The given polynomial function is:

$f(x) = x^4 - x^7$

We can factor this polynomial as follows:

$f(x) = x^4(1 - x^3)$

Now, set the equation equal to zero to find the zeros:

$x^4(1 - x^3) = 0$

This product will be zero if either $x^4 = 0$ or $1 - x^3 = 0$.

Case 1: $x^4 = 0$

This gives $x = 0$.

Case 2: $1 - x^3 = 0$

Solve for $x$:

$x^3 = 1$

Taking the cube root of both sides:

$x = 1$

The cubic equation $x^3 = 1$ also has two complex cube roots of unity:

$x = e^{2\pi i/3}, \quad x = e^{4\pi i/3}$

So, the zeros of the polynomial are:

$x = 0, \, 1, \, e^{2\pi i/3}, \, e^{4\pi i/3}$

Conclusion:

The function has 4 zeros: one real zero at $x = 0$, another real zero at $x = 1$, and two complex zeros $x = e^{2\pi i/3}$ and $x = e^{4\pi i/3}$.

Would you like further clarification or additional details?

Related questions:

1. What are the cube roots of unity and their significance in this problem?
2. How can we verify the zeros of a polynomial graphically?
3. Can you explain how the fundamental theorem of algebra applies to this problem?
4. What is the geometric interpretation of complex zeros?
5. How would the zeros change if we had a different power for $x^7$?

Tip: Complex roots often come in conjugate pairs, which means their geometric representations are symmetric in the complex plane.