Question 295815: The complex number 2+i is a root of z^3-z^2-pz+15=0, where p is real. Explain why 2-i is another root and find the third root. Found 2 solutions by stanbon, richwmiller:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The complex number 2+i is a root of z^3-z^2-pz+15=0, where p is real.
Explain why 2-i is another root and find the third root.
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If 2+i is a root (x-(2+i)) would be a factor.
So there must also be a factor p(x) such that (x-(2+i))p(x) = f(x)
Since the coefficients of f(x) are all Real, p(x) would have to
have complex number terms to form the product f(x) with Real
coefficients.
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Assuming that both (2+i) and (2-i) are roots
f(x) = p(x)(x-(2+i))(x-(2-i))
f(x) = p(x)((x-2)-i)((x-2)+i)
f(x) = p(x)((x-2)^2+1)
f(x) = p(x)(x^2-4x+5)
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Then f(x)/x^2-4x+5 = (x+3) with Remainder of -p+7
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If p = 7, x+3 is a factor of f(x) and x=-3 is a root.
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Cheers,
Stan H.
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