SOLUTION: I hope that someone can help me solve this problem: n=3; -5 and 4+3i are zeros; f(2)=273. Can you please help?

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Question 1114960: I hope that someone can help me solve this problem: n=3; -5 and 4+3i are zeros; f(2)=273. Can you please help?
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
I hope that someone can help me solve this problem: n=3; -5 and 4+3i are zeros; f(2)=273. Can you please help?
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f(n) = a(n-3)(n+5)(n-(4+3i))
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f(2) = a(-1)(7)(-2-3i) = 273
a = -39/(-2-3i) = 39/(2+3i)
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Equation:
f(n) = [(39/(2+3i))(n-3)(n+5)(n-(4+3i)]
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Cheers,
Stan H.
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Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

given: n=3, -5 and 4%2B3i are zeros; f%282%29=273
if n=3, we have a 3rd degree function
Since 4%2B3i is a root of f(x) and f(x) has real coefficients, +4-3i is also a root.
use zero product theorem:
f%28x%29+=+a%28x-%28-5%29%29%28x-%284%2B3i%29%29%28x-%284-3i+%29%29
f%28x%29+=+a%28x%2B5%29%28x-4-3i%29%28x-4%2B3i+%29....multiply all that and you will get
f%28x%29+=+a%28x%5E3+-+3x%5E2+-+15x+%2B+125%29 ....since f%282%29=273, use it to find a

273+=+a%282%5E3+-+3%2A2%5E2+-+15%2A2+%2B+125%29
273+=+91a
a=273%2F91
a=3
so, your function is:
f%28x%29+=+3%28x%5E3+-+3x%5E2+-+15x+%2B+125%29

f%28x%29+=+3x%5E3+-+9x%5E2+-+45x+%2B+375-> your function