Remainder Calculation for \(x^6\) Divided by \(x^2+1\)

This problem asks us to find the remainder when the polynomial \(x^6\) is divided by the polynomial \(x^2 + 1\). We can solve this using algebraic manipulation, specifically by leveraging the properties of polynomial division and substitution.

Understanding Polynomial Division and Remainders

When we divide a polynomial \(P(x)\) by another polynomial \(D(x)\), we get a quotient \(Q(x)\) and a remainder \(R(x)\) such that:

The degree of the remainder \(R(x)\) must be strictly less than the degree of the divisor \(D(x)\). In our case:

• \(P(x) = x^6\) (the dividend)
• \(D(x) = x^2 + 1\) (the divisor)

Since the degree of the divisor \(D(x)\) is 2 (the highest power of \(x\) is \(x^2\)), the remainder \(R(x)\) must have a degree less than 2. This means the remainder will be of the form \(ax + b\), where \(a\) and \(b\) are constants.

Method: Using Substitution

A quick way to find the remainder is to use the relationship derived from the divisor.

1. Set the divisor to zero to find the values of \(x\) that make it zero: \(x^2 + 1 = 0\)
2. Solve for \(x^2\): \(x^2 = -1\)
3. Now, we rewrite the dividend \(x^6\) in terms of \(x^2\): \(x^6 = (x^2)^3\)
4. Substitute the value \(x^2 = -1\) into the expression for \(x^6\): \(x^6 = (-1)^3\)
5. Calculate the result: \((-1)^3 = -1 \times -1 \times -1 = -1\)

This value, -1, represents the remainder when \(x^6\) is divided by \(x^2+1\). Because this result is a constant, it fits the requirement for the remainder \(R(x)\) (degree less than 2).

Step-by-Step Algebraic Approach

Alternatively, we can use algebraic steps:

1. Start with the dividend \(x^6\).
2. We know \(x^2 = -1\) implies \(x^2+1=0\). We want to express \(x^6\) using terms of \(x^2+1\).
3. Rewrite \(x^6\) as \((x^2)^3\).
4. Substitute \(x^2 = -1 + (x^2+1)\): \(x^6 = (x^2)^3 = (-1 + (x^2+1))^3\)
5. Expanding this using the binomial theorem would be complex. Instead, let's directly use \(x^2 \equiv -1 \pmod{x^2+1}\).
6. \(x^6 = (x^2)^3 \equiv (-1)^3 \pmod{x^2+1}\)
7. \(x^6 \equiv -1 \pmod{x^2+1}\)

This shows that \(x^6\) leaves a remainder of \(-1\) when divided by \(x^2+1\).

Conclusion

Both methods confirm that the remainder when \(x^6\) is divided by \(x^2 + 1\) is \(-1\).