Question 1181514: Find a polynomial function of lowest degree with rational coefficients that has the given numbers as some of its zeros. 5-i, SQUARE ROOT 11
Answer by Solver92311(821)   (Show Source):
You can put this solution on YOUR website!
Both complex and irrational zeros come in conjugate pairs. This means that if  is a zero, then so is  . And if  where  and  then  is also a zero.
Since the degree of a polynomial function is equal to the number of zeros, and since you are given one complex zero and one irrational zero there must, at a minimum, be four zeros, you are looking for a fourth-degree polynomial.
We also know that if  is a zero of a polynomial function, then  must be a factor of the polynomial.
You were given  as a zero, so ) is a factor of the desired polynomial. Likewise, ) is a factor. Also, you were given  as a zero so  is a factor of the desired polynomial and likewise ) is a factor.
Putting it all together, your polynomial function, in factored form, is:
\ =\ \(x\,-\,5\,+\,i\)\(x\,-\,5\,-\,i\)\(x\,-\,\sqrt{11}\)(x\,+\,\sqrt{11}\))
The only thing left for you to do is to multiply the four binomials and collect like terms. Hints: The product of two conjugates is the difference of two squares and  .
John
My calculator said it, I believe it, that settles it
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