Final answer:
A polynomial fulfilling the given criteria is p(x) = -(x^6 + 5x^4 + 5x^2 + 4), with an even degree of 6, 4 different zeros, and a reflection across the x-axis.
Step-by-step explanation:
To create a polynomial that meets the stated requirements, we must consider each condition carefully. The polynomial must:
- Have 4 different zeros.
- Have an even degree that is greater than 4.
- Cross through exactly 2 x-intercepts.
- Have been reflected across the x-axis.
Let's consider the most straightforward even degree that is greater than 4, which is 6. A degree 6 polynomial can be expressed with the general form p(x) = ax^6 + bx^5 + cx^4 + dx^3 + ex^2 + fx + g, where a, b, c, d, e, f, and g are constants. To ensure the polynomial crosses through exactly 2 x-intercepts and is reflected across the x-axis, we'll need to include complex conjugate pairs and pairs of imaginary roots. Hence, our polynomial could look like:
p(x) = a(x - r)(x + r)(x - (s + ti))(x - (s - ti))(x - (u + vi))(x - (u - vi))
Where r is a real number, s and u are real parts of the complex zeros, and t and v are their imaginary parts respectively. For simplicity, let's choose r = 1, s = 0, t = 1, u = 0, and v = 2. To ensure the reflection across the x-axis, we will need a to be negative (assuming the polynomial's leading coefficient was positive before reflection).
Therefore, we can have:
p(x) = -1(x - 1)(x + 1)(x - i)(x + i)(x - 2i)(x + 2i)
Expanding this, we have:
p(x) = -1(x^2 - 1)(x^2 + 1)(x^2 + 4)
Further expansion gives us a polynomial which satisfies all the conditions:
p(x) = -(x^6 + 5x^4 + 5x^2 + 4)
This polynomial has an even degree of 6, has 4 different zeros (namely 1, -1, i, -i, 2i, -2i), crosses at (1, 0) and (-1, 0) for x-intercepts and has been reflected across the x-axis due to the negative leading coefficient.