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 5i are zeros;
f (-1) = 78

f(x) = - (x - 2)(x - 5i)(x + 5i)

or

f(x) = - x³ + 2x² - 25x + 50

Step-by-step explanation:

Since n = 3, this is a degree 3 polynomial and has total of 3 zero's.

Two of zero's are given:

2 and 5i, the third one must be - 5i (conjugate of 5i).

So the function becomes:

f(x) = a(x - 2)(x - 5i)(x + 5i)

We have f(-1) = 78, using this find the value of a:

f(-1) = a(- 1 - 2)(-1 - 5i)(-1 + 5i) = a( - 3)(1 - 25i²) = -3a*26 = - 78a-78a = 78a = - 1

The function is:

f(x) = - (x - 2)(x - 5i)(x + 5i) ⇒f(x) = -(x - 2)(x² + 25)  ⇒f(x) = -(x³ - 2x² + 25x - 50) ⇒f(x) = - x³ + 2x² - 25x + 50

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