A real-coefficient polynomial with one complex root
A student is checking an algebraic model whose polynomial is p(x)=x³-5x²+11x-15. All coefficients are real. A symbolic calculation suggests that 1+2i is a root, but the student has not yet produced an exact factorization. The task is to establish the remaining roots without replacing the problem by a decimal graph. A classmate proposes using the real number 1 alone as a factor because it is the real part of the known root. Another proposes using the modulus of that root. Work over the complex numbers unless a task explicitly asks for real solutions. Every stated polynomial coefficient is part of this invented exercise, rather than a measurement from a physical device. Use multiplication to verify any factorization.
1. Find the other nonreal root and explain why it must occur.
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Because the coefficients are real, conjugating p(1+2i)=0 gives p(1-2i)=0. Thus 1-2i is also a root. Neither taking only the real part nor taking the modulus preserves this polynomial equation.
2. Find the real quadratic factor belonging to this conjugate pair.
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Multiply (x-(1+2i))(x-(1-2i))=(x-1)²+4=x²-2x+5. The imaginary terms cancel. Its discriminant is4-20=-16, so it has no real root.
3. Complete the factorization and identify all real solutions of p(x)=0.
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Polynomial division gives p(x)=(x-3)(x²-2x+5). Expanding verifies x³-5x²+11x-15. The only real solution is x=3; the two other solutions are1±2i.