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Find an nth degree polynomial function with real coefficients satisfying the given conditions..

N=3; -4 and 6+5i are zeros; f(1)=250

(if you know how to do this but dont feel like doing it for me; Could you please just inform me on how i's work while factoring, i.e. What do they affect, etc.

Verified Answer :

we
know that if we have a polynomial with real coefients and one root is
a+bi, another root is a-bi




so


some
roots are


3,-4,6+5i,
6-5i


for
roots, r1 and r2


teh
factors are


(x-r1)(x-r2)


so


we
can put them in and


(x-3)(x+4)(x-6-5i)(x-6+5i)


(x^2+x-12)(2x^2-12x+61)


x^4-11x^3+37x^2+205x-732


to
keep the same coefients, we multiply whole thing by c


c(x^4-11x^3+37x^2+205x-732)


we
don't know what c is


we
know that


f(1)=250


f(1)=c(x^4-11x^3+37x^2+205x-732)


find
c


f(1)=c(1^4-11(1^3)+37(1^2)+205(1)-732)=250


f(1)=c(-500)=250




-500c=250


divide
both sides by -500


c=-1/2




the
polynomial (factored with real coefients) is


f(x)=-0.5(x-3)(x+4)(2x^2-12x+61)