SOLUTION: find nth degree given n=3;3&4i are zeros; f(1)=68 68=nth degree of((1)^3-(1)^2+16(1)-16)

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Question 1027930: find nth degree given n=3;3&4i are zeros; f(1)=68
68=nth degree of((1)^3-(1)^2+16(1)-16)
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The assumption is that the coefficients of the cubic polynomial are real so the complex conjugate is also a root.
f%28x%29=a%28x-3%29%28x-4i%29%28x%2B4i%29
f%28x%29=a%28x-3%29%28x%5E2%2B16%29
Now use the point to solve for a.
f%281%29=a%281-3%29%281%5E2-16%29
68=a%28-2%29%28-15%29
a=68%2F30
a=34%2F15
.
.
f%28x%29=%2834%2F15%29%28x-3%29%28x%5E2%2B16%29