SOLUTION: Find an nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing utility, use it to graph the function and verify th

Question 1157875: Find an nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing utility, use it to graph the function and verify the real zeros and the given function value.
nequals3;
2 and 4 i are zeros;
f left parenthesis 1 right parenthesis equals negative 17
f left parenthesis x right parenthesisequals
nothing

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
2 and 4i and -4i are all zeros, since complex roots always occur in conjugate pairs.
f(1)=-17
f(x)=0 ? or f(x)=.
nothing can mean a blank, or it can mean a zero.
In any case, the factors are (x-2)(x-4i)and (x+4i)
(x^2-16i^2)=(x^2+16) is another factor, when I multiply the second and the third together.
So the polynomial is (x-2)(x^2+16)=x^3-2x^2+16x-32
put in the form a(x^3-2x^2+16x-32), since it may be a multiple of that.
f(1)=-17
so a(1-2+16-32)=-17
a(-17)=-17, so a=1
the answer is x^3-2x^2+16x-32

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SOLUTION: Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify th

SOLUTION: Find an nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing utility, use it to graph the function and verify th