Question 668545: Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros: 3i,2-i.
Found 2 solutions by swincher4391, solver91311:
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! We know that all imaginary roots come in conjugate pairs. 3i comes in the form 0 + 3i and so 0-3i is its conjugate. Similarly 2-i has conjugate pair 2+i.
This means we can form a product of linear factors like so:
(x+3i)(x-3i)(x-(2-i)(x-(2+i)
FOIL out each pair of binomials by their conjugates to get two nice easy quadratics.
(x^2+9)(x^2-4x+5) = x^4 - 4x^3 + 14x^2 - 36x + 45
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
If a polynomial equation has a zero at  , then  is a factor of the polynomial. Complex zeros of polynomials with rational coefficients ALWAYS come in conjugate pairs. That is to say, if  is a zero of a polynomial with rational coefficients then  is also a zero of that polynomial.
Given  and  as zeros means that you have at least four zeros, namely  and  . Also, since you want to derive the polynomial of least degree under the given circumstances, you cannot have any more than four zeros. Hence, the given values are all of the zeros.
That means that ) and ) are the four factors of your polynomial. Multiply them and collect like terms. Hints for easier calculation: First, remember that the product of two conjugates is the difference of two squares. Second,  . Taken together these facts mean that the product of two complex conjugates is the SUM of two squares.
John
My calculator said it, I believe it, that settles it
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