SOLUTION: One zero of f(x)=x^5+x^4+23x^3+23x^2−50x−50 is 5i where i=√−1. Completely factor f(x)over the set of complex numbers using this. All the values should be exact.

Question 1172472: One zero of f(x)=x^5+x^4+23x^3+23x^2−50x−50 is 5i where i=√−1. Completely factor f(x)over the set of complex numbers using this. All the values should be exact.
Answer by ikleyn(52817) (Show Source): You can put this solution on YOUR website!
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Since the coefficients of the polynomial are real numbers (they even are integers (!) ),
it implies that together with the imaginary root 5i, its complex conjugate -5i is also the root.


Hence, the polynimial f(x) is divisible by (x-5i)*(x+5i) = .


When you perform long division, you will get the quotient  q(x) =  = .


You can factor this quotient further using grouping/re-grouping


     =  -  =  -  = .


Therefore, the full decomposition of the given polynomial over complex number domain is


    f(x) = ,


and its roots are  5i, -5i, ,   and -1.    ANSWER

Solved.

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