find nth degree polynomial function with real coefficients with these conditions.

Question

n=3

-3 and 7+2i are zeros

f(-1)=136

f(x)=

Answer 1

Since f(x) has real coefficients and 7+2i is a root, so is 7- 2i

So, 3, 7+2i, and 7-2i are roots

Since f(x) has degree 3, there can't be more than 3 distinct roots.

f(x) = a(x - 3)[x - (7+2i)][x - (7-2i)] = a(x - 3)[(x-7) - 2i][(x-7) + 2i]

= a(x - 3)[(x-7)² - (2i)²]

= a(x - 3)(x² - 14x + 49 + 4)

= a(x - 3)(x² - 14x + 53)

f(-1) = 136, so a(-4)(68) = 136

--272a = 136

a = -1/2

f(x) = (-1/2)(x - 3)(x² - 14x + 53)