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=3 , 3 and 5i are zeros, and f(2)= 58

Answer 1

1. If 3 and 5i are zeros then the 3^rd degree polynomial has factors (x – 3) and (x – 5i)
2. Since the polynomial real coefficients it's other factor is (x + 5i), the complex conjugate of (x – 5i).
3. So the equation for our polynomial has the factored form f(x) = a(x – 3)(x – 5i)(x + 5i)
4. We are given f(2) = 58

This is everything we need. Expanding the complex part of the factored form equation gives

f(x) = a(x – 3)(x² + 25)

f(2) = 58 means that

a((2 – 3)((2)² + 25) = 58

So a(–1)(4 + 25) = a(–29) = =58

which means that a = 58/(–29) = –2

So f(x) = –2(x – 3)(x² + 25) = –2(x³ – 3x² + 25x – 75)

Answer

–2x³ + 6x² – 50x + 150

Answer 2

Since we are given that n = 3 and 3 and 5i are zeros, then -5i is also a zero.

So the equation has the form

f(x) = a (x - 3) (x - 5i) ( x + 5i) Multiplying the binomials gives

f(x) = a (x - 3) (x²+ 25)

Since we are also given that f(2) = 58, we can substitute those values in to find the value of a.

So a (2-3) ( 2² + 25) = 58

a (-1) (4 + 25) = 58

-29 a = 58

a = -2

So the function is f(x) = -2(x-3) (x² + 25). or

f(x) = -2 (x³ - 3x² + 25x - 75) = -2x³ + 6x² - 50x + 150