To determine which algebraic expressions are polynomials, we need to follow the rules that define a polynomial. A polynomial is an expression that consists of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents on the variables.
Let's evaluate each given expression according to these rules:
1. \( 2x^3 - \frac{1}{x} \)
This expression includes the term \(\frac{1}{x}\), which can be rewritten as \(x^{-1}\). Since \(x^{-1}\) has a negative exponent, this expression is not a polynomial.
2. \( x^3 y - 3x^2 + 6x \)
This expression includes terms \(x^3 y\), \(-3x^2\), and \(6x\). All variables (both \(x\) and \(y\)) are raised to non-negative integer exponents. Therefore, this expression is a polynomial.
3. \( y^2 + 5y - \sqrt{3} \)
This expression includes terms \(y^2\), \(5y\), and \(-\sqrt{3}\). Although \(\sqrt{3}\) is an irrational number, it is a constant term and the other terms have non-negative integer exponents. Therefore, this expression is a polynomial.
4. \( 2 - \sqrt{4x} \)
This expression includes the term \(\sqrt{4x}\). The term \(\sqrt{4x}\) can be rewritten as \(2\sqrt{x}\), and \(\sqrt{x}\) is equivalent to \(x^{1/2}\). Since \(x^{1/2}\) has a fractional exponent, this expression is not a polynomial.
5. \( -x + \sqrt{6} \)
This expression includes the terms \(-x\) and \(\sqrt{6}\). \(\sqrt{6}\) is a constant term and \(-x\) has a non-negative integer exponent. Therefore, this expression is a polynomial.
6. \( -\frac{1}{3} x^3 - \frac{1}{2} x^2 + \frac{1}{4} \)
This expression includes the terms \(-\frac{1}{3}x^3\), \(-\frac{1}{2}x^2\), and \(\frac{1}{4}\). All variables have non-negative integer exponents, and the coefficients are rational numbers. Therefore, this expression is a polynomial.
Based on the evaluation, the following expressions are polynomials:
- \( x^3 y - 3x^2 + 6x \)
- \( y^2 + 5y - \sqrt{3} \)
- \( -x + \sqrt{6} \)
- [tex]\( -\frac{1}{3} x^3 - \frac{1}{2} x^2 + \frac{1}{4} \)[/tex]
