Determine the Equation

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. If n=4 , 2i and 5i are zeros, and f(-1)= 130

Answer 1

1. If 2i and 5i are zeros then the 4th degree polynomial has factors (x – 2i) and (x – 5i)
2. Since the polynomial has real coefficients it's other two factors are (x + 2i) and (x + 5i), the complex conjugates of the first two factors.
3. So the equation for our polynomial has the factored form f(x) = a(x – 2i)(x + 2i)(x – 5i)(x + 5i)
4. We are given f(–1) = 130

This is everything we need. Expanding the two parts of the factored form equation gives

f(x) = a(x² + 4)(x² + 25)

f(–1) = 130 means that

a((–1)² + 4)((–1)² + 25) = 130

So a(1 + 4)(1 + 25) = a(5)(26) = a(130) = 130

which means that a =1

So f(x) = (x² + 4)(x² + 25)

Answer

f(x) = x⁴ + 29x² + 100