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

Answer 1

• f(x) = - (x - 2)(x - 5i)(x + 5i)

or

• f(x) = - x³ + 2x² - 25x + 50

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 = 78
• a = - 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