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.

Algebra ->  Equations -> 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.      Log On


   



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(52814) About Me  (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) = x%5E2+%2B+25.


When you perform long division, you will get the quotient  q(x) = f%28x%29%2F%28x%5E2%2B25%29 = x%5E3+%2B+x%5E2+-+2x+-+2.


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


    x%5E3+%2B+x%5E2+-+2x+-+2 = %28x%5E3+%2B+x%5E2%29 - %282x%2B2%29 = x%5E2%2A%28x%2B1%29 - 2%2A%28x%2B1%29 = %28x%5E2-2%29%2A%28x%2B1%29.


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


    f(x) = %28x-5i%29%2A%28x-5i%29%2A%28x%2B1%29%2A%28x-sqrt%282%29%29%2A%28x%2Bsqrt%282%29%29,


and its roots are  5i, -5i, sqrt%282%29, -sqrt%282%29  and -1.    ANSWER

Solved.