Explanation:
Okay, so I need to find the remainder when this polynomial f(x) is divided by x² - 1. The problem says not to use long division, which I guess is because it would be too tedious with such a high-degree polynomial. Hmm, so maybe there's another method? Let me recall...

I remember that when you divide a polynomial by a quadratic like x² - 1, the remainder should be a linear polynomial, right? Because the degree of the remainder has to be less than the divisor. So, the remainder will be of the form ax + b. My job is to find a and b such that f(x) = (x² - 1) * Q(x) + ax + b, where Q(x) is the quotient.

But how do I find a and b without doing long division? Maybe I can use the Remainder Factor Theorem or something related? Wait, the Remainder Factor Theorem usually applies to linear divisors, like x - c. But here the divisor is quadratic. However, maybe I can use the fact that if x² - 1 = 0, then x² = 1. So, if I plug x = 1 and x = -1 into f(x), I might get the remainder evaluated at those points?

Let me think. If I have f(x) = (x² - 1)Q(x) + ax + b, then if I plug in x = 1, the term (x² - 1) becomes zero, so f(1) = a*1 + b. Similarly, plugging in x = -1 gives f(-1) = a*(-1) + b. So then I have a system of two equations:

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

Then I can solve for a and b. That seems manageable! So, I need to compute f(1) and f(-1) and then solve the equations. Okay, let me do that.

First, compute f(1). Let me plug 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

Since any power of 1 is 1, this simplifies to:

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

Let me compute this 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

So f(1) = -10. Therefore, the first equation is:

a + b = -10. (Equation 1)

Now compute f(-1). Let me plug 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
5*(-1)^9 = 5*(-1) = -5
-8*(-1)^8 = -8*(1) = -8
7*(-1)^7 = 7*(-1) = -7
- (-1)^6 = - (1) = -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 -5.

Now let me write all these terms:

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

Let me compute step by step:

Start with 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.

Therefore, f(-1) = 16. Therefore, the second equation is:

- a + b = 16. (Equation 2)

Now we have:

Equation 1: a + b = -10
Equation 2: -a + b = 16

Let me solve this system. If I add Equation 1 and Equation 2:

(a + b) + (-a + b) = -10 + 16
Simplify left side: a - a + b + b = 0 + 2b
Right side: 6
So 2b = 6 => b = 3.

Then plug back b = 3 into Equation 1: a + 3 = -10 => a = -10 - 3 = -13.

Therefore, the remainder is ax + b = -13x + 3.

Wait, let me check if this is correct. Let me verify with another method or maybe plug back into the original equations.

Check Equation 1: a + b = -13 + 3 = -10. Correct.
Check Equation 2: -a + b = 13 + 3 = 16. Correct. Okay, seems right.

But let me verify if this remainder is correct by plugging in some value. For example, take x = 0. Let me compute f(0) and see if it equals (0^2 -1)*Q(0) + a*0 + b = -1*Q(0) + 0 + b. Therefore, f(0) should equal b. Let me compute f(0):

Original polynomial f(x): when x=0, all terms with x will be 0 except the constant term. The constant term is -5. Wait, f(0) = 0 + 0 - 0 + 0 - 0 -0 + 0 -0 +0 -0 -5 = -5. According to our remainder, when x=0, the remainder would be -13*0 + 3 = 3. But according to the division algorithm, f(0) should equal the remainder evaluated at 0. However, according to our calculation, f(0) is -5 and remainder at 0 is 3. That's a contradiction. Wait, that suggests there's a mistake. What did I do wrong?

Wait, no. Wait, the division algorithm says that f(x) = (x² - 1) * Q(x) + ax + b. Therefore, when x = 0, f(0) = (0 - 1)*Q(0) + a*0 + b = -Q(0) + b. Therefore, f(0) = -Q(0) + b. Therefore, we can't directly say that f(0) equals b. So my previous thought was wrong. Therefore, there is no contradiction here. So maybe the remainder is correct. Let me check another value.

Alternatively, maybe I can check with another value. Let me pick x = 2. Let me compute f(2) using the original polynomial and also compute (x² -1)*Q(x) + ax + b. But since Q(x) is unknown, it's hard. Alternatively, since we know the remainder is ax + b, then for any x, f(x) - ax - b should be divisible by x² -1. Therefore, if we plug in x = 1 and x = -1, it should be zero. Which we already did. But maybe another way is to use another x value?

Alternatively, let me check if my calculation of f(1) and f(-1) is correct.

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:

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

Let me recalculate:

Start from the beginning:

Term by term:

1 (x^10)

+5 (x^9) → total 6

-8 (x^8) → 6 - 8 = -2

+7 (x^7) → -2 +7 = 5

-1 (x^6) → 5 -1 = 4

-12 (x^5) → 4 -12 = -8

+4 (x^4) → -8 +4 = -4

-8 (x^3) → -4 -8 = -12

+12 (x^2) → -12 +12 = 0

-5 (x) → 0 -5 = -5

-5 (constant term) → -5 -5 = -10.

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

For f(-1):

Original polynomial:

(-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) = -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.

So adding up:

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

Let me recompute:

Start with 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. Correct. So f(-1) = 16.

Therefore, solving:

a + b = -10
- a + b = 16.

Adding equations: 2b = 6 ⇒ b = 3, then a = -13. Therefore, the remainder is -13x +3.

But when I checked f(0), I thought there was a contradiction. Let me recheck that.

The remainder is -13x +3. So when x = 0, remainder is 3. According to division algorithm, f(0) = (0^2 -1) * Q(0) + remainder(0). Therefore, f(0) = (-1)*Q(0) + 3 ⇒ -5 = -Q(0) + 3 ⇒ Q(0) = 3 +5 = 8. Which is possible. Since we don't know Q(x), it's okay. So there's no contradiction. Therefore, maybe the answer is correct. Let me check with another x value. Let me take x = 2.

Compute f(2) using original polynomial and also compute (x² -1)*Q(x) + (-13x + 3). But since I don't know Q(x), maybe I can compute f(2) and also compute (-13*2 +3) and see if f(2) - (-13*2 +3) is divisible by x² -1? Wait, how?

Alternatively, maybe since x² -1 divides f(x) - (-13x +3), then f(x) - (-13x +3) should be divisible by x² -1, so f(x) - (-13x +3) = (x² -1)*something. Therefore, if I evaluate f(x) - (-13x +3) at x =1 and x=-1, it should be 0, which we already checked.

But for x=1: f(1) - (-13*1 +3) = -10 - (-13 +3) = -10 - (-10) = 0. Correct. Similarly, for x=-1: f(-1) - (-13*(-1) +3) = 16 - (13 +3) = 16 -16 =0. Correct. So that is satisfied. Therefore, even though f(0) is not equal to the remainder at 0, that's okay because the division algorithm accounts for the other term. Therefore, maybe my answer is correct.

Wait, but let me check with another approach. Since x² ≡1 mod (x² -1). Therefore, in the ring of polynomials modulo x² -1, x² is equivalent to 1. So, we can replace every occurrence of x² with 1, x³ with x, x^4 with 1, etc. Therefore, to compute f(x) mod (x² -1), we can reduce the polynomial by replacing x² with 1, x³ with x, x^4 with 1, etc. Let me try this method.

So let's take f(x):

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

We need to reduce each term modulo x² -1.

First, note that x² ≡1, so x^k ≡ x^{k mod 2} if k is odd, or x^0 = 1 if k is even. Wait, more precisely, for any exponent:

x^0 = 1
x^1 = x
x^2 = 1
x^3 = x^2 * x ≡ 1 * x = x
x^4 = (x^2)^2 ≡ 1^2 = 1
x^5 = x^4 * x ≡ 1 * x = x
Similarly, x^6 = (x^2)^3 ≡1^3=1
x^7 = x^6 *x ≡1 *x =x
x^8 = (x^2)^4 ≡1
x^9 = x^8 *x ≡1 *x =x
x^10 = (x^2)^5 ≡1.

Therefore, for even exponents: x^{even} ≡1
For odd exponents: x^{odd} ≡x

Therefore, let's compute each term:

Term by term:

x^10 ≡1
5x^9 ≡5x
-8x^8 ≡-8*1 = -8
7x^7 ≡7x
- x^6 ≡-1
-12x^5 ≡-12x
4x^4 ≡4*1 =4
-8x^3 ≡-8x
12x^2 ≡12*1 =12
-5x ≡-5x
-5 ≡-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, combine the constant terms:

1 -8 -1 +4 +12 -5.

Compute step by step:

1 -8 = -7
-7 -1 = -8
-8 +4 = -4
-4 +12 = 8
8 -5 = 3.

Now, combine the x terms:

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

Compute step by step:

5x +7x =12x
12x -12x =0
0 -8x = -8x
-8x -5x = -13x.

Therefore, total is 3 -13x. Which is the same as -13x +3. Which matches the previous result. Therefore, that's correct. Therefore, the remainder is indeed -13x +3.

So my initial concern about f(0) was unfounded because the remainder isn't directly equal to f(0); it's only equal when you consider the division algorithm. So, after verifying with two methods—using the roots of the divisor and reducing the polynomial modulo x² -1—I can be confident that the remainder is -13x +3.

Therefore, the answer should be \boxed{-13x + 3} or in the form they might expect. Wait, the problem says "find the remainder", so it's acceptable to write it as a linear polynomial. Let me check if they want it in a specific form, but since they didn't specify, and the question says "the remainder when f(x) is divided by x² -1", which is ax + b, so -13x +3 is correct. But sometimes, they might prefer writing it as 3 -13x, but both are correct. Let me check if the problem has any preferences, but since they don't, both are okay. However, sometimes people prefer writing the x term first, so maybe -13x + 3 is better.

Alternatively, maybe even check with another method. For example, if I consider that the remainder is ax + b, then f(x) = (x² -1)Q(x) + ax + b. Then, taking derivatives? Not sure. Maybe not necessary. Given that two methods gave the same answer, and the check for f(1) and f(-1) worked, and the modulo approach worked, I think it's safe.

**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 use the **Remainder Theorem for higher-degree divisors**:

---

### **Step 1: Understand the Structure of the Remainder**

When dividing a polynomial by a quadratic polynomial $x^2 - 1$, the remainder must be a **linear polynomial** (degree 1 or less), because the remainder must be **less than the divisor in degree**.

Let the remainder be:
$$
r(x) = ax + b
$$

We are to find the coefficients $a$ and $b$ such that:
$$
f(x) = (x^2 - 1)Q(x) + ax + b
$$

---

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

The roots of $x^2 - 1 = 0$ are:
$$
x = 1, \quad x = -1
$$

Then:
$$
f(1) = a(1) + b = a + b \\
f(-1) = a(-1) + b = -a + b
$$

We solve this system of equations using the values of $f(1)$ and $f(-1)$.

---

### **Step 3: Evaluate $f(1)$ and $f(-1)$**

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

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

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

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

---

### **Step 4: Set Up and Solve the System**

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

Adding the equations:
$$
(a + b) + (-a + b) = 2b = 6 \Rightarrow b = 3
$$

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

---

### **Step 5: Write the Remainder**

Thus, the remainder is:
$$
r(x) = -13x + 3
$$

---

### **Final Answer**

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

[Assistant]
Explanation:
Okay, so I need to find the remainder when this polynomial f(x) is divided by x² - 1. The problem says not to use long division, which I guess is because it would be too tedious with such a high-degree polynomial. Hmm, so maybe there's another method? Let me recall...

I remember that when you divide a polynomial by a quadratic like x² - 1, the remainder should be a linear polynomial, right? Because the degree of the remainder has to be less than the divisor. So, the remainder will be of the form ax + b. My job is to find a and b such that f(x) = (x² - 1) * Q(x) + ax + b, where Q(x) is the quotient.

But how do I find a and b without doing long division? Maybe I can use the Remainder Factor Theorem or something related? Wait, the Remainder Factor Theorem usually applies to linear divisors, like x - c. But here the divisor is quadratic. However, maybe I can use the fact that if x² - 1 = 0, then x² = 1. So, if I plug x = 1 and x = -1 into f(x), I might get the remainder evaluated at those points?

Let me think. If I have f(x) = (x² - 1)Q(x) + ax + b, then if I plug in x = 1, the term (x² - 1) becomes zero, so f(1) = a*1 + b. Similarly, plugging in x = -1 gives f(-1) = a*(-1) + b. So then I have a system of two equations:

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

Then I can solve for a and b. That seems manageable! So, I need to compute f(1) and f(-1) and then solve the equations. Okay, let me do that.

First, compute f(1). Let me plug 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

Since any power of 1 is 1, this simplifies to:

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

Let me compute this 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

So f(1) = -10. Therefore, the first equation is:

a + b = -10. (Equation 1)

Now compute f(-1). Let me plug 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
5*(-1)^9 = 5*(-1) = -5
-8*(-1)^8 = -8*(1) = -8
7*(-1)^7 = 7*(-1) = -7
- (-1)^6 = - (1) = -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 -5.

Now let me write all these terms:

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

Let me compute step by step:

Start with 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.

Therefore, f(-1) = 16. Therefore, the second equation is:

- a + b = 16. (Equation 2)

Now we have:

Equation 1: a + b = -10
Equation 2: -a + b = 16

Let me solve this system. If I add Equation 1 and Equation 2:

(a + b) + (-a + b) = -10 + 16
Simplify left side: a - a + b + b = 0 + 2b
Right side: 6
So 2b = 6 => b = 3.

Then plug back b = 3 into Equation 1: a + 3 = -10 => a = -10 - 3 = -13.

Therefore, the remainder is ax + b = -13x + 3.

Wait, let me check if this is correct. Let me verify with another method or maybe plug back into the original equations.

Check Equation 1: a + b = -13 + 3 = -10. Correct.
Check Equation 2: -a + b = 13 + 3 = 16. Correct. Okay, seems right.

But let me verify if this remainder is correct by plugging in some value. For example, take x = 0. Let me compute f(0) and see if it equals (0^2 -1)*Q(0) + a*0 + b = -1*Q(0) + 0 + b. Therefore, f(0) should equal b. Let me compute f(0):

Original polynomial f(x): when x=0, all terms with x will be 0 except the constant term. The constant term is -5. Wait, f(0) = 0 + 0 - 0 + 0 - 0 -0 + 0 -0 +0 -0 -5 = -5. According to our remainder, when x=0, the remainder would be -13*0 + 3 = 3. But according to the division algorithm, f(0) should equal the remainder evaluated at 0. However, according to our calculation, f(0) is -5 and remainder at 0 is 3. That's a contradiction. Wait, that suggests there's a mistake. What did I do wrong?

Wait, no. Wait, the division algorithm says that f(x) = (x² - 1) * Q(x) + ax + b. Therefore, when x = 0, f(0) = (0 - 1)*Q(0) + a*0 + b = -Q(0) + b. Therefore, f(0) = -Q(0) + b. Therefore, we can't directly say that f(0) equals b. So my previous thought was wrong. Therefore, there is no contradiction here. So maybe the remainder is correct. Let me check another value.

Alternatively, maybe I can check with another value. Let me pick x = 2. Let me compute f(2) using the original polynomial and also compute (x² -1)*Q(x) + ax + b. But since Q(x) is unknown, it's hard. Alternatively, since we know the remainder is ax + b, then for any x, f(x) - ax - b should be divisible by x² -1. Therefore, if we plug in x = 1 and x = -1, it should be zero. Which we already did. But maybe another way is to use another x value?

Alternatively, let me check if my calculation of f(1) and f(-1) is correct.

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:

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

Let me recalculate:

Start from the beginning:

Term by term:

1 (x^10)

+5 (x^9) → total 6

-8 (x^8) → 6 - 8 = -2

+7 (x^7) → -2 +7 = 5

-1 (x^6) → 5 -1 = 4

-12 (x^5) → 4 -12 = -8

+4 (x^4) → -8 +4 = -4

-8 (x^3) → -4 -8 = -12

+12 (x^2) → -12 +12 = 0

-5 (x) → 0 -5 = -5

-5 (constant term) → -5 -5 = -10.

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

For f(-1):

Original polynomial:

(-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) = -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.

So adding up:

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

Let me recompute:

Start with 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. Correct. So f(-1) = 16.

Therefore, solving:

a + b = -10
- a + b = 16.

Adding equations: 2b = 6 ⇒ b = 3, then a = -13. Therefore, the remainder is -13x +3.

But when I checked f(0), I thought there was a contradiction. Let me recheck that.

The remainder is -13x +3. So when x = 0, remainder is 3. According to division algorithm, f(0) = (0^2 -1) * Q(0) + remainder(0). Therefore, f(0) = (-1)*Q(0) + 3 ⇒ -5 = -Q(0) + 3 ⇒ Q(0) = 3 +5 = 8. Which is possible. Since we don't know Q(x), it's okay. So there's no contradiction. Therefore, maybe the answer is correct. Let me check with another x value. Let me take x = 2.

Compute f(2) using original polynomial and also compute (x² -1)*Q(x) + (-13x + 3). But since I don't know Q(x), maybe I can compute f(2) and also compute (-13*2 +3) and see if f(2) - (-13*2 +3) is divisible by x² -1? Wait, how?

Alternatively, maybe since x² -1 divides f(x) - (-13x +3), then f(x) - (-13x +3) should be divisible by x² -1, so f(x) - (-13x +3) = (x² -1)*something. Therefore, if I evaluate f(x) - (-13x +3) at x =1 and x=-1, it should be 0, which we already checked.

But for x=1: f(1) - (-13*1 +3) = -10 - (-13 +3) = -10 - (-10) = 0. Correct. Similarly, for x=-1: f(-1) - (-13*(-1) +3) = 16 - (13 +3) = 16 -16 =0. Correct. So that is satisfied. Therefore, even though f(0) is not equal to the remainder at 0, that's okay because the division algorithm accounts for the other term. Therefore, maybe my answer is correct.

Wait, but let me check with another approach. Since x² ≡1 mod (x² -1). Therefore, in the ring of polynomials modulo x² -1, x² is equivalent to 1. So, we can replace every occurrence of x² with 1, x³ with x, x^4 with 1, etc. Therefore, to compute f(x) mod (x² -1), we can reduce the polynomial by replacing x² with 1, x³ with x, x^4 with 1, etc. Let me try this method.

So let's take f(x):

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

We need to reduce each term modulo x² -1.

First, note that x² ≡1, so x^k ≡ x^{k mod 2} if k is odd, or x^0 = 1 if k is even. Wait, more precisely, for any exponent:

x^0 = 1
x^1 = x
x^2 = 1
x^3 = x^2 * x ≡ 1 * x = x
x^4 = (x^2)^2 ≡ 1^2 = 1
x^5 = x^4 * x ≡ 1 * x = x
Similarly, x^6 = (x^2)^3 ≡1^3=1
x^7 = x^6 *x ≡1 *x =x
x^8 = (x^2)^4 ≡1
x^9 = x^8 *x ≡1 *x =x
x^10 = (x^2)^5 ≡1.

Therefore, for even exponents: x^{even} ≡1
For odd exponents: x^{odd} ≡x

Therefore, let's compute each term:

Term by term:

x^10 ≡1
5x^9 ≡5x
-8x^8 ≡-8*1 = -8
7x^7 ≡7x
- x^6 ≡-1
-12x^5 ≡-12x
4x^4 ≡4*1 =4
-8x^3 ≡-8x
12x^2 ≡12*1 =12
-5x ≡-5x
-5 ≡-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, combine the constant terms:

1 -8 -1 +4 +12 -5.

Compute step by step:

1 -8 = -7
-7 -1 = -8
-8 +4 = -4
-4 +12 = 8
8 -5 = 3.

Now, combine the x terms:

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

Compute step by step:

5x +7x =12x
12x -12x =0
0 -8x = -8x
-8x -5x = -13x.

Therefore, total is 3 -13x. Which is the same as -13x +3. Which matches the previous result. Therefore, that's correct. Therefore, the remainder is indeed -13x +3.

So my initial concern about f(0) was unfounded because the remainder isn't directly equal to f(0); it's only equal when you consider the division algorithm. So, after verifying with two methods—using the roots of the divisor and reducing the polynomial modulo x² -1—I can be confident that the remainder is -13x +3.

Therefore, the answer should be \boxed{-13x + 3} or in the form they might expect. Wait, the problem says "find the remainder", so it's acceptable to write it as a linear polynomial. Let me check if they want it in a specific form, but since they didn't specify, and the question says "the remainder when f(x) is divided by x² -1", which is ax + b, so -13x +3 is correct. But sometimes, they might prefer writing it as 3 -13x, but both are correct. Let me check if the problem has any preferences, but since they don't, both are okay. However, sometimes people prefer writing the x term first, so maybe -13x + 3 is better.

Alternatively, maybe even check with another method. For example, if I consider that the remainder is ax + b, then f(x) = (x² -1)Q(x) + ax + b. Then, taking derivatives? Not sure. Maybe not necessary. Given that two methods gave the same answer, and the check for f(1) and f(-1) worked, and the modulo approach worked, I think it's safe.

**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 use the **Remainder Theorem for higher-degree divisors**:

---

### **Step 1: Understand the Structure of the Remainder**

When dividing a polynomial by a quadratic polynomial $x^2 - 1$, the remainder must be a **linear polynomial** (degree 1 or less), because the remainder must be **less than the divisor in degree**.

Let the remainder be:
$$
r(x) = ax + b
$$

We are to find the coefficients $a$ and $b$ such that:
$$
f(x) = (x^2 - 1)Q(x) + ax + b
$$

---

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

The roots of $x^2 - 1 = 0$ are:
$$
x = 1, \quad x = -1
$$

Then:
$$
f(1) = a(1) + b = a + b \\
f(-1) = a(-1) + b = -a + b
$$

We solve this system of equations using the values of $f(1)$ and $f(-1)$.

---

### **Step 3: Evaluate $f(1)$ and $f(-1)$**

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

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

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

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

---

### **Step 4: Set Up and Solve the System**

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

Adding the equations:
$$
(a + b) + (-a + b) = 2b = 6 \Rightarrow b = 3
$$

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

---

### **Step 5: Write the Remainder**

Thus, the remainder is:
$$
r(x) = -13x + 3
$$

---

### **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, that seems like a lot of exponents. Direct computation would be impossible because raising a matrix to such high powers would take forever. There must be a smarter way. Maybe diagonalization? Or finding eigenvalues and eigenvectors? Or perhaps there's a pattern in the powers of A?

First, let me recall that if a matrix can be diagonalized, say A = PDP^{-1}, then A^n = PD^nP^{-1}. That might make computing high powers easier. So maybe I should try diagonalizing A?

To diagonalize A, I need to find its eigenvalues and eigenvectors. Let me start by finding the eigenvalues. The eigenvalues are the roots of the characteristic equation det(A - λI) = 0.

So, let's compute the characteristic polynomial:

det(A - λI) = det[[3 - λ, 7], [-1, -2 - λ]] = (3 - λ)(-2 - λ) - (-1)(7)

Let me compute that:

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

Then subtract (-1)(7) which is -(-7) = +7. So total determinant is λ² - λ - 6 + 7 = λ² - λ + 1.

Wait, let me check that again step by step:

First term: (3 - λ)(-2 - λ) = Multiply 3*(-2) + 3*(-λ) - λ*(-2) - λ*λ = -6 -3λ + 2λ + λ² = λ² - λ -6. Then the second term is - (-1)(7) which is +7. So total determinant is λ² - λ -6 +7 = λ² - λ +1. Yes, that seems right.

So the characteristic equation is λ² - λ +1 = 0. Let me solve for λ:

λ = [1 ± sqrt(1 - 4*1*1)] / 2 = [1 ± sqrt(-3)] / 2. So complex eigenvalues: (1 ± i√3)/2. Hmm, complex eigenvalues. That complicates things a bit. Maybe I can still diagonalize it over complex numbers? But if I need to work with real matrices, perhaps I should consider using Jordan form? Or maybe there's another approach?

Alternatively, maybe I can find a recurrence relation for the powers of A? Because if I can express A^n in terms of lower powers, maybe I can find a pattern or a formula?

Alternatively, perhaps using the Cayley-Hamilton theorem? Which says that a matrix satisfies its own characteristic equation. Since the characteristic equation is λ² - λ +1 = 0, then A² - A + I = 0. Therefore, A² = A - I. So that gives a recurrence relation for higher powers. That seems useful!

Let me verify Cayley-Hamilton here. For a 2x2 matrix, the characteristic equation is λ² - tr(A)λ + det(A) = 0. Wait, let me check the trace and determinant of A. The trace of A is 3 + (-2) = 1. The determinant is (3)(-2) - (7)(-1) = -6 +7 = 1. So the characteristic equation is λ² - tr(A)λ + det(A) = λ² - 1λ +1 = 0. So yes, that matches. Therefore, A² - A + I = 0, so A² = A - I. That's a key relation. So maybe I can use this recurrence to find a pattern for A^n?

Let me note that. So A² = A - I. Let me try to compute A³. A³ = A² * A = (A - I) * A = A² - A. But since A² = A - I, substitute: A³ = (A - I) - A = -I. Interesting! So A³ = -I. Let me check that again:

A³ = A² * A = (A - I) * A = A² - A. Then replace A² with A - I: (A - I) - A = -I. Yes, that's correct. So A³ = -I.

Now, that's a big simplification! If A³ = -I, then A^3 = -I, so A^6 = (A^3)^2 = (-I)^2 = I. Similarly, A^9 = (A^3)^3 = (-I)^3 = -I. So the powers of A cycle every 6 exponents? Let me see:

Let me check:

A^1 = A

A^2 = A - I

A^3 = -I

A^4 = A^3 * A = (-I) * A = -A

A^5 = A^4 * A = (-A) * A = -A² = -(A - I) = -A + I

A^6 = A^5 * A = (-A + I) * A = -A² + A = -(A - I) + A = -A + I + A = I

A^7 = A^6 * A = I * A = A

A^8 = A^7 * A = A^2 = A - I

So yes, the cycle repeats every 6 exponents. So the powers of A cycle with period 6. Therefore, A^n is periodic with period 6. So, if we can find n modulo 6, then we can express A^n in terms of A^0, A^1, ..., A^5.

Therefore, to compute A^27, A^31, A^40, we can find each exponent modulo 6.

Let me compute each exponent:

First, 27 divided by 6: 6*4=24, so 27-24=3. So 27 ≡ 3 mod 6. Therefore, A^27 = A^3 = -I.

Next, 31 divided by 6: 6*5=30, 31 -30=1. So 31 ≡1 mod 6. Thus A^31 = A^1 = A.

Then, 40 divided by 6: 6*6=36, 40-36=4. So 40 ≡4 mod 6. Therefore, A^40 = A^4 = -A, as computed earlier.

Therefore, the sum A^27 + A^31 + A^40 = (-I) + A + (-A) = -I + A - A = -I. Wait, that's interesting? Let me check again:

Wait, A^27 = A^3 = -I

A^31 = A^1 = A

A^40 = A^4 = -A

Therefore, adding them: (-I) + A + (-A) = -I + (A - A) = -I + 0 = -I

So the result is -I? That seems surprisingly simple. Let me confirm that with another approach, just to be safe.

Alternatively, since the cycle is 6, let me verify with specific exponents. Let me check A^3 = -I. Let me compute A^3 directly. Let me compute A^2 first:

A is [[3,7],[-1,-2]]

A^2 = A*A = [[3*3 + 7*(-1), 3*7 + 7*(-2)], [-1*3 + (-2)*(-1), -1*7 + (-2)*(-2)]]

Calculating entry by entry:

Top left: 9 -7 = 2

Top right: 21 -14 = 7

Bottom left: -3 + 2 = -1

Bottom right: -7 + 4 = -3

So A^2 = [[2,7],[-1,-3]]

Then A^3 = A^2 * A = [[2*3 +7*(-1), 2*7 +7*(-2)], [-1*3 + (-3)*(-1), -1*7 + (-3)*(-2)]]

Calculating:

Top left: 6 -7 = -1

Top right: 14 -14 = 0

Bottom left: -3 + 3 = 0

Bottom right: -7 + 6 = -1

So A^3 = [[-1, 0], [0, -1]] = -I. Correct! So A^3 = -I. Therefore, all the previous steps are correct. Then A^4 = A^3 * A = -I * A = -A. Similarly, A^5 = A^4 * A = -A * A = -A^2, which is -[[2,7],[-1,-3]] = [[-2,-7],[1,3]]. Then A^6 = A^5 * A = -A^2 * A = -A^3 = -(-I) = I. So yes, that works.

Therefore, the cycle repeats every 6. Thus:

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

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 = -A

Thus, adding them up: (-I) + A + (-A) = -I

So the answer is -I, which is [[-1, 0], [0, -1]].

Wait, but the problem says "compute A^27 + A^31 + A^40", so the answer is -I. Let me confirm once again.

Alternatively, maybe I made a mistake in adding?

Yes, A^27 is -I, A^31 is A, A^40 is -A. Therefore, adding them: (-I) + A - A = -I + 0 = -I. Yes. So the answer is -I. But let me write it as a matrix. So the final matrix is [[-1, 0], [0, -1]]. Therefore, boxed, it would be \boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}.

But wait, just to make sure there isn't a miscalculation in the exponents. Let me check again:

For 27: 27 divided by 6 is 4 with remainder 3, yes. 6*4=24, 27-24=3. So A^27 is A^3. Which is -I. Correct.

For 31: 6*5=30, 31-30=1. So A^1 is A. Correct.

For 40: 6*6=36, 40-36=4. So A^4 is -A. Correct.

Therefore, sum is -I. So I think that is correct.

Alternatively, maybe there is a mistake in the Cayley-Hamilton application? Let me check again:

We had A² = A - I. Then A³ = A² * A = (A - I) * A = A² - A. Then substitute A² again: (A - I) - A = -I. Correct. Then A^4 = A³ * A = (-I)*A = -A. Correct. A^5 = A^4 * A = -A * A = -A² = -(A - I) = -A + I. A^6 = A^5 * A = (-A + I)*A = -A² + A = -(A - I) + A = -A + I + A = I. Correct. Therefore, the cycle is 6. Therefore, the reasoning holds.

Therefore, the answer is indeed -I. Therefore, the answer is [[-1, 0], [0, -1]]. So boxed as requested. But let me check once more if there's any possible mistake in the modular arithmetic.

27 divided by 6: 6*4=24. 27-24=3. Correct. 31-6*5=31-30=1. Correct. 40-6*6=40-36=4. Correct. So exponents 3,1,4. Then A^3=-I, A^1=A, A^4=-A. So sum is -I + A - A = -I. Yes.

Alternatively, perhaps the problem is expecting a different answer? Maybe I made a mistake in the cycle?

Wait, let me check A^6. Earlier, I said A^6=I, but let me compute A^3 * A^3 = (-I)*(-I) = I. Yes. So A^6 is I, yes. So exponents:

A^6 = I

A^7 = A^6 * A = I*A = A

A^8 = A^7 * A = A^2

So the cycle is indeed 6. Therefore, everything checks out. Therefore, the conclusion is correct.

Therefore, the answer is -I. Therefore, the final answer is the matrix with -1 on the diagonal and 0 elsewhere. Therefore, boxed as:

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

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

To compute $\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40}$, where
$$
\mathbf{A} = \begin{pmatrix} 3 & 7 \\ -1 & -2 \end{pmatrix},
$$
we begin by leveraging the **Cayley-Hamilton theorem**, which says that a matrix satisfies its own characteristic equation.

---

### Step 1: Apply the Cayley-Hamilton Theorem

The characteristic equation for $\mathbf{A}$ is:
$$
\det(\mathbf{A} - \lambda \mathbf{I}) = \lambda^2 - \lambda + 1 = 0.
$$
This gives the identity:
$$
\mathbf{A}^2 - \mathbf{A} + \mathbf{I} = \mathbf{0} \Rightarrow \mathbf{A}^2 = \mathbf{A} - \mathbf{I}.
$$

We can use this recurrence to find a pattern for higher powers of $\mathbf{A}$. Let's compute the powers of $\mathbf{A}$ to identify a cycle.

---

### Step 2: Compute a Few Powers of A Using Recurrence

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

So the powers of $\mathbf{A}$ cycle with **period 6**:
$$
\mathbf{A}^1 = \mathbf{A}, \quad
\mathbf{A}^2 = \mathbf{A} - \mathbf{I}, \quad
\mathbf{A}^3 = -\mathbf{I}, \quad
\mathbf{A}^4 = -\mathbf{A}, \quad
\mathbf{A}^5 = -\mathbf{A} + \mathbf{I}, \quad
\mathbf{A}^6 = \mathbf{I}.
$$

---

### Step 3: Reduce Each Exponent Modulo 6

We now compute the exponents modulo 6:

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

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

---

### 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, that seems like a lot of exponents. Direct computation would be impossible because raising a matrix to such high powers would take forever. There must be a smarter way. Maybe diagonalization? Or finding eigenvalues and eigenvectors? Or perhaps there's a pattern in the powers of A?

First, let me recall that if a matrix can be diagonalized, say A = PDP^{-1}, then A^n = PD^nP^{-1}. That might make computing high powers easier. So maybe I should try diagonalizing A?

To diagonalize A, I need to find its eigenvalues and eigenvectors. Let me start by finding the eigenvalues. The eigenvalues are the roots of the characteristic equation det(A - λI) = 0.

So, let's compute the characteristic polynomial:

det(A - λI) = det[[3 - λ, 7], [-1, -2 - λ]] = (3 - λ)(-2 - λ) - (-1)(7)

Let me compute that:

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

Then subtract (-1)(7) which is -(-7) = +7. So total determinant is λ² - λ - 6 + 7 = λ² - λ + 1.

Wait, let me check that again step by step:

First term: (3 - λ)(-2 - λ) = Multiply 3*(-2) + 3*(-λ) - λ*(-2) - λ*λ = -6 -3λ + 2λ + λ² = λ² - λ -6. Then the second term is - (-1)(7) which is +7. So total determinant is λ² - λ -6 +7 = λ² - λ +1. Yes, that seems right.

So the characteristic equation is λ² - λ +1 = 0. Let me solve for λ:

λ = [1 ± sqrt(1 - 4*1*1)] / 2 = [1 ± sqrt(-3)] / 2. So complex eigenvalues: (1 ± i√3)/2. Hmm, complex eigenvalues. That complicates things a bit. Maybe I can still diagonalize it over complex numbers? But if I need to work with real matrices, perhaps I should consider using Jordan form? Or maybe there's another approach?

Alternatively, maybe I can find a recurrence relation for the powers of A? Because if I can express A^n in terms of lower powers, maybe I can find a pattern or a formula?

Alternatively, perhaps using the Cayley-Hamilton theorem? Which says that a matrix satisfies its own characteristic equation. Since the characteristic equation is λ² - λ +1 = 0, then A² - A + I = 0. Therefore, A² = A - I. So that gives a recurrence relation for higher powers. That seems useful!

Let me verify Cayley-Hamilton here. For a 2x2 matrix, the characteristic equation is λ² - tr(A)λ + det(A) = 0. Wait, let me check the trace and determinant of A. The trace of A is 3 + (-2) = 1. The determinant is (3)(-2) - (7)(-1) = -6 +7 = 1. So the characteristic equation is λ² - tr(A)λ + det(A) = λ² - 1λ +1 = 0. So yes, that matches. Therefore, A² - A + I = 0, so A² = A - I. That's a key relation. So maybe I can use this recurrence to find a pattern for A^n?

Let me note that. So A² = A - I. Let me try to compute A³. A³ = A² * A = (A - I) * A = A² - A. But since A² = A - I, substitute: A³ = (A - I) - A = -I. Interesting! So A³ = -I. Let me check that again:

A³ = A² * A = (A - I) * A = A² - A. Then replace A² with A - I: (A - I) - A = -I. Yes, that's correct. So A³ = -I.

Now, that's a big simplification! If A³ = -I, then A^3 = -I, so A^6 = (A^3)^2 = (-I)^2 = I. Similarly, A^9 = (A^3)^3 = (-I)^3 = -I. So the powers of A cycle every 6 exponents? Let me see:

Let me check:

A^1 = A

A^2 = A - I

A^3 = -I

A^4 = A^3 * A = (-I) * A = -A

A^5 = A^4 * A = (-A) * A = -A² = -(A - I) = -A + I

A^6 = A^5 * A = (-A + I) * A = -A² + A = -(A - I) + A = -A + I + A = I

A^7 = A^6 * A = I * A = A

A^8 = A^7 * A = A^2 = A - I

So yes, the cycle repeats every 6 exponents. So the powers of A cycle with period 6. Therefore, A^n is periodic with period 6. So, if we can find n modulo 6, then we can express A^n in terms of A^0, A^1, ..., A^5.

Therefore, to compute A^27, A^31, A^40, we can find each exponent modulo 6.

Let me compute each exponent:

First, 27 divided by 6: 6*4=24, so 27-24=3. So 27 ≡ 3 mod 6. Therefore, A^27 = A^3 = -I.

Next, 31 divided by 6: 6*5=30, 31 -30=1. So 31 ≡1 mod 6. Thus A^31 = A^1 = A.

Then, 40 divided by 6: 6*6=36, 40-36=4. So 40 ≡4 mod 6. Therefore, A^40 = A^4 = -A, as computed earlier.

Therefore, the sum A^27 + A^31 + A^40 = (-I) + A + (-A) = -I + A - A = -I. Wait, that's interesting? Let me check again:

Wait, A^27 = A^3 = -I

A^31 = A^1 = A

A^40 = A^4 = -A

Therefore, adding them: (-I) + A + (-A) = -I + (A - A) = -I + 0 = -I

So the result is -I? That seems surprisingly simple. Let me confirm that with another approach, just to be safe.

Alternatively, since the cycle is 6, let me verify with specific exponents. Let me check A^3 = -I. Let me compute A^3 directly. Let me compute A^2 first:

A is [[3,7],[-1,-2]]

A^2 = A*A = [[3*3 + 7*(-1), 3*7 + 7*(-2)], [-1*3 + (-2)*(-1), -1*7 + (-2)*(-2)]]

Calculating entry by entry:

Top left: 9 -7 = 2

Top right: 21 -14 = 7

Bottom left: -3 + 2 = -1

Bottom right: -7 + 4 = -3

So A^2 = [[2,7],[-1,-3]]

Then A^3 = A^2 * A = [[2*3 +7*(-1), 2*7 +7*(-2)], [-1*3 + (-3)*(-1), -1*7 + (-3)*(-2)]]

Calculating:

Top left: 6 -7 = -1

Top right: 14 -14 = 0

Bottom left: -3 + 3 = 0

Bottom right: -7 + 6 = -1

So A^3 = [[-1, 0], [0, -1]] = -I. Correct! So A^3 = -I. Therefore, all the previous steps are correct. Then A^4 = A^3 * A = -I * A = -A. Similarly, A^5 = A^4 * A = -A * A = -A^2, which is -[[2,7],[-1,-3]] = [[-2,-7],[1,3]]. Then A^6 = A^5 * A = -A^2 * A = -A^3 = -(-I) = I. So yes, that works.

Therefore, the cycle repeats every 6. Thus:

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

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 = -A

Thus, adding them up: (-I) + A + (-A) = -I

So the answer is -I, which is [[-1, 0], [0, -1]].

Wait, but the problem says "compute A^27 + A^31 + A^40", so the answer is -I. Let me confirm once again.

Alternatively, maybe I made a mistake in adding?

Yes, A^27 is -I, A^31 is A, A^40 is -A. Therefore, adding them: (-I) + A - A = -I + 0 = -I. Yes. So the answer is -I. But let me write it as a matrix. So the final matrix is [[-1, 0], [0, -1]]. Therefore, boxed, it would be \boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}.

But wait, just to make sure there isn't a miscalculation in the exponents. Let me check again:

For 27: 27 divided by 6 is 4 with remainder 3, yes. 6*4=24, 27-24=3. So A^27 is A^3. Which is -I. Correct.

For 31: 6*5=30, 31-30=1. So A^1 is A. Correct.

For 40: 6*6=36, 40-36=4. So A^4 is -A. Correct.

Therefore, sum is -I. So I think that is correct.

Alternatively, maybe there is a mistake in the Cayley-Hamilton application? Let me check again:

We had A² = A - I. Then A³ = A² * A = (A - I) * A = A² - A. Then substitute A² again: (A - I) - A = -I. Correct. Then A^4 = A³ * A = (-I)*A = -A. Correct. A^5 = A^4 * A = -A * A = -A² = -(A - I) = -A + I. A^6 = A^5 * A = (-A + I)*A = -A² + A = -(A - I) + A = -A + I + A = I. Correct. Therefore, the cycle is 6. Therefore, the reasoning holds.

Therefore, the answer is indeed -I. Therefore, the answer is [[-1, 0], [0, -1]]. So boxed as requested. But let me check once more if there's any possible mistake in the modular arithmetic.

27 divided by 6: 6*4=24. 27-24=3. Correct. 31-6*5=31-30=1. Correct. 40-6*6=40-36=4. Correct. So exponents 3,1,4. Then A^3=-I, A^1=A, A^4=-A. So sum is -I + A - A = -I. Yes.

Alternatively, perhaps the problem is expecting a different answer? Maybe I made a mistake in the cycle?

Wait, let me check A^6. Earlier, I said A^6=I, but let me compute A^3 * A^3 = (-I)*(-I) = I. Yes. So A^6 is I, yes. So exponents:

A^6 = I

A^7 = A^6 * A = I*A = A

A^8 = A^7 * A = A^2

So the cycle is indeed 6. Therefore, everything checks out. Therefore, the conclusion is correct.

Therefore, the answer is -I. Therefore, the final answer is the matrix with -1 on the diagonal and 0 elsewhere. Therefore, boxed as:

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

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

To compute $\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40}$, where
$$
\mathbf{A} = \begin{pmatrix} 3 & 7 \\ -1 & -2 \end{pmatrix},
$$
we begin by leveraging the **Cayley-Hamilton theorem**, which says that a matrix satisfies its own characteristic equation.

---

### Step 1: Apply the Cayley-Hamilton Theorem

The characteristic equation for $\mathbf{A}$ is:
$$
\det(\mathbf{A} - \lambda \mathbf{I}) = \lambda^2 - \lambda + 1 = 0.
$$
This gives the identity:
$$
\mathbf{A}^2 - \mathbf{A} + \mathbf{I} = \mathbf{0} \Rightarrow \mathbf{A}^2 = \mathbf{A} - \mathbf{I}.
$$

We can use this recurrence to find a pattern for higher powers of $\mathbf{A}$. Let's compute the powers of $\mathbf{A}$ to identify a cycle.

---

### Step 2: Compute a Few Powers of A Using Recurrence

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

So the powers of $\mathbf{A}$ cycle with **period 6**:
$$
\mathbf{A}^1 = \mathbf{A}, \quad
\mathbf{A}^2 = \mathbf{A} - \mathbf{I}, \quad
\mathbf{A}^3 = -\mathbf{I}, \quad
\mathbf{A}^4 = -\mathbf{A}, \quad
\mathbf{A}^5 = -\mathbf{A} + \mathbf{I}, \quad
\mathbf{A}^6 = \mathbf{I}.
$$

---

### Step 3: Reduce Each Exponent Modulo 6

We now compute the exponents modulo 6:

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

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

---

### 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 1, 3, 5,...,21) such that t + 2 is the inverse of t modulo 23. Let me recall what an inverse means here.

In modular arithmetic, the inverse of a number a modulo m is a number b such that (a * b) ≡ 1 mod m. So, if t + 2 is the inverse of t modulo 23, then multiplying t by (t + 2) should give me 1 modulo 23. So, mathematically, this means:

t * (t + 2) ≡ 1 mod 23

So, expanding that equation:

t² + 2t ≡ 1 mod 23

Which can be rewritten as:

t² + 2t - 1 ≡ 0 mod 23

So, I need to solve this quadratic congruence equation for t. Since we're working modulo 23, which is a prime number, maybe I can use some methods from number theory to solve this?

Alternatively, maybe I can just try plugging in the possible odd values of t between 1 and 21 and check which one satisfies the condition. Since there aren't that many possibilities, maybe that's feasible. Let me try that approach first, just to see if I can find t without too much trouble.

First, let me list all odd integers between 1 and 21:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21.

Now, for each of these t, I need to compute t * (t + 2) mod 23 and check if it equals 1. Let's start:

Starting with t = 1:

1 * (1 + 2) = 1 * 3 = 3 mod 23. 3 ≠ 1. Not good.

t = 3:

3 * (3 + 2) = 3 * 5 = 15 mod 23. 15 ≠ 1.

t = 5:

5 * 7 = 35. 35 mod 23 is 35 - 23 = 12. 12 ≠ 1.

t = 7:

7 * 9 = 63. 63 divided by 23: 23*2=46, 63 - 46=17. 17 ≠ 1.

t = 9:

9 * 11 = 99. Let's divide 99 by 23. 23*4=92, 99 - 92 = 7. 7 ≠ 1.

t = 11:

11 * 13 = 143. Let's divide by 23. 23*6 = 138, 143 - 138 = 5. 5 ≠ 1.

t = 13:

13 * 15 = 195. Let me divide by 23. 23*8 = 184, 195 - 184 = 11. 11 ≠ 1.

t = 15:

15 * 17 = 255. Dividing by 23. Let's see: 23*11 = 253. 255 - 253 = 2. 2 ≠ 1.

t = 17:

17 * 19 = 323. Dividing by 23: Let me compute 23*14 = 322. 323 - 322 = 1. Oh! That's 1 mod 23! So t=17 gives us 1. Wait, is that right?

Wait, let me double-check that calculation. 17 * 19 is indeed 323. Then 23*14 is 322, so 323 - 322 = 1. Therefore, 323 ≡ 1 mod 23. Therefore, t=17 satisfies the equation. But let me check if that's the only one. Let me continue checking the rest to see if there's another one.

t = 19:

19 * (19 + 2) = 19 * 21. Let's compute that. 19*20=380, plus 19 is 399. Now, 399 divided by 23. Let me calculate 23*17= 391. 399 - 391=8. 8 ≠ 1.

t = 21:

21 * (21 + 2) = 21 * 23. Well, 21*23 mod 23 is 0, since 23 is a multiple of 23. So 0 ≠ 1. So, nope.

So, the only one that works is t=17. And since the question says "unique odd integer t", so that's probably the answer. But let me just make sure that I didn't make any mistakes in my calculations for t=17.

Wait, t=17 is odd, and between 0 and 23. Also, t + 2 is 19. So, does 17 * 19 ≡ 1 mod 23? As above, 17*19=323. Let me compute 23*14=322, so 323 - 322=1. Yes, exactly. So that works. And since we checked all other values, and only t=17 worked, so that's the answer.

But just to be thorough, maybe I should also try solving the quadratic equation algebraically, just to confirm that there isn't another solution or that maybe I missed something. Let me try that.

So, starting with t² + 2t - 1 ≡ 0 mod 23. Let's rewrite that as t² + 2t ≡ 1 mod 23. To solve quadratic congruences, sometimes completing the square helps.

So, t² + 2t ≡ 1 mod 23. Let me complete the square. The coefficient of t is 2, so half of that is 1, square is 1. Therefore, add 1 to both sides:

t² + 2t + 1 ≡ 1 + 1 mod 23 => (t + 1)^2 ≡ 2 mod 23.

So, now, we have (t + 1)^2 ≡ 2 mod 23. Therefore, solving for t, we need to find t such that t + 1 is a square root of 2 modulo 23. So first, find x such that x² ≡ 2 mod 23. Then t + 1 ≡ x mod 23, so t ≡ x - 1 mod 23.

So, first step: find square roots of 2 mod 23.

How do I find square roots modulo a prime? Well, since 23 is a prime, we can use Tonelli-Shanks or other methods, but maybe here we can just try numbers.

Let me check squares mod 23:

Compute x² mod 23 for x from 0 to 11 (since after that it repeats):

0²=0

1=1

2=4

3=9

4=16

5=25 mod23=2

Oh! x=5: 5²=25≡2 mod23. So 5 is a square root. Then also -5 mod23 is 23-5=18. So 18² ≡ (-5)²=25≡2 mod23. Therefore, the square roots of 2 mod23 are 5 and 18.

Therefore, x ≡ 5 or 18 mod23. Therefore, going back:

t + 1 ≡ 5 mod23 ⇒ t ≡ 5 -1 = 4 mod23

or t +1 ≡ 18 mod23 ⇒ t ≡ 18 -1 =17 mod23.

So, solutions are t ≡4 or 17 mod23. But the problem says that t is an odd integer between 0 and 23. So, 4 is even, so we discard it. 17 is odd, so that's the solution. Therefore, t=17. Which matches our previous result.

So that confirms that t=17 is the unique solution. Since 4 is even, and the problem says "unique odd integer t", so 17 is the answer.

Therefore, the answer is 17.

But let me double-check once more to be safe. So, t=17. Let me check if 17 + 2 = 19 is the inverse of 17 mod23. So, does 17*19 ≡1 mod23?

Yes, as we saw earlier, 17*19=323, 323 divided by 23 is 14*23=322, remainder 1. Therefore, 17*19 ≡1 mod23. Therefore, 19 is the inverse of 17 mod23. So that works.

Also, just to check that there's no other odd solution. The other solution is t=4. Which is even. So since 4 is even, we can ignore it. Hence, t=17 is unique odd solution.

Therefore, I think that is the answer. Seems solid. Both methods give the same result, and checking each value also gives the same answer. So, confident that 17 is correct.

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

To solve the problem, we are given that $ t $ is an **odd** integer satisfying:

$$
0 < t < 23
$$
and
$$
t + 2 \text{ is the inverse of } t \text{ modulo } 23.
$$

---

### Step 1: Understand the Condition

In modular arithmetic, the **inverse** of $ t $ modulo $ 23 $ is a number $ x $ such that:

$$
t \cdot x \equiv 1 \pmod{23}
$$

So, the condition tells us:

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

Expanding this gives:

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

Or:

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

---

### Step 2: Solve the Congruence

We rewrite the quadratic congruence:

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

To complete the square, we add 1 to both sides:

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

So:

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

Now, we must find values $ x $ such that:

$$
x^2 \equiv 2 \pmod{23}
$$

Trying small integers modulo 23, we find:

- $ 5^2 = 25 \equiv 2 \pmod{23} $
- So, $ x \equiv 5 \pmod{23} $

Also, since $ (-5)^2 = 25 \equiv 2 \pmod{23} $, then:

- $ x \equiv -5 \equiv 18 \pmod{23} $

Therefore, the square roots of 2 modulo 23 are $ 5 $ and $ 18 $. So:

- $ t + 1 \equiv 5 \pmod{23} \Rightarrow t \equiv 4 $
- $ t + 1 \equiv 18 \pmod{23} \Rightarrow t \equiv 17 $

---

### Step 3: Check the Conditions

We are told that $ t $ must be **odd** and between 0 and 23. From our results:

- $ t = 4 \Rightarrow \text{even} $ → reject
- $ t = 17 \Rightarrow \text{odd} $ and within range → accept

Now, verify that $ t + 2 = 19 $ is the inverse of $ t = 17 $ modulo 23:

$$
17 \cdot 19 = 323
$$
$$
323 \mod 23 = 1
$$

Since $ 323 \equiv 1 \mod 23 $, it confirms that $ t = 17 $ is indeed the unique value that satisfies the condition.

---

### Final Answer

$$
\boxed{17}
$$

[Assistant]
Explanation:
Okay, so I need to find an odd integer t between 0 and 23 (so t is 1, 3, 5,...,21) such that t + 2 is the inverse of t modulo 23. Let me recall what an inverse means here.

In modular arithmetic, the inverse of a number a modulo m is a number b such that (a * b) ≡ 1 mod m. So, if t + 2 is the inverse of t modulo 23, then multiplying t by (t + 2) should give me 1 modulo 23. So, mathematically, this means:

t * (t + 2) ≡ 1 mod 23

So, expanding that equation:

t² + 2t ≡ 1 mod 23

Which can be rewritten as:

t² + 2t - 1 ≡ 0 mod 23

So, I need to solve this quadratic congruence equation for t. Since we're working modulo 23, which is a prime number, maybe I can use some methods from number theory to solve this?

Alternatively, maybe I can just try plugging in the possible odd values of t between 1 and 21 and check which one satisfies the condition. Since there aren't that many possibilities, maybe that's feasible. Let me try that approach first, just to see if I can find t without too much trouble.

First, let me list all odd integers between 1 and 21:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21.

Now, for each of these t, I need to compute t * (t + 2) mod 23 and check if it equals 1. Let's start:

Starting with t = 1:

1 * (1 + 2) = 1 * 3 = 3 mod 23. 3 ≠ 1. Not good.

t = 3:

3 * (3 + 2) = 3 * 5 = 15 mod 23. 15 ≠ 1.

t = 5:

5 * 7 = 35. 35 mod 23 is 35 - 23 = 12. 12 ≠ 1.

t = 7:

7 * 9 = 63. 63 divided by 23: 23*2=46, 63 - 46=17. 17 ≠ 1.

t = 9:

9 * 11 = 99. Let's divide 99 by 23. 23*4=92, 99 - 92 = 7. 7 ≠ 1.

t = 11:

11 * 13 = 143. Let's divide by 23. 23*6 = 138, 143 - 138 = 5. 5 ≠ 1.

t = 13:

13 * 15 = 195. Let me divide by 23. 23*8 = 184, 195 - 184 = 11. 11 ≠ 1.

t = 15:

15 * 17 = 255. Dividing by 23. Let's see: 23*11 = 253. 255 - 253 = 2. 2 ≠ 1.

t = 17:

17 * 19 = 323. Dividing by 23: Let me compute 23*14 = 322. 323 - 322 = 1. Oh! That's 1 mod 23! So t=17 gives us 1. Wait, is that right?

Wait, let me double-check that calculation. 17 * 19 is indeed 323. Then 23*14 is 322, so 323 - 322 = 1. Therefore, 323 ≡ 1 mod 23. Therefore, t=17 satisfies the equation. But let me check if that's the only one. Let me continue checking the rest to see if there's another one.

t = 19:

19 * (19 + 2) = 19 * 21. Let's compute that. 19*20=380, plus 19 is 399. Now, 399 divided by 23. Let me calculate 23*17= 391. 399 - 391=8. 8 ≠ 1.

t = 21:

21 * (21 + 2) = 21 * 23. Well, 21*23 mod 23 is 0, since 23 is a multiple of 23. So 0 ≠ 1. So, nope.

So, the only one that works is t=17. And since the question says "unique odd integer t", so that's probably the answer. But let me just make sure that I didn't make any mistakes in my calculations for t=17.

Wait, t=17 is odd, and between 0 and 23. Also, t + 2 is 19. So, does 17 * 19 ≡ 1 mod 23? As above, 17*19=323. Let me compute 23*14=322, so 323 - 322=1. Yes, exactly. So that works. And since we checked all other values, and only t=17 worked, so that's the answer.

But just to be thorough, maybe I should also try solving the quadratic equation algebraically, just to confirm that there isn't another solution or that maybe I missed something. Let me try that.

So, starting with t² + 2t - 1 ≡ 0 mod 23. Let's rewrite that as t² + 2t ≡ 1 mod 23. To solve quadratic congruences, sometimes completing the square helps.

So, t² + 2t ≡ 1 mod 23. Let me complete the square. The coefficient of t is 2, so half of that is 1, square is 1. Therefore, add 1 to both sides:

t² + 2t + 1 ≡ 1 + 1 mod 23 => (t + 1)^2 ≡ 2 mod 23.

So, now, we have (t + 1)^2 ≡ 2 mod 23. Therefore, solving for t, we need to find t such that t + 1 is a square root of 2 modulo 23. So first, find x such that x² ≡ 2 mod 23. Then t + 1 ≡ x mod 23, so t ≡ x - 1 mod 23.

So, first step: find square roots of 2 mod 23.

How do I find square roots modulo a prime? Well, since 23 is a prime, we can use Tonelli-Shanks or other methods, but maybe here we can just try numbers.

Let me check squares mod 23:

Compute x² mod 23 for x from 0 to 11 (since after that it repeats):

0²=0

1=1

2=4

3=9

4=16

5=25 mod23=2

Oh! x=5: 5²=25≡2 mod23. So 5 is a square root. Then also -5 mod23 is 23-5=18. So 18² ≡ (-5)²=25≡2 mod23. Therefore, the square roots of 2 mod23 are 5 and 18.

Therefore, x ≡ 5 or 18 mod23. Therefore, going back:

t + 1 ≡ 5 mod23 ⇒ t ≡ 5 -1 = 4 mod23

or t +1 ≡ 18 mod23 ⇒ t ≡ 18 -1 =17 mod23.

So, solutions are t ≡4 or 17 mod23. But the problem says that t is an odd integer between 0 and 23. So, 4 is even, so we discard it. 17 is odd, so that's the solution. Therefore, t=17. Which matches our previous result.

So that confirms that t=17 is the unique solution. Since 4 is even, and the problem says "unique odd integer t", so 17 is the answer.

Therefore, the answer is 17.

But let me double-check once more to be safe. So, t=17. Let me check if 17 + 2 = 19 is the inverse of 17 mod23. So, does 17*19 ≡1 mod23?

Yes, as we saw earlier, 17*19=323, 323 divided by 23 is 14*23=322, remainder 1. Therefore, 17*19 ≡1 mod23. Therefore, 19 is the inverse of 17 mod23. So that works.

Also, just to check that there's no other odd solution. The other solution is t=4. Which is even. So since 4 is even, we can ignore it. Hence, t=17 is unique odd solution.

Therefore, I think that is the answer. Seems solid. Both methods give the same result, and checking each value also gives the same answer. So, confident that 17 is correct.

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

To solve the problem, we are given that $ t $ is an **odd** integer satisfying:

$$
0 < t < 23
$$
and
$$
t + 2 \text{ is the inverse of } t \text{ modulo } 23.
$$

---

### Step 1: Understand the Condition

In modular arithmetic, the **inverse** of $ t $ modulo $ 23 $ is a number $ x $ such that:

$$
t \cdot x \equiv 1 \pmod{23}
$$

So, the condition tells us:

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

Expanding this gives:

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

Or:

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

---

### Step 2: Solve the Congruence

We rewrite the quadratic congruence:

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

To complete the square, we add 1 to both sides:

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

So:

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

Now, we must find values $ x $ such that:

$$
x^2 \equiv 2 \pmod{23}
$$

Trying small integers modulo 23, we find:

- $ 5^2 = 25 \equiv 2 \pmod{23} $
- So, $ x \equiv 5 \pmod{23} $

Also, since $ (-5)^2 = 25 \equiv 2 \pmod{23} $, then:

- $ x \equiv -5 \equiv 18 \pmod{23} $

Therefore, the square roots of 2 modulo 23 are $ 5 $ and $ 18 $. So:

- $ t + 1 \equiv 5 \pmod{23} \Rightarrow t \equiv 4 $
- $ t + 1 \equiv 18 \pmod{23} \Rightarrow t \equiv 17 $

---

### Step 3: Check the Conditions

We are told that $ t $ must be **odd** and between 0 and 23. From our results:

- $ t = 4 \Rightarrow \text{even} $ → reject
- $ t = 17 \Rightarrow \text{odd} $ and within range → accept

Now, verify that $ t + 2 = 19 $ is the inverse of $ t = 17 $ modulo 23:

$$
17 \cdot 19 = 323
$$
$$
323 \mod 23 = 1
$$

Since $ 323 \equiv 1 \mod 23 $, it confirms that $ t = 17 $ is indeed the unique value that satisfies the condition.

---

### 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 to get a + b + m. Let me think about how to approach this step by step.

First, I remember that the x-intercept of a line is the point where the line crosses the x-axis. At that point, the y-coordinate is zero. Similarly, the y-intercept is where the line crosses the y-axis, so the x-coordinate is zero there. The slope is the steepness of the line, which can be found if I can rewrite the equation into slope-intercept form (y = mx + b).

But let me start with the intercepts. The given equation is (x/4) + (y/12) = 1. To find the x-intercept, I can set y = 0 and solve for x. Let me do that.

Setting y = 0:

x/4 + 0/12 = 1 → x/4 = 1 → x = 4. So the x-intercept is at (4, 0), so a = 4. Got that.

Similarly, to find the y-intercept, set x = 0:

0/4 + y/12 = 1 → y/12 = 1 → y = 12. Therefore, the y-intercept is at (0, 12), so b = 12. Okay, so far so good.

Now, the slope. Hmm. The original equation is in the form of intercepts. Maybe I can convert it into slope-intercept form. Let me recall that the standard form of a line is usually written as Ax + By = C, and slope-intercept form is y = mx + b. Alternatively, there's also the intercept form, which is x/a + y/b = 1, where a is the x-intercept and b is the y-intercept. Wait, isn't that exactly the form we have here? The equation given is (x/4) + (y/12) = 1. So comparing to x/a + y/b = 1, that would mean a = 4 and b = 12, which matches what I found earlier. So that's consistent.

But how do I get the slope from this? Well, if I want to find the slope, maybe I can rearrange the equation into slope-intercept form. Let me try that.

Starting with (x/4) + (y/12) = 1. Let me subtract (x/4) from both sides:

(y/12) = - (x/4) + 1

Now multiply both sides by 12 to solve for y:

y = -3x + 12

Wait, let me check that:

Multiplying -x/4 by 12: (-x/4)*12 = -3x. And 1*12 = 12. Yes, so y = -3x + 12. Therefore, the slope m is -3, and the y-intercept is 12. Which matches the previous value of b, which is 12. That makes sense because in the slope-intercept form, the y-intercept is b, so that's consistent.

So then, the slope m is -3. Therefore, now I can find a + b + m. Let me recall:

a = 4

b = 12

m = -3

Adding them together: 4 + 12 + (-3) = 4 + 12 - 3. Let me compute that: 4 + 12 is 16, 16 - 3 is 13. So the total is 13.

Wait, that seems straightforward. But let me verify to make sure I didn't make a mistake in finding the slope. Another way to find the slope is using the intercepts. Since we have two points on the line: the x-intercept (4, 0) and the y-intercept (0, 12). The slope between these two points can be calculated as (y2 - y1)/(x2 - x1).

So taking (x1, y1) as (4, 0) and (x2, y2) as (0, 12). Then slope m = (12 - 0)/(0 - 4) = 12 / (-4) = -3. Yep, that's the same as before. So slope is indeed -3. Therefore, a + b + m = 4 + 12 + (-3) = 13. That seems correct.

But wait, let me check if I had miscalculated anything. Let me go through each step again.

Original equation: x/4 + y/12 = 1. For the x-intercept, set y = 0: x/4 = 1 ⇒ x = 4. Correct. For y-intercept, set x = 0: y/12 = 1 ⇒ y = 12. Correct. Then converting to slope-intercept: subtract x/4: y/12 = -x/4 + 1. Multiply both sides by 12: y = -3x + 12. So slope is -3. Then adding a + b + m: 4 + 12 + (-3) = 13. All steps check out.

Alternatively, maybe I can write the equation in the form y = mx + b to confirm again. Let me do that once more:

Starting with x/4 + y/12 = 1. Multiply both sides by 12 to eliminate denominators:

12*(x/4) + 12*(y/12) = 12*1 ⇒ 3x + y = 12. Then rearrange to solve for y:

y = -3x + 12. Yep, same result. So slope is -3. So same conclusion.

Another way to think about the intercept form: intercept form is x/a + y/b = 1, so slope can be calculated as -b/a. Wait, let me recall. If I have x/a + y/b = 1, then solving for y gives y = (-b/a)x + b. So the slope is -b/a. Therefore, in this case, a is 4, b is 12, so slope is -12/4 = -3. Exactly. So that formula gives the slope as -b/a. So that's another way to find m without going through the entire process. So that formula is useful.

Therefore, using that formula, m = -b/a = -12/4 = -3. Therefore, same result. So a is 4, b is 12, m is -3. Sum is 4 + 12 + (-3) = 13. So I think this answer is correct.

But just to be thorough, let me check using another method. Let's take the general form of a line: Ax + By + C = 0. The slope is -A/B. Wait, but the given equation is x/4 + y/12 = 1. Let me write it in Ax + By + C = 0 form. Subtract 1 from both sides: x/4 + y/12 - 1 = 0. Multiply both sides by 12 to eliminate denominators: 3x + y - 12 = 0. So that's 3x + y - 12 = 0. Comparing to Ax + By + C = 0, A = 3, B = 1, so slope is -A/B = -3/1 = -3. Yep, same result. So that's another confirmation.

So, all methods lead to the same conclusion. Therefore, the value of a + b + m is 4 + 12 + (-3) = 13. Therefore, the answer is 13.

But just to make sure, let me think if there's any chance I mixed up a and b? Wait, in the intercept form x/a + y/b = 1, so x-intercept is a, y-intercept is b. So the given equation is x/4 + y/12 = 1, so yes, a is 4, b is 12. So no mix-up there. Therefore, adding them up with m = -3 is correct.

Alternatively, if someone thought that the slope was positive 3, but according to all calculations, it's negative. Let me double-check the slope between (4, 0) and (0, 12). So change in y is 12 - 0 = 12. Change in x is 0 - 4 = -4. Therefore, slope is 12 / (-4) = -3. So positive over negative is negative. So slope is negative, which matches. So that's correct.

So, all steps check out. Therefore, I think I can be confident that a + b + m = 13.

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

To solve the problem, we are given a linear equation:

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

We are asked to:

- Find the **x-intercept** $a$,
- Find the **y-intercept** $b$,
- Find the **slope** $m$, and
- Compute $a + b + m$.

---

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

The x-intercept occurs where $y = 0$. Substitute $y = 0$ 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: Find the y-intercept $b$

The y-intercept occurs when $x = 0$. Substitute $x = 0$ 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 $m$

The given equation is in **intercept form**:

$$
\frac{x}{a} + \frac{y}{b} = 1
$$

This is equivalent to:

$$
x/4 + y/12 = 1
$$

We recognize this form, where:
- $a = 4$ is the x-intercept,
- $b = 12$ is the y-intercept,
- The slope $m$ can be derived from the relationship $m = -\frac{b}{a} = -\frac{12}{4} = -3$.

---

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

Now sum up the values:

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

---

### 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 to get a + b + m. Let me think about how to approach this step by step.

First, I remember that the x-intercept of a line is the point where the line crosses the x-axis. At that point, the y-coordinate is zero. Similarly, the y-intercept is where the line crosses the y-axis, so the x-coordinate is zero there. The slope is the steepness of the line, which can be found if I can rewrite the equation into slope-intercept form (y = mx + b).

But let me start with the intercepts. The given equation is (x/4) + (y/12) = 1. To find the x-intercept, I can set y = 0 and solve for x. Let me do that.

Setting y = 0:

x/4 + 0/12 = 1 → x/4 = 1 → x = 4. So the x-intercept is at (4, 0), so a = 4. Got that.

Similarly, to find the y-intercept, set x = 0:

0/4 + y/12 = 1 → y/12 = 1 → y = 12. Therefore, the y-intercept is at (0, 12), so b = 12. Okay, so far so good.

Now, the slope. Hmm. The original equation is in the form of intercepts. Maybe I can convert it into slope-intercept form. Let me recall that the standard form of a line is usually written as Ax + By = C, and slope-intercept form is y = mx + b. Alternatively, there's also the intercept form, which is x/a + y/b = 1, where a is the x-intercept and b is the y-intercept. Wait, isn't that exactly the form we have here? The equation given is (x/4) + (y/12) = 1. So comparing to x/a + y/b = 1, that would mean a = 4 and b = 12, which matches what I found earlier. So that's consistent.

But how do I get the slope from this? Well, if I want to find the slope, maybe I can rearrange the equation into slope-intercept form. Let me try that.

Starting with (x/4) + (y/12) = 1. Let me subtract (x/4) from both sides:

(y/12) = - (x/4) + 1

Now multiply both sides by 12 to solve for y:

y = -3x + 12

Wait, let me check that:

Multiplying -x/4 by 12: (-x/4)*12 = -3x. And 1*12 = 12. Yes, so y = -3x + 12. Therefore, the slope m is -3, and the y-intercept is 12. Which matches the previous value of b, which is 12. That makes sense because in the slope-intercept form, the y-intercept is b, so that's consistent.

So then, the slope m is -3. Therefore, now I can find a + b + m. Let me recall:

a = 4

b = 12

m = -3

Adding them together: 4 + 12 + (-3) = 4 + 12 - 3. Let me compute that: 4 + 12 is 16, 16 - 3 is 13. So the total is 13.

Wait, that seems straightforward. But let me verify to make sure I didn't make a mistake in finding the slope. Another way to find the slope is using the intercepts. Since we have two points on the line: the x-intercept (4, 0) and the y-intercept (0, 12). The slope between these two points can be calculated as (y2 - y1)/(x2 - x1).

So taking (x1, y1) as (4, 0) and (x2, y2) as (0, 12). Then slope m = (12 - 0)/(0 - 4) = 12 / (-4) = -3. Yep, that's the same as before. So slope is indeed -3. Therefore, a + b + m = 4 + 12 + (-3) = 13. That seems correct.

But wait, let me check if I had miscalculated anything. Let me go through each step again.

Original equation: x/4 + y/12 = 1. For the x-intercept, set y = 0: x/4 = 1 ⇒ x = 4. Correct. For y-intercept, set x = 0: y/12 = 1 ⇒ y = 12. Correct. Then converting to slope-intercept: subtract x/4: y/12 = -x/4 + 1. Multiply both sides by 12: y = -3x + 12. So slope is -3. Then adding a + b + m: 4 + 12 + (-3) = 13. All steps check out.

Alternatively, maybe I can write the equation in the form y = mx + b to confirm again. Let me do that once more:

Starting with x/4 + y/12 = 1. Multiply both sides by 12 to eliminate denominators:

12*(x/4) + 12*(y/12) = 12*1 ⇒ 3x + y = 12. Then rearrange to solve for y:

y = -3x + 12. Yep, same result. So slope is -3. So same conclusion.

Another way to think about the intercept form: intercept form is x/a + y/b = 1, so slope can be calculated as -b/a. Wait, let me recall. If I have x/a + y/b = 1, then solving for y gives y = (-b/a)x + b. So the slope is -b/a. Therefore, in this case, a is 4, b is 12, so slope is -12/4 = -3. Exactly. So that formula gives the slope as -b/a. So that's another way to find m without going through the entire process. So that formula is useful.

Therefore, using that formula, m = -b/a = -12/4 = -3. Therefore, same result. So a is 4, b is 12, m is -3. Sum is 4 + 12 + (-3) = 13. So I think this answer is correct.

But just to be thorough, let me check using another method. Let's take the general form of a line: Ax + By + C = 0. The slope is -A/B. Wait, but the given equation is x/4 + y/12 = 1. Let me write it in Ax + By + C = 0 form. Subtract 1 from both sides: x/4 + y/12 - 1 = 0. Multiply both sides by 12 to eliminate denominators: 3x + y - 12 = 0. So that's 3x + y - 12 = 0. Comparing to Ax + By + C = 0, A = 3, B = 1, so slope is -A/B = -3/1 = -3. Yep, same result. So that's another confirmation.

So, all methods lead to the same conclusion. Therefore, the value of a + b + m is 4 + 12 + (-3) = 13. Therefore, the answer is 13.

But just to make sure, let me think if there's any chance I mixed up a and b? Wait, in the intercept form x/a + y/b = 1, so x-intercept is a, y-intercept is b. So the given equation is x/4 + y/12 = 1, so yes, a is 4, b is 12. So no mix-up there. Therefore, adding them up with m = -3 is correct.

Alternatively, if someone thought that the slope was positive 3, but according to all calculations, it's negative. Let me double-check the slope between (4, 0) and (0, 12). So change in y is 12 - 0 = 12. Change in x is 0 - 4 = -4. Therefore, slope is 12 / (-4) = -3. So positive over negative is negative. So slope is negative, which matches. So that's correct.

So, all steps check out. Therefore, I think I can be confident that a + b + m = 13.

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

To solve the problem, we are given a linear equation:

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

We are asked to:

- Find the **x-intercept** $a$,
- Find the **y-intercept** $b$,
- Find the **slope** $m$, and
- Compute $a + b + m$.

---

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

The x-intercept occurs where $y = 0$. Substitute $y = 0$ 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: Find the y-intercept $b$

The y-intercept occurs when $x = 0$. Substitute $x = 0$ 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 $m$

The given equation is in **intercept form**:

$$
\frac{x}{a} + \frac{y}{b} = 1
$$

This is equivalent to:

$$
x/4 + y/12 = 1
$$

We recognize this form, where:
- $a = 4$ is the x-intercept,
- $b = 12$ is the y-intercept,
- The slope $m$ can be derived from the relationship $m = -\frac{b}{a} = -\frac{12}{4} = -3$.

---

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

Now sum up the values:

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

---

### 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 about how to approach this.

First, I remember that the volume of a cube is calculated by cubing the length of one of its sides. The formula for volume (V) is V = s³, where s is the side length. Since the volume is given as 1 cubic foot, I can set up the equation:

s³ = 1 cubic foot.

To find the side length, I need to take the cube root of both sides. The cube root of 1 is 1, so s = 1 foot. Wait, so each side of the cube is 1 foot long? That seems straightforward. But the question asks for the surface area in square inches. Oh, right! I need to convert the units from 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 indeed 1 foot. Now, converting feet to inches. I know that 1 foot is equal to 12 inches. Therefore, each side of the cube is 12 inches long.

Now, moving on to the surface area. The total surface area (SA) of a cube is calculated by the formula SA = 6s², where s is the side length. Since we have the side length in inches now, we can plug that into the formula.

So, substituting s = 12 inches into the formula:

SA = 6 * (12 inches)².

Let me calculate that. First, square 12 inches: 12 squared is 144. Then multiply by 6: 6 * 144. Hmm, 6 times 144... Let me do that step by step.

Well, 144 times 6. Let me break it down: 100 * 6 = 600, 40 * 6 = 240, and 4 * 6 = 24. Adding those up: 600 + 240 is 840, plus 24 is 864. So, 6 * 144 = 864. Therefore, the total surface area is 864 square inches.

Wait, let me double-check my calculations to make sure I didn't make any mistakes. So, starting again, if the side length is 1 foot, which is 12 inches, then each face of the cube is 12 inches by 12 inches. The area of one face is 12*12 = 144 square inches. Since a cube has 6 faces, multiplying 144 by 6 gives 864. Yep, that seems right.

Alternatively, maybe I can approach this using unit conversions earlier on? Let me try that method to cross-verify.

Starting from volume: 1 cubic foot. Since 1 foot is 12 inches, then 1 cubic foot is equal to (12 inches)^3. Let me calculate that. 12 inches * 12 inches * 12 inches. Which is 12³ = 1728 cubic inches. Wait, but the volume is given as 1 cubic foot, which is 1728 cubic inches. However, the volume of the cube is also equal to s³, where s is in inches. Therefore, s³ = 1728 cubic inches.

Wait, hold on. This seems conflicting with my previous calculation. Earlier, I thought that since the volume is 1 cubic foot, then s = 1 foot, which is 12 inches. But if I convert the volume to cubic inches first, then s³ = 1728 cubic inches. Therefore, solving for s, we have s = cube root of 1728. What is the cube root of 1728? Let me recall. 12 cubed is 12*12*12 = 144*12. Wait, 144*12 is 1728. Yes, so the cube root of 1728 is 12. So, s = 12 inches. Which matches my previous result.

Therefore, regardless of whether I convert the volume to cubic inches first or convert the side length first, I end up with the side length of 12 inches. So then, the surface area is 6*(12)^2 = 864 square inches. So that's consistent.

Wait, but why did I think there was a conflict? Because I initially thought of the volume as 1 cubic foot, so s is 1 foot, but if I convert the volume to cubic inches, it's 1728, so s is cube root of 1728 which is 12 inches. So, both approaches are consistent.

Therefore, the total surface area is 864 square inches.

But let me check another way. Maybe by using unit conversion for the surface area.

Alternatively, if I didn't convert the side length, but instead tried to compute the surface area in square feet and then convert that to square inches. Let's try that.

The surface area in square feet would be 6*(1 foot)^2 = 6 square feet. Then, converting square feet to square inches. Since 1 square foot is 12 inches * 12 inches = 144 square inches. Therefore, 6 square feet is 6 * 144 = 864 square inches. Again, same answer.

So, all methods converge to 864. Therefore, that must be the correct answer.

But just to make sure, let me think if there is any chance I made a unit conversion mistake. For instance, sometimes when converting units, there's a possibility of mixing up linear vs. area vs. volume conversions. But in this case, since I converted the side length from feet to inches, then calculated the area, or converted the volume to cubic inches then found the side length, or converted surface area from square feet to square inches. All roads lead to 864.

Alternatively, if someone mistakenly thought that since the volume is 1 cubic foot, the surface area is 6 square feet, and then thought to convert that to square inches by multiplying by 12, which would be wrong, because square feet to square inches is multiplying by 144, not 12. But if I did that mistake, I would get 6*12=72 square inches, which is clearly wrong. But since I know that 1 square foot is 144 square inches, then 6 square feet is 864 square inches.

So, the key point here is that when converting areas, you have to square the conversion factor. Similarly, for volumes, you cube it. So, just to recall: 1 foot = 12 inches, so 1 square foot = 12^2 square inches = 144, and 1 cubic foot = 12^3 cubic inches = 1728.

Therefore, in this problem, since the volume is given in cubic feet, but the answer needs to be in square inches, we need to convert the side length to inches first, then calculate the surface area.

Alternatively, we could calculate the surface area in square feet and then convert to square inches. Either way, as shown, the answer is 864.

So, I think I've verified this multiple times. Therefore, 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 in a structured, step-by-step manner.

---

### Step 1: Understand the Volume Formula for a Cube

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

$$
V = s^3
$$

Where $ s $ is the length of one of the cube's sides.

Given:
$$
V = 1 \text{ cubic foot}
$$

Solving for $ s $:
$$
s^3 = 1 \implies s = \sqrt[3]{1} = 1 \text{ foot}
$$

So, each edge of the cube is **1 foot** long.

---

### Step 2: Convert Side Length to Inches

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

$$
1 \text{ foot} = 12 \text{ inches}
$$

$$
s = 1 \text{ foot} = 12 \text{ inches}
$$

---

### Step 3: Use the Surface Area Formula for a Cube

The **total surface area (SA)** of a cube is given by:

$$
\text{SA} = 6s^2
$$

Substituting $ s = 12 $ inches:

$$
\text{SA} = 6 \times (12)^2 = 6 \times 144 = 864 \text{ square inches}
$$

---

### Step 4: Final Answer

Thus, the **total surface area of the cube** is:

$$
\boxed{864}
$$

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 about how to approach this.

First, I remember that the volume of a cube is calculated by cubing the length of one of its sides. The formula for volume (V) is V = s³, where s is the side length. Since the volume is given as 1 cubic foot, I can set up the equation:

s³ = 1 cubic foot.

To find the side length, I need to take the cube root of both sides. The cube root of 1 is 1, so s = 1 foot. Wait, so each side of the cube is 1 foot long? That seems straightforward. But the question asks for the surface area in square inches. Oh, right! I need to convert the units from 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 indeed 1 foot. Now, converting feet to inches. I know that 1 foot is equal to 12 inches. Therefore, each side of the cube is 12 inches long.

Now, moving on to the surface area. The total surface area (SA) of a cube is calculated by the formula SA = 6s², where s is the side length. Since we have the side length in inches now, we can plug that into the formula.

So, substituting s = 12 inches into the formula:

SA = 6 * (12 inches)².

Let me calculate that. First, square 12 inches: 12 squared is 144. Then multiply by 6: 6 * 144. Hmm, 6 times 144... Let me do that step by step.

Well, 144 times 6. Let me break it down: 100 * 6 = 600, 40 * 6 = 240, and 4 * 6 = 24. Adding those up: 600 + 240 is 840, plus 24 is 864. So, 6 * 144 = 864. Therefore, the total surface area is 864 square inches.

Wait, let me double-check my calculations to make sure I didn't make any mistakes. So, starting again, if the side length is 1 foot, which is 12 inches, then each face of the cube is 12 inches by 12 inches. The area of one face is 12*12 = 144 square inches. Since a cube has 6 faces, multiplying 144 by 6 gives 864. Yep, that seems right.

Alternatively, maybe I can approach this using unit conversions earlier on? Let me try that method to cross-verify.

Starting from volume: 1 cubic foot. Since 1 foot is 12 inches, then 1 cubic foot is equal to (12 inches)^3. Let me calculate that. 12 inches * 12 inches * 12 inches. Which is 12³ = 1728 cubic inches. Wait, but the volume is given as 1 cubic foot, which is 1728 cubic inches. However, the volume of the cube is also equal to s³, where s is in inches. Therefore, s³ = 1728 cubic inches.

Wait, hold on. This seems conflicting with my previous calculation. Earlier, I thought that since the volume is 1 cubic foot, then s = 1 foot, which is 12 inches. But if I convert the volume to cubic inches first, then s³ = 1728 cubic inches. Therefore, solving for s, we have s = cube root of 1728. What is the cube root of 1728? Let me recall. 12 cubed is 12*12*12 = 144*12. Wait, 144*12 is 1728. Yes, so the cube root of 1728 is 12. So, s = 12 inches. Which matches my previous result.

Therefore, regardless of whether I convert the volume to cubic inches first or convert the side length first, I end up with the side length of 12 inches. So then, the surface area is 6*(12)^2 = 864 square inches. So that's consistent.

Wait, but why did I think there was a conflict? Because I initially thought of the volume as 1 cubic foot, so s is 1 foot, but if I convert the volume to cubic inches, it's 1728, so s is cube root of 1728 which is 12 inches. So, both approaches are consistent.

Therefore, the total surface area is 864 square inches.

But let me check another way. Maybe by using unit conversion for the surface area.

Alternatively, if I didn't convert the side length, but instead tried to compute the surface area in square feet and then convert that to square inches. Let's try that.

The surface area in square feet would be 6*(1 foot)^2 = 6 square feet. Then, converting square feet to square inches. Since 1 square foot is 12 inches * 12 inches = 144 square inches. Therefore, 6 square feet is 6 * 144 = 864 square inches. Again, same answer.

So, all methods converge to 864. Therefore, that must be the correct answer.

But just to make sure, let me think if there is any chance I made a unit conversion mistake. For instance, sometimes when converting units, there's a possibility of mixing up linear vs. area vs. volume conversions. But in this case, since I converted the side length from feet to inches, then calculated the area, or converted the volume to cubic inches then found the side length, or converted surface area from square feet to square inches. All roads lead to 864.

Alternatively, if someone mistakenly thought that since the volume is 1 cubic foot, the surface area is 6 square feet, and then thought to convert that to square inches by multiplying by 12, which would be wrong, because square feet to square inches is multiplying by 144, not 12. But if I did that mistake, I would get 6*12=72 square inches, which is clearly wrong. But since I know that 1 square foot is 144 square inches, then 6 square feet is 864 square inches.

So, the key point here is that when converting areas, you have to square the conversion factor. Similarly, for volumes, you cube it. So, just to recall: 1 foot = 12 inches, so 1 square foot = 12^2 square inches = 144, and 1 cubic foot = 12^3 cubic inches = 1728.

Therefore, in this problem, since the volume is given in cubic feet, but the answer needs to be in square inches, we need to convert the side length to inches first, then calculate the surface area.

Alternatively, we could calculate the surface area in square feet and then convert to square inches. Either way, as shown, the answer is 864.

So, I think I've verified this multiple times. Therefore, 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 in a structured, step-by-step manner.

---

### Step 1: Understand the Volume Formula for a Cube

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

$$
V = s^3
$$

Where $ s $ is the length of one of the cube's sides.

Given:
$$
V = 1 \text{ cubic foot}
$$

Solving for $ s $:
$$
s^3 = 1 \implies s = \sqrt[3]{1} = 1 \text{ foot}
$$

So, each edge of the cube is **1 foot** long.

---

### Step 2: Convert Side Length to Inches

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

$$
1 \text{ foot} = 12 \text{ inches}
$$

$$
s = 1 \text{ foot} = 12 \text{ inches}
$$

---

### Step 3: Use the Surface Area Formula for a Cube

The **total surface area (SA)** of a cube is given by:

$$
\text{SA} = 6s^2
$$

Substituting $ s = 12 $ inches:

$$
\text{SA} = 6 \times (12)^2 = 6 \times 144 = 864 \text{ square inches}
$$

---

### Step 4: Final Answer

Thus, the **total surface area of the cube** is:

$$
\boxed{864}
$$

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 $\boxed{864 \mbox{ inches}^2}$