math (precalc)
posted by Kyle .
Find a polynomial with integer coefficients that satisfies the given conditions.
R has degree 4 and zeros 3 − 4i and 5, with 5 a zero of multiplicity 2.

x = 3  4i
x = 3 + 4i
x = 5
x =5
two 5's because multiplicity 2
You always have to have the conjugate of an imaginary number.
(x (34i))(x(3+4i)(x5)(x5) = 0
Easiest way is to multiply the first two and the second two factors.
Then multiply those answers together to get the polynomial. 
(x5)(x5)(x3+4i)(x34i) = 0
(x^210x+25)(x^26x+25)
multiply that out 
complex values come in pairs (why?), so the roots are
34i, 3+4i, 5, 5
R(x) = (x(34i))(x(3+4i))(x5)^2
= ((x3)+4i)((x3)4i)(x5)^2
= ((x3)^2+4^2)(x5)^2
= (x^26x+25)(x5)^2
= x^4  16x^3 + 110x^2  400x + 625
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