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an equation with a degree of 5 can have 3 real solutions and 2 non-real solutions
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Yes — that's possible.
By the Fundamental Theorem of Algebra a degree-5 polynomial has 5 complex roots counting multiplicity. If the polynomial has real coefficients, nonreal complex roots occur in conjugate pairs, so the number of nonreal roots must be even (0, 2, or 4). Thus a real-coefficient quintic can have 5 real roots, 3 real + 2 nonreal, or 1 real + 4 nonreal.
Example: p(x) = (x−1)(x−2)(x+3)(x^2+1).
Roots: 1, 2, −3 (real) and i, −i (nonreal).
By the Fundamental Theorem of Algebra a degree-5 polynomial has 5 complex roots counting multiplicity. If the polynomial has real coefficients, nonreal complex roots occur in conjugate pairs, so the number of nonreal roots must be even (0, 2, or 4). Thus a real-coefficient quintic can have 5 real roots, 3 real + 2 nonreal, or 1 real + 4 nonreal.
Example: p(x) = (x−1)(x−2)(x+3)(x^2+1).
Roots: 1, 2, −3 (real) and i, −i (nonreal).
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