Answer :

To determine if the expression \(4n^3 + 5n^2 - 3nt + 6\sqrt{n} + 8\) is a polynomial, we need to consider the fundamental definition of what constitutes a polynomial.

A polynomial in one variable \(n\) is an expression consisting of terms in the form \(a_n n^k\), where \(a_n\) is a coefficient, \(n\) is the variable, and \(k\) is a non-negative integer (0, 1, 2, 3, ...). Additionally, every term should consist of the variable raised only to a non-negative integer power, and operations involved should only be addition, subtraction, and multiplication by a scalar.

Let’s break down the given expression term by term:

1. Term 1: \(4n^3\)
- This term is in the proper form \(a_n n^k\) where \(a_n = 4\) and \(k = 3\). This is a valid polynomial term.

2. Term 2: \(5n^2\)
- Similarly, this is in the form \(a_n n^k\) where \(a_n = 5\) and \(k = 2\). This is also a valid polynomial term.

3. Term 3: \(-3nt\)
- Here, we have the term \(nt\). For this term to be a part of a polynomial in \(n\), the exponent of \(n\) must be a non-negative integer, and the coefficient should not contain any variables. However, the presence of another variable \(t\) violates this rule, suggesting it may not be part of a single-variable polynomial.

4. Term 4: \(6\sqrt{n}\)
- This term is written as \(6n^{1/2}\), where the exponent is \(1/2\). Polynomial terms must have integer exponents, and since \(1/2\) is not an integer, this term does not qualify as part of a polynomial.

5. Term 5: \(8\)
- This is a constant term which is valid in any polynomial.

Reviewing each term, we note that the presence of \(nt\) and \(6n^{1/2}\) disqualifies the expression from being a polynomial. Specifically:
- The term \(-3nt\) includes a variable \(t\) that is not part of a univariate expression.
- The term \(6\sqrt{n}\) introduces a fractional exponent, which is not allowed in a polynomial definition.

Thus, based on these observations, we conclude that the given expression:

[tex]\[4n^3 + 5n^2 - 3nt + 6\sqrt{n} + 8\][/tex]

is not a polynomial.

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