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If f(x) has 2+3i as a root, then 2+3i could be the only non-real root of f(x).

a-true
b-false

User Uncoke
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1 Answer

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Final answer:

The statement that 2+3i could be the only non-real root of f(x) is false. Non-real roots of polynomials with real coefficients must come in complex conjugate pairs.

Step-by-step explanation:

If f(x) has 2+3i as a root, the question is whether 2+3i could be the only non-real root of f(x). The answer to this is false. In complex analysis, roots of polynomials appear in complex conjugate pairs. This means that if a polynomial has real coefficients and a non-real root A = a + ib, where a and b are real numbers, then its conjugate A* = a - ib must also be a root. By multiplying A and A*, we get A * A* = (a + ib) * (a - ib) = a² + b², where the complex parts cancel out, preserving the polynomial's real coefficients.

User Whygee
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