<Question>
Let $f(x) = x^{10}+5x^9-8x^8+7x^7-x^6-12x^5+4x^4-8x^3+12x^2-5x-5$.

Without using long division (which would be horribly nasty!), find the remainder when $f(x)$ is divided by $x^2-1$.
</Question>

<think>
<Plan_1>
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 because it would be so nasty. Hmm, right, long division would take forever with such a high-degree polynomial. Let me think of another way.

Wait, I remember something about the Remainder Theorem. But the standard Remainder Theorem applies when dividing by linear polynomials, right? Like (x - a). But here we're dividing by a quadratic, x² - 1. Maybe there's a generalized version of the Remainder Theorem for higher degrees?

Let me recall. If we divide a polynomial f(x) by another polynomial g(x), the remainder has a degree less than the degree of g(x). Since we're dividing by x² - 1, which is degree 2, the remainder should be a first-degree polynomial, so something like ax + b. So the remainder will be R(x) = ax + b, and we need to find a and b.

But how to find a and b without long division? Maybe by using the roots of the divisor polynomial? Because if x² - 1 = 0, then x = 1 or x = -1. So, if we plug x = 1 and x = -1 into f(x), we can get two equations to solve for a and b.

Let me check this idea. If we evaluate f(1), that should equal R(1), since when x = 1, the divisor x² - 1 is zero, so the division equation becomes f(1) = R(1). Similarly, f(-1) = R(-1).
Then we can set up two equations:

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

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

Then solve for a and b. That sounds manageable. Let me try that.

First, compute f(1).

<prompt_1.1> Compute f(1) by substituting x = 1 into the polynomial f(x). </prompt_1.1>

Now compute f(-1).

<prompt_1.2> Compute f(-1) by substituting x = -1 into the polynomial f(x). </prompt_1.2>
-----

Is there another method to verify? Maybe using polynomial interpolation. Since the remainder is a linear polynomial, determined by its values at two points.

<prompt_1.3> Consider using polynomial interpolation as another method to verify the remainder. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
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

Simplify each term:

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

Let me compute step by step:

Start with 1.

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.
</execution_1.1>

<execution_1.2>
Let's do that step by step.

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) = -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 is just -5.

Now add them all 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

So f(-1) = 16.
</execution_1.2>

<execution_1.3>
Since x=1 and x=-1 are roots of x² -1, then f(1) = R(1) and f(-1) = R(-1). Which is exactly the same as above. So that's the same method, just phrased differently.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2:
Now set up the equations:

From x=1: -10 = a*1 + b ⇒ a + b = -10

From x=-1: 16 = -a + b ⇒ -a + b = 16

Now, we can solve these two equations:

<prompt_2.1> Solve the equations a + b = -10 and -a + b = 16 to find the values of a and b. </prompt_2.1>
-----

Based on execution_1.1:
Let me verify this. Let's take the remainder R(x) = 7x -17. Then, when we evaluate at x=1, 7*1 -17 = -10, which matches f(1).
At x=-1, 7*(-1) -17 = -7 -17 = -24. Wait, but f(-1) was 16. Wait, that's a problem. -24 is not equal to 16. Did I make a mistake here?

<prompt_2.2> Verify the calculations of f(1) and f(-1) to ensure they are correct. </prompt_2.2>
-----

Based on execution_1.2:
Therefore, the remainder is 7x -17. So when f(x) is divided by x² -1, the remainder is 7x -17.

But just to be thorough, let me check another point. For example, take x=0.

<prompt_2.3> Evaluate f(0) and the remainder at x=0 to verify consistency. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
Let's subtract the two equations:

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

Simplify left side: a + b + a - b = 2a

Right side: -26

So 2a = -26 ⇒ a = -13

Then plug a = -13 into the first equation:

-13 + b = -10 ⇒ b = -10 +13 = 3

So the remainder is R(x) = -13x + 3.
</execution_2.1>

<execution_2.2>
Wait, that can't be. There must be a mistake here. Let me check my calculations again.

First, f(1):

1 (from x^10) +5 (from 5x^9) -8 (from -8x^8) +7 (from 7x^7) -1 (from -x^6) -12 (from -12x^5) +4 (from 4x^4) -8 (from -8x^3) +12 (from 12x^2) -5 (from -5x) -5 (constant term).

Adding those 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. That seems correct.

f(-1):

1 (from x^10, since even exponent, positive) +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.

Adding those:

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. That also checks out.
</execution_2.2>

<execution_2.3>
Compute f(0) from original polynomial: all terms except the constant term. The constant term is -5. So f(0) = -5.

Compute remainder at x=0: R(0) = 7*0 -17 = -17. But f(0) is -5. So -17 vs -5. That's not equal. Wait, but this is expected because the remainder is only defined modulo x² -1. So for any x, f(x) = (x² -1)*q(x) + R(x). So when x is not 1 or -1, R(x) doesn't have to equal f(x). However, when we evaluate at x=1 and x=-1, then (x² -1) is zero, so f(1) = R(1) and f(-1) = R(-1).

But if we evaluate at x=0, then f(0) = R(0) + (0² -1)*q(0) = -17 + (-1)*q(0). So unless we know q(0), we can't say. But since we only need the remainder, which is defined for all x, but the remainder must satisfy R(1)=f(1) and R(-1)=f(-1). The other values are adjusted by the multiple of (x² -1). So even though R(0) doesn't equal f(0), that's okay because the equality f(x) = (x² -1)q(x) + R(x) holds for all x, but when x is not 1 or -1, (x² -1) isn't zero, so R(x) can differ from f(x). So the key is that at x=1 and x=-1, R(x) must equal f(x). So our solution seems correct because it satisfies those two equations.
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So R(x) = -13x + 3. Let's check this.

But wait, let me verify with another method. Suppose I use the concept that any polynomial f(x) can be written as (x² -1)*q(x) + ax + b. Then, when we divide by x² -1, the remainder is ax + b. To find a and b, we can use the Remainder Factor Theorem for higher degrees, which states that the remainder is the same as evaluating the polynomial at the roots of the divisor, but since there are two roots, we can set up a system of equations.

Alternatively, another approach is to express f(x) as (x² -1)*q(x) + ax + b, and then plug in x=1 and x=-1 to find a and b. Which is exactly what we did. So that seems correct.

But let me just cross-verify by plugging in another value.

<prompt_3.1> Cross-verify the remainder by evaluating f(2) and the remainder at x=2. </prompt_3.1>
-----

Based on execution_2.3:
But to be absolutely sure, let's try another value. Let's take x=2.

<prompt_3.2> Evaluate f(2) and the remainder at x=2 to further verify the solution. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Let's pick x=0. If we plug x=0 into the equation f(x) = (x² -1)q(x) + ax + b, we get f(0) = (-1)q(0) + a*0 + b ⇒ f(0) = -q(0) + b. But we don't know q(0). However, our remainder is -13x +3. So at x=0, remainder is 3. Therefore, f(0) = -q(0) + 3. From earlier, f(0) is -5. Therefore, -5 = -q(0) +3 ⇒ q(0) = 8. But we don't really care about q(x). Since we can't compute q(0) without knowing q(x), this check isn't helpful. But since we used the method of plugging in x=1 and x=-1, which are the roots of the divisor, that's sufficient.
</execution_3.1>

<execution_3.2>
Compute f(2):

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

Calculate each term:

1024 + 5*512=2560; -8*256=-2048; 7*128=896; -64; -12*32=-384; 4*16=64; -8*8=-64; 12*4=48; -10; -5.

Add them up:

1024 +2560 = 3584; 3584 -2048 = 1536; 1536 +896 = 2432; 2432 -64 = 2368; 2368 -384 = 1984; 1984 +64 = 2048; 2048 -64 = 1984; 1984 +48 = 2032; 2032 -10 = 2022; 2022 -5 = 2017.

So f(2) = 2017.

Now compute R(2) = -13*2 +3 = -26 +3 = -23.

Compute (x² -1)*q(x) at x=2: (4 -1)*q(2) = 3*q(2). So f(2) = 3*q(2) + (-23). Therefore, 3*q(2) = 2017 +23 = 2040. So q(2) = 2040 /3 = 680. So if we compute q(2) as (f(2) - R(2))/3, which would be (2017 - (-23))/3 = 2040/3=680. But since we don't know q(x), this is just a consistency check. It shows that the numbers are consistent, but since we can't compute q(x) without division, this doesn't help us find the remainder, but confirms that the remainder is correct.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But let me check again the steps to make sure I didn't make a mistake.

<prompt_4.1> Recheck the steps and calculations to ensure the remainder is correct. </prompt_4.1>
-----

Based on execution_3.2:
Therefore, after careful consideration and checking, the remainder is indeed -13x +3.

But just to make 100% sure, let me check once more with another value, say x=3.

<prompt_4.2> Perform a final verification using x=3 to ensure the remainder is correct. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
First, confirming f(1) and f(-1):

Yes, for f(1), the calculation step by step gives -10. For f(-1), adding all the terms gives 16. Then setting up the equations:

a + b = -10

-a + b = 16

Subtracting the equations: 2a = -26 ⇒ a = -13. Then b = -10 -a = -10 -(-13) = 3. So R(x) = -13x +3.
</execution_4.1>

<execution_4.2>
Compute f(3):

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

That's going to be a big number, but let's compute step by step:

3^10 = 59049

5*3^9 =5*19683=98415

-8*3^8 =-8*6561=-52488

7*3^7=7*2187=15309

-3^6=-729

-12*3^5=-12*243=-2916

4*3^4=4*81=324

-8*3^3=-8*27=-216

12*3^2=12*9=108

-5*3=-15

-5

Now sum all these:

Start with 59049 +98415 = 157464

157464 -52488 = 104976

104976 +15309 = 120285

120285 -729 = 119556

119556 -2916 = 116640

116640 +324 = 116964

116964 -216 = 116748

116748 +108 = 116856

116856 -15 = 116841

116841 -5 = 116836

So f(3) = 116836.

Compute R(3) = -13*3 +3 = -39 +3 = -36.

Then, (x² -1)q(x) at x=3 is (9 -1)q(3) =8*q(3). So f(3) =8*q(3) + (-36). Therefore, 8*q(3) =116836 +36=116872. Therefore, q(3)=116872 /8=14609. So q(3)=14609. If we compute (f(3) - R(3))/8, we get (116836 - (-36))/8=116872/8=14609. Which is consistent.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.2:
**Final Answer**
The remainder when \( f(x) \) is divided by \( x^2 - 1 \) is \boxed{-13x + 3}.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, the remainder when $f(x)$ is divided by $x^2-1$ is $\boxed{-13x+3}$.
</Final_answer>
</think>