The question asks for a polynomial $p(x)$ with integer coefficients such that $2^{3/5}$ is a root of the polynomial.

Solution:

Let $\alpha = 2^{3/5}$. We need to find a polynomial with integer coefficients that has $\alpha$ as a root.

To eliminate the fractional exponent, we raise both sides of the equation to the power of 5: $\alpha^5 = 2^3 = 8$ This implies: 2^{3/5}^5 = 8 Thus, $\alpha$ satisfies the equation: $x^5 - 8 = 0$ Therefore, the polynomial with integer coefficients that has $2^{3/5}$ as a zero is: $p(x) = x^5 - 8$

This polynomial has the root $2^{3/5}$.

Do you have any further questions, or would you like more details on any step? Here are a few related questions:

1. How can we verify if a given number is a root of a polynomial?
2. Can every irrational root be turned into a polynomial with integer coefficients?
3. What role do exponents play in constructing polynomials from roots?
4. How do we construct polynomials for other types of roots, like square or cube roots?
5. What is the importance of having integer coefficients in a polynomial?

Tip: When dealing with roots involving fractional exponents, raising both sides to eliminate the fraction often simplifies the problem.