## Finding nth-degree polynomial

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
Cody
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Joined: Fri Mar 12, 2010 1:45 am
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### Finding nth-degree polynomial

This is my first post on here so let me begin by giving a hearty Hello to all!
On to the problem:
Find a nth-degree polynomial function w/real coefficients given the following conditions:
n=3 | 4 and 2i are zeros | f(2)=-16

I begin by creating a function from the zeros that are given and working backwards. Like this:

$f(x)=a^n(x-4)(x-2i)(x+2i)$

I included (x+2i) as well because if complex numbers appear in the zeros of the polynomial then they also have the complex conjugate.
Then I expand like so:

$f(x)=a^n(x-4)(x^2+4)$

$f(x)=a^n(x^3-4x^2+4x-16)$

So now I just need to find $a^n$... I think I am suppose to plug 3 in for $a^n$, since it is a 3rd degree polynomial, and then solve. I am not sure though. Any help would be appreciated.

stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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You're on the right track. (And thank you for showing your work and reasoning so clearly!)

The "n = 3" part means that you're looking for a cubic polynomial; the highest degree on x will be 3. You have correctly found the factors, and multiplying them together (if that's required) will give you the right power on the variable.

The "a" part is the unknown multiplier, and it has no power on it. You know that "3x - 6 = 0" and "x - 2 = 0" have the same solution, but "y = 3x - 6" and "y = x - 2" do not have the same graphs. They have the same x-intercept, but are different functions. In the same way, any cubic polynomial with the factors you've found would have the same solutions. But the graphs for "a = 1" and "a = -1" would be upside-down from each other.

You are given the point, (x, f(x)) = (2, -16) so that you can figure out what "a" is. So plug "2" in for x, set y equal to -16, and solve for the value of "a".