Find an nth-degree polynomial function with real coefficients satisfying the given conditions.

Question

N=3

2 and 4i are zeros

F(-1)=51

My instructor explained it but I still don't understand what to do

Answer 1

The key is to interpret what each of these limitations means for the polynomial.

To begin with, N=3, implies that it its of the form f(x) = ax³ + bx² + cx + d, where a, b, c, and d are all real numbers. Also, by the fundamental theorem of algebra, we know that a 3rd degree polynomial should have three zeros (although they can either repeat or be imaginary or irrational). Then, we are given two of the function's zeros. If 2 is a zero of the function, then (x-2) has to be a factor of that polynomial. Also we are given that 4i is a zero, so another factor has to be (x-4i). But we need to find one more zero, and we can find it because imaginary and irrational zeros always come in pairs because we have to have real coefficients. And the pair are always conjugates of each other, or (x+4i) and (x-4i). Thus, we now have all three zeros we expect to have and all three factors (x-2)(x-4i)(x+4i). But we also have to account for the possibility that all the factors are multiplied by some constant (which I will label "a"), so the polynomial has to be like this f(x)=a(x-2)(x-4i)(x+4i).

Now, we can use the fact that that f(-1)=51 to solve for a, i.e.

51=a(-1-2)(-1-4i)(-1+4i). Multiply the terms with "i" first to get rid of the imaginary numbers

(-1-4i)(-1+4i)=1+16=17, so 51=a(-3)(17) or 51=a(-51), or a=-1

so the polynomial is -(x-2)(x-4i)(x+4i)=f(x). You can foil out the final form the the polynomial, but to do so, be sure to multiply the imaginary terms first as it will make things easier.

I hope this helps. John