Question

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

n = 3;
2 and 2 i are zeros;
f(-1) = 30

Answer 1

`\displaystyle f(x) = -2(x-2)(x^2+4)`

Explanation:

We want to find a third degree polynomial with zeros x = 2 and x = 2i and f(-1) = 30.

First, note that by the Complex Root Theorem, since x = 2i is a root, x = -2i must also be a root.

Hence, we will have the three factors:

`\displaystyle f(x) = a(x-(2))(x-(2i))(x-(-2i))`

Where a is the leading coefficient.

Expand and simplify the second and third factors:

`\displaystyle \begin{aligned} (x-(2i))(x-(-2i)) &= (x-2i)(x+2i) \\ \\ &= x(x-2i)+2i(x-2i) \\ \\ &= (x^2 - 2ix) + (2ix - 4i^2) \\ \\ &=x^2 + 4\end{aligned}`

Hence:

`\displaystyle f(x) = a(x-2)(x^2+4)`

Since f(-1) = 30:

`\displaystyle \begin{aligned} f(x) &= a(x-2)(x^2+4) \\ \\ (30) &= a((-1)-2)((-1)^2+4) \\ \\ 30 &= -15a \\ \\ a&= -2\end{aligned}`

In conclusion, third degree polynomial function is:

`\displaystyle f(x) = -2(x-2)(x^2+4)`