Question 1159703: What is the simplest polynomial function with the given zeros, 2i, -3?
Found 2 solutions by ikleyn, solver91311:
Answer by ikleyn(52814)   (Show Source):
You can put this solution on YOUR website! .
p(x) = (x-2i)*(x+3)
Notice that the word "simplest" is not a Math term.
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Tutor @solver91311 teaches you that
"Complex zeros always appear in conjugate pairs, therefore if is a zero, must also be a zero."
It is NOT ALWAYS TRUE.
It is true, if the polynomial has real coefficients - - - but the problems SAYS NOTHING about it;
so, there is NO RATIONALE for such a statement.
* * * " incontrovertibly correct answer to the question " ? * * * - - - Ha - ha - ha. * * *
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
Complex zeros always appear in conjugate pairs, therefore if  is a zero,  must also be a zero. Real zeros have no such restriction, so the simplest polynomial with the given zeros has exactly three zeros, and is a 3rd-degree polynomial.
If  is a zero of a polynomial function then  is a factor of the polynomial. Hence, your desired polynomial function, in factored form is:
\ =\ (x\,+\,3)(x\,+\,2i)(x\,-\,2i))
While the above is incontrovertibly a correct answer to the question, I suspect that you will need to present the answer in standard form, namely:
\ =\ Ax^3\ +\ Bx^2\ +\ Cx\ +\ D)
And this will require that you multiply the three factors together and collect like terms. Hint: The product of two conjugate binomials is the difference of two squares. Don't forget that  . I'll leave the rest in your capable hands.
John
My calculator said it, I believe it, that settles it
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