To determine why the given expression is not a polynomial, we need to closely analyze its components. The expression in question is:
[tex]\[
\frac{33}{16} - 62 y^2 xy - 35 z^{\frac{1}{3}} y^2
\][/tex]
A polynomial expression has the following characteristics:
1. The exponents of the variables must be non-negative integers (0, 1, 2, etc.).
2. The expression cannot have variables in the denominator.
3. The coefficients of the terms can be any real numbers (they can be negative or positive).
Let’s analyze each term in the given expression:
1. [tex]\(\frac{33}{16}\)[/tex]: This is a constant term and it fits within the definition of a polynomial since it does not involve variables and hence doesn't have an exponent issue.
2. [tex]\(-62 y^2 xy\)[/tex]: This term involves the variables [tex]\(y\)[/tex] and [tex]\(x\)[/tex]. It can be simplified as [tex]\(-62 y^3 x\)[/tex] by combining the exponents of [tex]\(y\)[/tex]. Here, the exponents of [tex]\(y\)[/tex] and [tex]\(x\)[/tex] (3 and 1 respectively) are non-negative integers, so this term also fits within the definition of a polynomial.
3. [tex]\(-35 z^{\frac{1}{3}} y^2\)[/tex]: This term involves the variables [tex]\(z\)[/tex] and [tex]\(y\)[/tex]. The exponent of [tex]\(y\)[/tex] is 2, which is a non-negative integer. However, the exponent of [tex]\(z\)[/tex] is [tex]\(\frac{1}{3}\)[/tex], which is a fraction.
The key point is in the third term: [tex]\(z^{\frac{1}{3}}\)[/tex]. The exponent [tex]\(\frac{1}{3}\)[/tex] is a fractional value.
Since polynomials can only have variables raised to non-negative integer exponents, the presence of the term [tex]\(z^{\frac{1}{3}}\)[/tex] makes the entire expression a non-example of a polynomial.
Therefore, the best statement to demonstrate why the given expression is not a polynomial is:
The expression has a variable raised to a fraction.
