Find an nth-degree polynomial function with real coefficients, a leading coefficient of 1, and satisfying the given conditions n = 4; - 3, 5 and 4 + 3 are zeros

Answer 1

Considering the "4 + 3" should be "4 + 3i"

P(x) = (x + 3)(x - 5)(x - (4 + 3i))(x - (4 - 3i))

Answer 2

Hi Myles,

Simplest function is:

f(x)=(x+3)(x-4)(x-5)(x-3)

"Zeros" mean that when x is the given value, f(x)=0. The above satisfies all four of the zeros you were given--x=-3, 5, 4, 3. Plug any of those values into the above, and you will get f(x)=0. Now, you were also told it was a fourth-degree polynomial (n=4). From that, you know you will need four total x terms, which we also have here. Now, two of those terms, (x+3)(x-3) can be combined to:

(x²-9)

I do not know how much simplification your instructor wants you to do. You could go with the function as written above or:

f(x)= (x²-9)(x-4)(x-5)

Those two expressions are mathematically equivalent. I hope this helps.