The complex zeros of a polynomial with complex coefficients always occur in conjugate pairs? Why?

Answer 1

That doesn't sound right.

It should say, "The complex zeros of a polynomial with *real* coefficients always occur in conjugate pairs."

Can you make sure you're reading your question correctly?

Example:

f(x) = x^2 + ix + 2 factors as (x - i)(x + 2i), and the zeros are
i and -2i, which are NOT conjugate pairs.

The complex zeros of a polynomial with REAL coefficients always occur in conjugate pairs because that's how complex numbers get eliminated. In the end, the polynomial should have real coefficients, so if the complex zero does not have a conjugate pair, the i will not cancel out.

Example:

f(x) = x^2 + 4

This factors as

f(x) = (x - 2i)(x + 2i)

Conjugate pairs, real coefficients.

Answer 2

first be conscious that one given root is complicated and u would desire to recognize that complicated and irrational roots constantly take place in conjugate pairs so 3 - i visit additionally be the muse of this polynomial f(x)=x(x+4)(x^2-(3+i))(x-(3-i)) =x(x+4)(x^2-6x+10) this in linear and quadratic type

Answer 3

because that way when you mulitple the factors, all of the i cancells out, if they were different, there would be some i left over

Answer 4

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