Question 1052703: I need help finding the polynomial with the given zeros of 0, 8, and 1+sqrt2i with real coefficients please! I've tried and so far all I've gotten is x^2-8x-x+sqrt2i+8+sqrt2i
Any help is appreciated, thanks!
Found 2 solutions by stanbon, Boreal:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I need help finding the polynomial with the given zeros of 0, 8, and 1+sqrt2i with real coefficients please! I've tried and so far all I've gotten is x^2-8x-x+sqrt2i+8+sqrt2i
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If the coefficients are to be Real, 1-sqrt(2)i must also be a zero
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Zeroes:: 0, 8, 1+sqrt(2)i, 1-sqrt(2)i
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f(x) = x(x-8)(x-(1+sqrt(2)i)(x-(1-sqrt(2)i)
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f(x) = x(x-8)((x-1)-sqrt(2)i)((x-1)+sqrt(2)i)
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f(x) = x(x-8)((x-1)^2 +2)
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f(x) = x(x-8)(x^2 -2x + 3)
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Cheers,
Stan H.
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! with zeros of 0 and 8, the factors are x and (x-8)
1+/-sqrt2i needs to be found.
[-b +/- sqrt (b^2-4ac)](1/2) needs to be evaluated
if a is 1 and b is -2, x=(1/2)(2 +/- sqrt (4-4(1)(c))
This gives me 1 +/- sqrt (4-4c), and that has to become negative 16, because then it will be 4i, and it is divided by 2 in the formula, and that will be 2i.
c has to be 5, then sqrt (4-4c)=sqrt(-16)
The third factor is (x^2-2x+5), and x=(1/2)+/- (2 +/- sqrt (4-20))
=(1/2)(2+/-4i)=1+/-2i
The polynomial is x(x-8)(x^2-2x+5)=x^4-10x^3+21x^2-40x.
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