Given the question "<span>Which algebraic expression is a polynomial with a degree of 2?" and the options:
1).
2).
3).
4).
A polynomial </span><span>is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.
</span><span>The degree of a polynomial is the highest exponent of the terms of the polynomial.
For option 1: </span><span>It contains no fractional or negative exponent, hence it is a polynomial. But the highest exponent of the terms is 3, hence it is not of degree 2.
For opton 2: It contains a fractional exponent which violates the definition of a polynomial, hence, it is not a polynomial.
i.e.
For option 3: </span><span>It contains a negative exponent which violates the definition of a polynomial, hence, it is not a polynomial.
i.e.
For option 4: It contains no fractional or negative exponent, hence it is a polynomial. Also, the highest exponent of the terms is 2, hence it is of degree 2.
</span>
Therefore, <span> s a polynomial with a degree of 2. [option 4]</span>
1).
2).
3).
4).
A polynomial </span><span>is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.
</span><span>The degree of a polynomial is the highest exponent of the terms of the polynomial.
For option 1: </span><span>It contains no fractional or negative exponent, hence it is a polynomial. But the highest exponent of the terms is 3, hence it is not of degree 2.
For opton 2: It contains a fractional exponent which violates the definition of a polynomial, hence, it is not a polynomial.
i.e.
For option 3: </span><span>It contains a negative exponent which violates the definition of a polynomial, hence, it is not a polynomial.
i.e.
For option 4: It contains no fractional or negative exponent, hence it is a polynomial. Also, the highest exponent of the terms is 2, hence it is of degree 2.
</span>
Therefore, <span>