Find nth degree polynomial function

Question

Find an nth degree polynomial function with real coefficients satisfying the given conditions. Of you are using a graphing utility, use it to graph the function and verify the real zeros and the given function value

N=3

-4 and 6+5i are zeros;

f(2)=246

Answer 1

If -4 is a zero, then (x + 4) is a factor of f(x).

If (6+5i) is a zero, then [x - (6+5i)] is a factor. But so is its conjugate (6 - 5i), so...

[(x - (6 - 5i)] is a factor of f(x) too.

f(x) = A (x + 4) [(x - (6 + 5i)][x - ( 6 - 5i)], where A is a real number, but not zero.

There is a formula for multiplying those last two factors together, but here it is in some detail:

[(x - 6) - 5i][(x - 6) + 5i] ; remove the inside parentheses and regroup into a product of conjugates, resulting in a difference of squares:

(x - 6)² - (5i)²

x² - 12x + 36 - 25i²

x² - 12x + 61

Multiply that trinomial by the binomial factor (x + 4):

x³ - 12x² + 61x

+4x² -48x + 244

--------------------------------

x³ - 8x² + 13x + 244

Now use f(2) = 246 to determine the value for A:

246 = A(8 - 32 + 26 +244)

246 = 246A

A = 1

So:

f(x) = x³ - 8x² + 13x + 244 is the function that has zeros: -4, 6+5i, 6 - 5i such that f(2) = 246.

This Desmos graph is in complex mode,:

desmos.com/calculator/b8jzyxb1os