Zeros of Polynomial Functions

Question

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.

n=3

2 and 4i are zeros;

f(1)=-17

Answer 1

through (1,-17) given f(1) = minus 17

n=3

roots 2, +/-4i

3 degree poly

y= a(x-2)(x-4i)(x+4i) = a(x-2)(x^2+16) = a(x^3 -2x^2+16x-32), plug in point

-17=a(-1)(17)

a=1

a+f(x)=x^3 -2x^2+16x -32

Answer 2

The equation can be expressed as the product of the zero factors time a constant (this can be solved for f(1) = 17

(x-2)(x+4i)(x-4i)k = f(x) which is guaranteed to be 0 at 2, +/- 4i (complex conjugate roots)

(x-2)(x²+16)k = f(x)

(x³-2x²+16x-32)k = f(x)

f(1) = -32k = -17 so k = 17/32

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