## Problema Solution

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

N=4;

2i and 3i are zeros;

F(-1)=100

## Answer provided by our tutors

The complex conjugate root theorem states that if F is a polynomial in one variable with real coefficients, and a + bi is a root of F with a and b real numbers, then its complex conjugate a − bi is also a root of F.

This means that if 2i and 3i are roots then so are -2i and -3i.

We can write the polynomial as:

F(x) = C(x - 2i)(x - 3i)(x + 2i)(x + 3i), where C is constant that we need to find

Since,

F(-1) = 100

we have:

C((-1) - 2i)((-1) - 3i)((-1) + 2i)((-1) + 3i) = 100

........

Click here to see the step by step solution for 'C'

........

C = 2

The polynomial is:

F(x) = 2(x - 2i)(x - 3i)(x + 2i)(x + 3i)

F(x) = 2(x^4 + 13x^2 + 36)

F(x) = 2x^4 + 26x^2 + 72