: x/4 + y/12 = 1. Let me plug in the intercepts to check. For x-intercept, y=0: x/4 + 0 = 1 => x=4. Correct. For y-intercept, x=0: 0 + y/12 = 1 => y=12. Correct. Then slope: between the two intercepts, from (4, 0) to (0, 12). The slope is (12 - 0)/(0 - 4) = 12 / (-4) = -3. Correct. So slope is -3. Therefore, a + b + m = 4 + 12 + (-3) = 13. So yes, that's correct.

Therefore, I think the answer is 13. I don't see any mistakes in my reasoning. So I think that's the correct answer.

**Final Answer**
The value of $a + b + m$ is \boxed{13}.

To solve the problem, we are given the equation of a line:

$$
\frac{x}{4} + \frac{y}{12} = 1
$$

We are asked to find the **x-intercept** $ a $, the **y-intercept** $ b $, and the **slope** $ m $, and then compute the sum $ a + b + m $.

---

### Step 1: Identify the x-intercept $ a $

The **x-intercept** occurs when $ y = 0 $. Substituting into the equation:

$$
\frac{x}{4} + \frac{0}{12} = 1 \Rightarrow \frac{x}{4} = 1 \Rightarrow x = 4
$$

So, the x-intercept is $ a = 4 $.

---

### Step 2: Identify the y-intercept $ b $

The **y-intercept** occurs when $ x = 0 $. Substituting into the equation:

$$
\frac{0}{4} + \frac{y}{12} = 1 \Rightarrow \frac{y}{12} = 1 \Rightarrow y = 12
$$

So, the y-intercept is $ b = 12 $.

---

### Step 3: Find the slope $ m $

We can rewrite the equation in **slope-intercept form** $ y = mx + c $ to identify the slope.

Start with the original equation:

$$
\frac{x}{4} + \frac{y}{12} = 1
$$

Multiply both sides by 12 to eliminate denominators:

$$
3x + y = 12
$$

Now solve for $ y $:

$$
y = -3x + 12
$$

From this, we see that the **slope** $ m = -3 $.

---

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

Now that we have:

- $ a = 4 $
- $ b = 12 $
- $ m = -3 $

We compute:

$$
a + b + m = 4 + 12 + (-3) = 13
$$

---

### Final Answer

$$
\boxed{13}
$$

Solution: We can find the $x$-intercept by setting $y=0$ in the equation.  This gives us $\frac{x}{4} = 1$, so $x =4$, which means $a=4$.  Similarly, letting $x=0$ gives $\frac{y}{12} = 1$, so $y=12$, which means $b=12$.  There are several ways we can find the slope.  First, we could put the equation in slope-intercept form by subtracting $\frac{x}{4}$ from both sides, and then multiplying by 12.  This gives $y = -3x +12$, which tells us that the slope is $-3$ (and confirms our solution for the $y$-intercept).  We also could have noted that since we have already shown that $(4,0)$ and $(0,12)$ are on the line, the slope of the line is $\frac{12 -0}{0-4} = -3$.  Therefore, the desired sum is $4+12 -3 = \boxed{13}$.


Explanation:
Okay, so I need to find the total surface area of a cube that has a volume of 1 cubic foot. Hmm, let me think. First, I remember that the volume of a cube is calculated by cubing the length of one of its sides. So, if the volume is 1 cubic foot, then the length of each side must be the cube root of 1. Let me write that down.

Volume of a cube = side³. So, if Volume = 1 ft³, then side = ∛1. The cube root of 1 is 1, right? So each side of the cube is 1 foot long. Wait, but the question asks for the surface area in square inches. Oh, right! I need to convert feet to inches because the answer needs to be in square inches.

Alright, so first, let me confirm the side length. If the volume is 1 cubic foot, then the side length is 1 foot. Now, converting feet to inches. Since 1 foot is 12 inches, the side length is 12 inches. Got it.

Now, the surface area of a cube. I remember that a cube has 6 faces, each face being a square. The area of one face is side length squared. So, total surface area is 6 times the area of one face. Let me write that formula down.

Total Surface Area = 6 × (side length)².

So, substituting the side length in inches, which is 12 inches. Let me calculate that.

First, calculate the area of one face: 12 inches × 12 inches = 144 square inches. Then multiply by 6 for all six faces. So, 6 × 144. Let me do that multiplication.

Hmm, 6 × 144. Well, 144 × 6. Let me break it down. 100 × 6 = 600, 40 × 6 = 240, 4 × 6 = 24. Adding those together: 600 + 240 = 840, plus 24 is 864. So, 864 square inches.

Wait, let me check that again. Maybe there's a simpler way. 144 × 6. Since 144 × 5 = 720, and then add 144 more, so 720 + 144 = 864. Yep, same result. So the total surface area is 864 square inches.

But let me just make sure I didn't skip any steps or make any conversion errors. Let me go through it again.

Original volume is 1 cubic foot. Convert that to cubic inches? Wait, maybe another way to approach it is to convert the volume to cubic inches first, then find the side length in inches, then compute the surface area. Let me try that method to cross-verify.

1 cubic foot is equal to how many cubic inches? Since 1 foot is 12 inches, then 1 cubic foot is 12 inches × 12 inches × 12 inches. Let me calculate that. 12 × 12 is 144, and 144 × 12. Let me compute that. 144 × 10 = 1440, 144 × 2 = 288, so total is 1440 + 288 = 1728. So, 1 cubic foot is 1728 cubic inches.

So, the volume is 1728 cubic inches. Then, the side length is the cube root of 1728. Let me calculate that. Hmm, 12³ is 1728, right? Because 12 × 12 = 144, 144 × 12 = 1728. Yes, so the side length is 12 inches. Which matches what I had before. So, then the surface area is 6 × (12²) = 6 × 144 = 864 square inches.

Okay, so both methods give me the same answer. That seems consistent. So, I think that's correct.

But just to make sure, let me recall the formulas again. Volume of a cube is s³, surface area is 6s². So, if I have volume in cubic feet, converting to cubic inches is necessary because the answer needs to be in square inches.

Alternatively, if I had kept the side length in feet, then calculated the surface area in square feet, and then converted that to square inches. Let me try that approach as well.

Original side length is 1 foot. Surface area is 6 × (1 foot)² = 6 square feet. Then, converting square feet to square inches. Since 1 foot is 12 inches, 1 square foot is 12 inches × 12 inches = 144 square inches. Therefore, 6 square feet is 6 × 144 = 864 square inches.

Same answer again. So, regardless of the method, I get 864 square inches. Therefore, I think that's the correct answer.

I don't see any mistakes in my reasoning. All three methods—converting volume to cubic inches first, converting side length to inches first, and converting surface area from square feet to square inches—lead me to the same result. So, I feel confident that the total surface area is 864 square inches.

**Final Answer**
The total surface area of the cube is \boxed{864} square inches.

To determine the total surface area of a cube with a volume of 1 cubic foot, we proceed step by step, ensuring all units are consistent and the calculations are accurate.

---

### Step 1: Understand the Volume and Side Length

The volume $ V $ of a cube is given by the formula:

$$
V = s^3
$$

where $ s $ is the length of one side of the cube. Given that the volume is 1 cubic foot:

$$
s^3 = 1 \text{ ft}^3 \Rightarrow s = \sqrt[3]{1} = 1 \text{ ft}
$$

So, the side length of the cube is 1 foot.

---

### Step 2: Convert Side Length to Inches

Since the final answer must be in square inches, we convert the side length from feet to inches. Recall:

$$
1 \text{ ft} = 12 \text{ in}
$$

Thus, the side length in inches is:

$$
s = 12 \text{ in}
$$

---

### Step 3: Calculate the Surface Area

The surface area $ A $ of a cube is given by:

$$
A = 6s^2
$$

Substituting $ s = 12 $ inches:

$$
A = 6 \times (12)^2 = 6 \times 144 = 864 \text{ in}^2
$$

---

### Final Answer

$$
\boxed{864}
$$

The total surface area of the cube is **864 square inches**.

[Assistant]
Explanation:
Okay, so I need to find the total surface area of a cube that has a volume of 1 cubic foot. Hmm, let me think. First, I remember that the volume of a cube is calculated by cubing the length of one of its sides. So, if the volume is 1 cubic foot, then the length of each side must be the cube root of 1. Let me write that down.

Volume of a cube = side³. So, if Volume = 1 ft³, then side = ∛1. The cube root of 1 is 1, right? So each side of the cube is 1 foot long. Wait, but the question asks for the surface area in square inches. Oh, right! I need to convert feet to inches because the answer needs to be in square inches.

Alright, so first, let me confirm the side length. If the volume is 1 cubic foot, then the side length is 1 foot. Now, converting feet to inches. Since 1 foot is 12 inches, the side length is 12 inches. Got it.

Now, the surface area of a cube. I remember that a cube has 6 faces, each face being a square. The area of one face is side length squared. So, total surface area is 6 times the area of one face. Let me write that formula down.

Total Surface Area = 6 × (side length)².

So, substituting the side length in inches, which is 12 inches. Let me calculate that.

First, calculate the area of one face: 12 inches × 12 inches = 144 square inches. Then multiply by 6 for all six faces. So, 6 × 144. Let me do that multiplication.

Hmm, 6 × 144. Well, 144 × 6. Let me break it down. 100 × 6 = 600, 40 × 6 = 240, 4 × 6 = 24. Adding those together: 600 + 240 = 840, plus 24 is 864. So, 864 square inches.

Wait, let me check that again. Maybe there's a simpler way. 144 × 6. Since 144 × 5 = 720, and then add 144 more, so 720 + 144 = 864. Yep, same result. So the total surface area is 864 square inches.

But let me just make sure I didn't skip any steps or make any conversion errors. Let me go through it again.

Original volume is 1 cubic foot. Convert that to cubic inches? Wait, maybe another way to approach it is to convert the volume to cubic inches first, then find the side length in inches, then compute the surface area. Let me try that method to cross-verify.

1 cubic foot is equal to how many cubic inches? Since 1 foot is 12 inches, then 1 cubic foot is 12 inches × 12 inches × 12 inches. Let me calculate that. 12 × 12 is 144, and 144 × 12. Let me compute that. 144 × 10 = 1440, 144 × 2 = 288, so total is 1440 + 288 = 1728. So, 1 cubic foot is 1728 cubic inches.

So, the volume is 1728 cubic inches. Then, the side length is the cube root of 1728. Let me calculate that. Hmm, 12³ is 1728, right? Because 12 × 12 = 144, 144 × 12 = 1728. Yes, so the side length is 12 inches. Which matches what I had before. So, then the surface area is 6 × (12²) = 6 × 144 = 864 square inches.

Okay, so both methods give me the same answer. That seems consistent. So, I think that's correct.

But just to make sure, let me recall the formulas again. Volume of a cube is s³, surface area is 6s². So, if I have volume in cubic feet, converting to cubic inches is necessary because the answer needs to be in square inches.

Alternatively, if I had kept the side length in feet, then calculated the surface area in square feet, and then converted that to square inches. Let me try that approach as well.

Original side length is 1 foot. Surface area is 6 × (1 foot)² = 6 square feet. Then, converting square feet to square inches. Since 1 foot is 12 inches, 1 square foot is 12 inches × 12 inches = 144 square inches. Therefore, 6 square feet is 6 × 144 = 864 square inches.

Same answer again. So, regardless of the method, I get 864 square inches. Therefore, I think that's the correct answer.

I don't see any mistakes in my reasoning. All three methods—converting volume to cubic inches first, converting side length to inches first, and converting surface area from square feet to square inches—lead me to the same result. So, I feel confident that the total surface area is 864 square inches.

**Final Answer**
The total surface area of the cube is \boxed{864} square inches.

To determine the total surface area of a cube with a volume of 1 cubic foot, we proceed step by step, ensuring all units are consistent and the calculations are accurate.

---

### Step 1: Understand the Volume and Side Length

The volume $ V $ of a cube is given by the formula:

$$
V = s^3
$$

where $ s $ is the length of one side of the cube. Given that the volume is 1 cubic foot:

$$
s^3 = 1 \text{ ft}^3 \Rightarrow s = \sqrt[3]{1} = 1 \text{ ft}
$$

So, the side length of the cube is 1 foot.

---

### Step 2: Convert Side Length to Inches

Since the final answer must be in square inches, we convert the side length from feet to inches. Recall:

$$
1 \text{ ft} = 12 \text{ in}
$$

Thus, the side length in inches is:

$$
s = 12 \text{ in}
$$

---

### Step 3: Calculate the Surface Area

The surface area $ A $ of a cube is given by:

$$
A = 6s^2
$$

Substituting $ s = 12 $ inches:

$$
A = 6 \times (12)^2 = 6 \times 144 = 864 \text{ in}^2
$$

---

### Final Answer

$$
\boxed{864}
$$

The total surface area of the cube is **864 square inches**.

Solution:
Recall that  $$1 \mbox{ foot} = 12 \mbox{ inches}$$

Hence $$1 \mbox{ foot}^2 = 12^2 \mbox{ inches}^2 = 144 \mbox{ inches}^2$$

Finally, remember the formula $V = l \times w \times h$, that is, Volume is the product of length, width, and height.

The length, height, and width for a cube are equal, so the cube we are given has sides of length $1 \mbox{ feet}$. Now, a cube has 6 faces, so the surface area of our cube is $$6 \times (1 \mbox{ foot} \times 1 \mbox{ foot}) = 6 \mbox{ feet}^2$$

Now, convert: $$ 6 \mbox{ feet}^2 \frac{144 \mbox{ inches}^2}{1 \mbox{ foot}^2} = 864 \mbox{ inches}^2$$

So, our final answer is $\boxed{864 \mbox{ inches}^2}$