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Question 1095813: Find a polynomial function of lowest degree with rational coefficients that has -4i, 3 as some of its zeros.
The possible answers are:
A) x^4-7x^2+144
B)x^3-3x^2+16x-48
C)x^3-4x^2+16x+48
D)x^4+7x^2-144
I am struggling with this concept in class, if anyone could possibly help that would be amazing!!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i believe you're looking at selection B.
that would be x^3 - 3x^2 + 16x - 48
the trick here is to know that complex roots always come in pairs.
the roots shown are -4i and 3.
the factors from those roots would be (x+4i) and (x-3)
basically your root says that x = -4i
add 4i to both sides of that to get x + 4i = 0
x = -4i is your root
x + 4i is your factor.
same with x = 3
subtract 3 from both sides of that to get x = 3 = 0
x = 3 is your root.
x - 3 is your factor.
simce complex roots always come in pairs, then your other factor has to be x-4i
that's because the roots are plus or minus 4i.
from that you get factors of (x + 4i) and (x - 4i)
the rest is just multiplying your roots to see what equation they become.
(x + 4i) * (x = 4i) results in x^2 - 16i^2 which becomes x^2 + 16 because i^2 is equal to -1 and -16 * -1 = 16.
when you multiply (x^2 + 16) * (x-3), you get (x^3 - 3x^2 + 16x - 48).
that's selection B.
x+4i and x-4i are called conjugate pairs.
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