A real polynomial #P(x)# is the product of real linear and real quadratic factors. Can you that if a real quadratic factor has one zero, #a+bi#, then the other zero must be the complex conjugate, #a-bi#?
Hint: Let the quadratic be #alphax^2 + betax + gamma# and use the properties of complex conjugates
Hint: Let the quadratic be
2 Answers
See below.
Explanation:
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Another version.
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The feasible solution is
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So if
Use the quadratic formula...
Explanation:
If one of the real quadratic factors is:
#alphax^2+betax+gamma#
then using the quadratic formula, it has zeros:
#(-beta+-sqrt(beta^2-4alphagamma))/(2alpha)#
If
Otherwise,
#sqrt(beta^2-4alphagamma) = (sqrt(4alphagamma-beta^2))i#
resulting in a complex conjugate pair of non-real zeros.