Finding the Remainder using the Polynomial Remainder Theorem

The question asks about the remainder obtained when the polynomial \(x^n + m^n\) is divided by \((x - m)\). Here, \(m\) is a non-zero integer and \(n\) is a positive integer.

We can find the remainder easily using the Remainder Theorem.

Understanding the Remainder Theorem

The Remainder Theorem states that if a polynomial \(P(x)\) is divided by a linear factor \((x - a)\), the remainder is equal to \(P(a)\).

Applying the Remainder Theorem

In this problem, our polynomial is \(P(x) = x^n + m^n\). The divisor is \((x - m)\). Comparing this to the standard form \((x - a)\), we see that \(a = m\).

According to the Remainder Theorem, the remainder \(R\) is \(P(m)\). Let's substitute \(x = m\) into the polynomial \(P(x)\):

\(R = P(m) = m^n + m^n\)

Combining the terms, we get:

\(R = 2 \cdot m^n\)

Analyzing the Properties of the Remainder R

We have found that the remainder \(R = 2m^n\), where \(m\) is a non-zero integer and \(n\) is a positive integer. Now let's examine the given options based on this result.

• Option 1: R is a non-zero even integer

  We have \(R = 2m^n\). Since \(m\) is a non-zero integer, \(m^n\) is also a non-zero integer (as \(n\) is a positive integer). When any integer (\(m^n\)) is multiplied by 2, the result (\(2m^n\)) is always an even integer. Since \(m^n\) is non-zero, \(2m^n\) is also non-zero. Therefore, \(R\) is indeed a non-zero even integer.

• Option 2: R = s² for some integer s, if n is even

  Let's test this. If \(n\) is even, \(R = 2m^n\). For \(R\) to be a perfect square \(s^2\), the prime factorization of \(2m^n\) must have all exponents even. The prime factorization of \(2m^n\) includes a factor of \(2^1\). If \(m\) is odd, the exponent of 2 in \(R\) is just 1, which is odd. So, \(R\) is not a perfect square. If \(m\) is even, say \(m = 2^k \cdot j\) where \(j\) is odd and \(k \ge 1\), then \(R = 2 \cdot (2^k j)^n = 2 \cdot 2^{kn} j^n = 2^{kn+1} j^n\). If \(n\) is even, \(kn\) is even, so \(kn+1\) is odd. The exponent of 2 in \(R\) is odd. Thus, \(R\) cannot be a perfect square if \(n\) is even and \(m\) is a non-zero integer.

• Option 3: R = s² for some integer s, if n is odd

  If \(n\) is odd, \(R = 2m^n\). Let's take an example: If \(m=1\) and \(n=3\) (odd), \(R = 2 \cdot 1^3 = 2\). 2 is not a perfect square. So, \(R\) is not always a perfect square when \(n\) is odd.

• Option 4: R = t³ for some integer t, if 3 divides n

  If 3 divides \(n\), let \(n = 3k\) for some positive integer \(k\). Then \(R = 2m^n = 2m^{3k}\). For \(R\) to be a perfect cube \(t^3\), the exponents of all prime factors in its prime factorization must be multiples of 3. The prime factorization of \(R = 2m^{3k}\) includes a factor of \(2^1\). The exponent of 2 is 1, which is not a multiple of 3. If \(m\) contains factors other than 2, say \(p\), then \(R = 2 \cdot (\dots p^e \dots)^{3k} = 2 \dots p^{3ke} \dots\). The exponent of \(p\) is \(3ke\), which is a multiple of 3. However, the exponent of 2 remains 1 (if \(m\) is odd) or \(1 + e_{m} \cdot 3k\) where \(e_m\) is the exponent of 2 in \(m\). \(e_{m} \cdot 3k\) is a multiple of 3, so \(1 + e_{m} \cdot 3k\) is not a multiple of 3. Thus, \(R\) cannot be a perfect cube when 3 divides \(n\) and \(m\) is non-zero.

Based on the analysis, only the first option holds true for all non-zero integers \(m\) and positive integers \(n\).

Conclusion on Remainder Properties

The remainder \(R\) obtained by dividing \(x^n + m^n\) by \((x - m)\) is consistently \(2m^n\). For any non-zero integer \(m\) and positive integer \(n\), \(m^n\) is a non-zero integer. Multiplying it by 2 guarantees that the result \(R\) is a non-zero even integer.

Revision Table: Remainder Calculation

Additional Information: Polynomial Division

Polynomial division is a fundamental concept in algebra. The Remainder Theorem is a shortcut derived from the polynomial long division process. It is particularly useful for finding the remainder quickly when the divisor is a linear term of the form \((x-a)\).

For a general polynomial \(P(x)\) and a divisor \(D(x)\), we can write \(P(x) = Q(x)D(x) + R(x)\), where \(Q(x)\) is the quotient and \(R(x)\) is the remainder. The degree of \(R(x)\) must be less than the degree of \(D(x)\). When \(D(x) = x - a\), its degree is 1, so the degree of \(R(x)\) must be 0. This means \(R(x)\) is a constant, which we simply denote as \(R\). Substituting \(x=a\) into the division equation gives \(P(a) = Q(a)(a-a) + R\), which simplifies to \(P(a) = Q(a) \cdot 0 + R\), so \(P(a) = R\). This confirms the Remainder Theorem.