SOLUTION: Hello, I have the following multiple choice question: Find an nth degree polynomial function with real coefficients satisfying the given conditions. n = 3; - 4 and i are ze

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hello, I have the following multiple choice question: Find an nth degree polynomial function with real coefficients satisfying the given conditions. n = 3; - 4 and i are ze      Log On


   



Question 104483: Hello,
I have the following multiple choice question:
Find an nth degree polynomial function with real coefficients satisfying the given conditions.
n = 3; - 4 and i are zeros; f(-3) = 60

A) f(x) = 6 + 24 + 6x + 24

B)
f(x) = -6 - 24 + 6x + 24
C)
f(x) = 6 + 24 - 6x - 24

D)
f(x) = -6 - 24 - 6x - 24
I'm not sure if I'm supposed to substitute the values into the multiple choices to see which one works out. Regardless, I would like to find the answer on my own without having to plug the values in. Could you help me? I looked in my algebra book but I couldn't find a section for this. I would greately appreciate it if someone could give me some direction.
Thank you,
EJ

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
EJ, let's start with what you know.
Nth degree polynomial with n=3 means
f%28x%29=ax%5E3%2Bbx%5E2%2Bcx%2Bd
Your job is to find a,b,c,d with the rest of the information provided.
f%28-4%29=0
f%28i%29=0
f%28-3%29=60
The problems is you have 4 unknowns (a,b,c,d) and only three equations to solve them with. Let's plug in and see what we find.
1.f%28-4%29=a%28-4%29%5E3%2Bb%28-4%29%5E2%2Bc%28-4%29%2Bd=0
2.f%28i%29=a%28i%29%5E3%2Bb%28i%29%5E2%2Bc%28i%29%2Bd=0
3.f%28-3%29=a%28-3%29%5E3%2Bb%28-3%29%5E2%2Bc%28-3%29%2Bd=0
Simplifying the equations, gives you
1.-64a%2B16b-4c%2Bd=0
2. -ia-b%2Bic%2Bd=0
3.-27a%2B9b-3c%2Bd=60
where i%5E2=-1 and i%5E3=-i.
Since you have a complex expression in your second equation, it actually acts as an additional equation. Since both the real part and the complex part of equation 2 have to equal zero, you now have 4 equations with four unknowns.
1.-64a%2B16b-4c%2Bd=0
2.(Real)-b%2Bd=0
3.(Complex)-a%2Bc=0
4.-27a%2B9b-3c%2Bd=60
With those 4 equations, you are able to substitute and find the values for a,b,c,d.
Good luck and post another question if you get stuck.