Question 825390: Write a polynomial function F of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.
 , 3i
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! There are three keys to this problem:- For each zero of the function, (x-z), where "z" is a zero, will be a factor of the function:
 ...
where "a" is the leading coefficient and the z's are the various zeros. - If zero of a polynomial function with rational coefficients has a square root for a zero, then the negative of that square root will also be a zero.
- If zero of a polynomial function with rational coefficients has a complex number (like 3i) for a zero, then the complex conjugate of that complex zero will also be a zero. (a+bi is the standard form for a complex number. a-bi is the complex conjugate of a+bi. In words: The complex conjugate of a complex number has the same real part, "a", but the imaginary part is the opposite.)
So if is a zero then so is  . And if 3i (or 0 + 3i) is a zero then 0 - 3i (or just -3i) is also a zero.
Altogether then we have 4 zeros: , , 3i and -3i. Each zero is a factor:
Since we're told that the desired leading coefficient is 1, this becomes:
Simplifying the 2nd and 4th factors:
All that is left is to multiply this out. Since multiplication is commutative (i.e. the order in which the multiplication is done does not matter) then we can multiply these factors in any order we choose. (Hint: It's easiest to start by using the  pattern to multiply the factors with "pairs" of zeros.) Multiplying the first two factors with this pattern:
which simplifies to:
Using the pattern on the last two factors:
which simplifies as follows:
Since  :
(Note how the square roots and the i's are all gone. This is due to the order of multiplication we choose. If we had done it differently, we would still have square roots and/or i's.)
Last we use FOIL to multiply the remaining factors:
which simplifies as follows:
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