Question

Find the nth degree polynomial function with real coefficients satisfying the given conditions. n=3; -2 and 2+3i are zeros; leading coefficient is 1.

Answer 1

To find the nth degree polynomial function with real coefficients satisfying the given conditions, we consider the zeros and the leading coefficient. By multiplying the factors (x+2)(x-(2+3i))(x-(2-3i)), we get the polynomial function f(x) = x^3 - 2x^2 + 5x + 26.

Step-by-step explanation:

To find the nth degree polynomial function with real coefficients, we need to consider the given zeros and leading coefficient. Since -2 and 2+3i are zeros, the associated conjugate 2-3i will also be a zero. This means that the polynomial will have a factor of (x+2)(x-(2+3i))(x-(2-3i)).

Now we can multiply these factors to find the polynomial function. Start by multiplying (x-(2+3i))(x-(2-3i)). This can be simplified using the difference of squares: (x-2-3i)(x-2+3i) = (x-2)^2 - (3i)^2 = x^2 - 4x + 4 - 9i^2 = x^2 - 4x + 13.

Multiplying this by (x+2), we get the final polynomial function: f(x) = (x+2)(x^2 - 4x + 13) = x^3 - 2x^2 + 5x + 26.

Answer 2

f(x) = x^3 -2x^2 +5x +26

Step-by-step explanation:

When p is a root, (x -p) is a factor.

If the polynomial has real coefficients, its complex roots come in conjugate pairs. Then the linear factorization for the given roots is ...

f(x) = (x +2)(x -2-3i)(x -2+3i) = (x +2)(x^2 -4x +13)

f(x) = x^3 -2x^2 +5x +26