Okay, so I need to find the remainder when this polynomial f(x) is divided by x² - 1 without using long division. Hmm, long division with such a high-degree polynomial would be really tedious. Let me think of another approach.

I remember that when dividing a polynomial by another polynomial, say of degree 2 like x² - 1, the remainder will be of degree less than 2. So the remainder should be a linear polynomial, right? That means the remainder should look like ax + b, where a and b are constants that I need to find.

Wait, how do I find a and b? Maybe I can use the Remainder Theorem or some substitution method. The Remainder Theorem says that the remainder when dividing by (x - c) is f(c), but here we're dividing by a quadratic. Maybe there's a generalized version of the Remainder Theorem for higher-degree divisors?

I think so. If we divide by a quadratic, the remainder is of degree 1, so we can write f(x) as (x² - 1) * Q(x) + ax + b, where Q(x) is the quotient. Then, if we plug in the roots of x² - 1 into f(x), we can set up equations to solve for a and b.

The roots of x² - 1 are x = 1 and x = -1. So, if I substitute x = 1 into the equation, I get f(1) = a*1 + b. Similarly, substituting x = -1, I get f(-1) = a*(-1) + b. That gives me a system of two equations with two variables, which I can solve.

Let me calculate f(1) first. Plugging x = 1 into f(x):

f(1) = 1¹⁰ + 5*1⁹ - 8*1⁸ + 7*1⁷ - 1*1⁶ - 12*1⁵ + 4*1⁴ - 8*1³ + 12*1² - 5*1 - 5.

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

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

Let me compute this step by step:

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.

Now, let's 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.

Calculating 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) = -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 -5.

Now add all these together:

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

Let's compute step by step:

1 -5 = -4

-4 -8 = -12

-12 -7 = -19

-19 -1 = -20

-20 +12 = -8

-8 +4 = -4

-4 +8 = 4

4 +12 = 16

16 +5 = 21

21 -5 = 16.

So f(-1) = 16. Therefore, the second equation is -a + b = 16.

Now we have the system:

1. a + b = -10

2. -a + b = 16

Let me solve this system. If I add both equations together, the a terms will cancel:

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

Which simplifies to:

2b = 6

Therefore, b = 3.

Substituting b = 3 into the first equation:

a + 3 = -10

So a = -13.

Therefore, the remainder is -13x + 3.

Wait, let me check my calculations again to make sure I didn't make a mistake.

First, f(1):

1^10 is 1.

5*1^9 is 5.

-8*1^8 is -8.

7*1^7 is 7.

-1*1^6 is -1.

-12*1^5 is -12.

4*1^4 is 4.

-8*1^3 is -8.

12*1^2 is 12.

-5*1 is -5.

-5.

Adding those:

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. Correct.

f(-1):

(-1)^10=1.

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

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

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

-1*(-1)^6=-1*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. Correct.

So equations:

a + b = -10

-a + b =16

Adding gives 2b=6 => b=3. Then a = -13.

Thus, remainder is -13x +3.

Wait, but let me check if this is correct by plugging in x=1 and x=-1 into the remainder.

If remainder is -13x +3, then at x=1: -13(1)+3 = -10, which matches f(1)=-10. At x=-1: -13*(-1)+3=13+3=16, which matches f(-1)=16. So that works.

Therefore, the remainder is -13x +3. But the question says "the remainder", so I think that's the answer. However, sometimes people prefer writing the remainder with positive coefficients, but in this case, since it's a linear term, it's already in the simplest form.

Alternatively, maybe I can write it as -13x +3, or factor out a negative sign: - (13x -3). But the standard form is usually written with the coefficient of x first, so -13x +3 is fine.

But let me verify once more by another method, just to be sure.

Another approach: Since x² ≡1 mod (x² -1), so any higher power of x can be reduced modulo x² -1. For example, x³ =x*x² ≡x*1 =x. Similarly, x⁴ =x²*x² ≡1*1=1. So in general, x^(2k) ≡1, x^(2k+1)≡x.

Therefore, we can write f(x) as the sum of terms, each of which is either x or 1, depending on the exponent.

Let me try that.

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

Let's break down each term:

x^{10}: exponent 10 is even, so x^{10} ≡ (x²)^5 ≡1^5=1.

5x^9: exponent 9 is odd, so x^9 ≡x*(x²)^4 ≡x*1^4=x. So 5x^9 ≡5x.

-8x^8: exponent 8 is even, so x^8 ≡1. Thus, -8x^8≡-8*1=-8.

7x^7: exponent 7 is odd, so x^7≡x*(x²)^3≡x*1^3=x. So 7x^7≡7x.

-x^6: exponent 6 is even, so x^6≡1. Thus, -x^6≡-1.

-12x^5: exponent 5 is odd, so x^5≡x*(x²)^2≡x*1^2=x. So -12x^5≡-12x.

4x^4: exponent 4 is even, so x^4≡1. Thus, 4x^4≡4*1=4.

-8x^3: exponent 3 is odd, so x^3≡x*(x²)^1≡x*1=x. So -8x^3≡-8x.

12x^2: exponent 2 is even, so x^2≡1. Thus, 12x^2≡12*1=12.

-5x: exponent 1 is odd, so x≡x. Thus, -5x remains -5x.

-5: constant term, remains -5.

Now, let's collect all the terms:

Constants: 1 (from x^{10}) -8 (from x^8) -1 (from x^6) +4 (from x^4) +12 (from x^2) -5 (constant term).

Variables: 5x (from x^9) +7x (from x^7) -12x (from x^5) -8x (from x^3) -5x (from x).

Let's compute the constants first:

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

1 -8 = -7

-7 -1 = -8

-8 +4 = -4

-4 +12 = 8

8 -5 = 3.

Now the variables:

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

Combine like terms:

5 +7 =12

12 -12 =0

0 -8 =-8

-8 -5 =-13.

So variables: -13x.

Therefore, the remainder is -13x +3, which matches the previous result.

Therefore, the remainder is indeed -13x +3.

I think that's solid. Both methods lead to the same answer, so I'm confident it's correct.
Therefore, the remainder when $f(x)$ is divided by $x^2 - 1$ is $\boxed{-13x + 3}$.
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