We are to divide \( u^4 + 3 \) by \( u^2 - 2u + 2 \) and find the remainder. Since the divisor is quadratic, the remainder must be linear or constant: let \( R(u) = au + b \).

["Dividing ( u^4 + 3 ) by ( u^2 - 2u + 2 ): Finding the Remainder", "When dividing polynomials, understanding the structure of the dividend, divisor, and remainder is essential—especially when the divisor is quadratic. In this article, we explore how to divide ( u^4 + 3 ) by ( u^2 - 2u + 2 ) and find the remainder using polynomial division and algebraic reasoning.", "---", "### Setting Up the Division", "We seek to divide:\n[\n\ ext{Dividend: } u^4 + 0u^3 + 0u^2 + 0u + 3\n]\nby\n[\n\ ext{Divisor: } u^2 - 2u + 2\n]\nand determine the remainder ( R(u) ), which, by the division algorithm for polynomials, will be of degree less than 2. Therefore, we assume:\n[\nR(u) = au + b\n]\nOur goal is to express:\n[\nu^4 + 3 = (u^2 - 2u + 2) \cdot Q(u) + au + b\n]\nfor some quotient polynomial ( Q(u) ).", "---", "### Use Polynomial Long Division (or Equating Coefficients)", "Rather than full long division, we use the fact that the degree of the remainder is less than 2, and substitute convenient values of ( u ) to form equations—or equate coefficients after expressing the equation above. Here, equating coefficients is efficient.", "Let’s write:\n[\nu^4 + 3 = (u^2 - 2u + 2)(qu^2 + ru + s) + au + b\n]\nOur unknowns are the coefficients ( q, r, s ) (from the quotient) and ( a, b ) (from the remainder). Expand the right-hand side:", "[\n(u^2 - 2u + 2)(qu^2 + ru + s) = \nu^2(qu^2 + ru + s) - 2u(qu^2 + ru + s) + 2(qu^2 + ru + s)\n]", "Compute each term:\n- ( u^2(qu^2 + ru + s) = qu^4 + ru^3 + su^2 )\n- ( -2u(qu^2 + ru + s) = -2qu^3 - 2ru^2 - 2su )\n- ( 2(qu^2 + ru + s) = 2qu^2 + 2ru + 2s )", "Add them together:\n[\nqu^4 + (r - 2q)u^3 + (s - 2r + 2q)u^2 + (-2s + 2r)u + 2s\n]", "Now add the remainder ( au + b ):\nFull expansion:\n[\nqu^4 + (r - 2q)u^3 + (s - 2r + 2q)u^2 + (-2s + 2r + a)u + (2s + b)\n]", "Now match this with the left-hand side:\n[\nu^4 + 0u^3 + 0u^2 + 0u + 3\n]", "Matching coefficients term by term:", "| Power of ( u ) | LHS Coefficient | Equation |\n|------------------|----------------|---------|\n| ( u^4 ) | 1 | ( q = 1 ) |\n| ( u^3 ) | 0 | ( r - 2q = 0 \Rightarrow r = 2q = 2 ) |\n| ( u^2 ) | 0 | ( s - 2r + 2q = 0 \Rightarrow s - 4 + 2 = 0 \Rightarrow s = 2 ) |\n| ( u^1 ) | 0 | ( -2s + 2r + a = 0 \Rightarrow -4 + 4 + a = 0 \Rightarrow a = 0 ) |\n| Constant | 3 | ( 2s + b = 3 \Rightarrow 4 + b = 3 \Rightarrow b = -1 ) |", "---", "### Final Remainder", "We now have:\n[\nR(u) = au + b = 0\cdot u - 1 = -1\n]", "So the remainder is a constant: ( -1 )", "---", "### Conclusion", "When dividing ( u^4 + 3 ) by ( u^2 - 2u + 2 ), the quotient is a quadratic polynomial and the remainder is a constant:\n[\n\boxed{R(u) = -1}\n]", "This result confirms that:\n[\nu^4 + 3 = (u^2 - 2u + 2)(u^2 + 2u + 2) - 1\n]\nand verifies via polynomial identity.", "---", "Key Takeaways:\n- For divisors of degree 2, the remainder is linear at most — but here it’s constant.\n- Equating coefficients after expansion provides a reliable alternative to long division.\n- This method works for any polynomial divided by a quadratic.", "Use this approach when finding remainders in polynomial division—especially useful in roots of unity, Taylor expansions, and symbolic computation."]