Definition of a Polynomial

(\sqrt{3}) is a real coefficient and is allowed. The powers of the variable are (2), (1), and (0), so it is a polynomial.

A quadratic polynomial has highest degree 2, and every exponent of the variable must be a non-negative integer. In \(5x^2-4x+9\), the highest power of \(x\) is 2, so it is quadratic. \(x^3+x+1\) is cubic, while \(x^{-2}+5\) is not a polynomial because it contains a negative exponent. Exam tip: identify the highest valid exponent of the variable to find the degree of a polynomial.

The degree of a polynomial is the highest power of the variable among terms with non-zero coefficients. Here, 0x^7 is a zero term, so it does not contribute to the degree. Among the remaining terms, 3x^5 has the highest power; therefore, the degree is 5. Choosing 7 would be incorrect because the coefficient of x^7 is zero. Exam tip: Remove terms with zero coefficients before finding the degree.

In (x^{2/3}), the power is (\frac{2}{3}), which is not an integer. Polynomial powers must be non-negative integers.

A polynomial whose highest power of the variable is 3 is called a cubic polynomial. In \(4x^3-6x+1\), the highest power of \(x\) is 3, so it is cubic. \(x^2+8\) is quadratic because its degree is 2. Exam tip: To find the degree of a polynomial, identify the greatest exponent of the variable.

The direct answer is C, constant polynomial. The expression is the number 12 alone, so it has no variable. A constant polynomial has a fixed value and can be understood as 12x^0; its degree is 0 because the highest non-zero power is 0. Option A, zero polynomial, is incorrect because the zero polynomial is 0, not 12. Option B, linear polynomial, is incorrect because a linear polynomial must have degree 1 and normally contains a term such as 5x. Option C is correct: 12 is a non-zero constant polynomial, and its value remains 12 for every permitted value of x. Option D, cubic polynomial, is incorrect because a cubic polynomial has degree 3, with a non-zero term such as x^3. The absence of a variable is the important clue, not the size of the number. Exam reminder: a non-zero number written alone is constant of degree 0; do not call it zero merely because it has no variable.

The governing concept is the definition of a polynomial: the variable may have only non-negative integer powers, and it cannot occur in a denominator after the expression is written in standard form. Option B gives (x + 5)/x = 1 + 5/x = 1 + 5x⁻¹. The exponent −1 is not a non-negative integer, so this expression is not a polynomial. In the other displayed options, the same non-zero expression appears in numerator and denominator, so each simplifies to the constant 1 on its domain and is therefore a polynomial expression after simplification. Thus B is the intended and unambiguous answer.

In \(3z^4-z^2+8\), the powers of \(z\) are \(4\), \(2\), and \(0\), all of which are non-negative integers. Therefore, it is a polynomial in \(z\). In \(z^{-1}+2\) and \(\frac{6}{z}+1\), \(z\) has a negative power, while \(\sqrt{z}+5\) contains \(z^{\frac12}\); hence, these are not polynomials. Exam tip: powers of the variable in a polynomial must be \(0,1,2,\ldots\).

In a polynomial, terms are separated by plus (+) or minus (−) signs. Here the terms are \(5x^3\), \(-2x^2\), \(7x\), and \(-1\). Therefore, there are 4 terms. The negative sign is part of the terms \(-2x^2\) and \(-1\); they are not omitted. Exam tip: Count each expression separated by a + or − sign as one term.

The degree of a polynomial is the highest exponent of the variable with a non-zero coefficient. Since the highest power is given as 4, its degree is 4. The degree does not become 8; it is determined only by the greatest exponent present in the polynomial. Exam tip: Ignore terms with zero coefficient and select the largest remaining exponent.

The governing concept is the definition of a polynomial in x. Every exponent of x must be a non-negative integer, although coefficients may be irrational constants. In option C, √x can be written as x^(1/2), and 1/2 is not an integer. Therefore √x + 2x is not a polynomial in x, making option C correct. In option A, √2 is only a constant coefficient, so x² + √2 is a valid polynomial of degree 2. Similarly, √5 is a constant coefficient in option B, which is a valid linear polynomial. Option D, 7x³ − 1, is also valid and has degree 3. Thus the radical involving the variable, rather than a radical constant, causes the failure.

A linear polynomial has degree 1, meaning that the highest power of its variable is 1. In 6x-11, the highest power of x is 1, so it is linear. In contrast, 2x^2+x is quadratic, x^3+1 is cubic, and 15 is a constant polynomial of degree 0. Exam tip: To find the degree of a polynomial, identify the highest power of the variable.

In a polynomial, every exponent of the variable must be a non-negative integer such as \(0,1,2,\ldots\). Here, the exponent of \(x\) in \(x^{1/3}\) is \(1/3\), which is not an integer, so the expression is not a polynomial. The term \(x^2\) and the constant \(4\) are valid in a polynomial, and the number of terms does not matter. Exam tip: first check the exponents of variables when identifying a polynomial.

A non-zero constant polynomial has degree 0 because it contains no variable term. Since 13 is a non-zero constant polynomial, its degree is 0. The degrees of x+5 and x^2+1 are 1 and 2 respectively. The degree of the zero polynomial is not defined. Exam tip: identify the highest power of the variable; for a non-zero constant, it is 0.

Treating (a) as a constant, (ax^2) has (x)-power (2). All powers of (x) are non-negative integers.

\(4x^2+\frac{3}{7}x-6\) is a polynomial because the powers of \(x\) are \(2\), \(1\), and \(0\), all of which are non-negative integers. Polynomial coefficients may be fractional or negative, so neither \(\frac{3}{7}\) nor the constant term \(-6\) makes it invalid. A zero polynomial has every coefficient equal to zero, which is not the case here. Exam tip: to identify a polynomial, first check the exponents of the variable.

The power in (x^{-2}) is negative, so it is not a polynomial. (7) is only a constant term, and the issue is the negative power.

In a polynomial, the powers of a variable must be zero or positive integers. Here, the powers of x are 0, 1, and 2, so the expression is a polynomial. Also, x^0 = 1, so 9x^0 = 9 is a constant term; therefore, option C is incorrect. Option A is also wrong because x^0 = 1. Exam tip: while identifying a polynomial, check that no variable has a negative, fractional, or variable exponent.

The degree of a polynomial is the highest exponent of the variable with a non-zero coefficient. In \(6x^3-x+5\), the highest power of \(x\) is \(3\), so its degree is not \(2\). In all the other options, the highest power is \(2\). Exam tip: To find a polynomial's degree, identify the term with the greatest exponent of the variable.

In a polynomial, every exponent of the variable must be a non-negative integer. Here, \(\sqrt{x}=x^{1/2}\), so the expression is not a polynomial. An integer exponent in \(x^3\) alone is not enough. Exam tip: rewrite radicals as fractional powers before checking.

In 3x^2-5, the highest power of x is 2, so its degree is quadratic. It has two non-zero terms, 3x^2 and -5, making it a binomial. Hence, it is a quadratic binomial. A linear binomial would have highest power 1. Exam tip: identify the highest exponent first, then count the non-zero terms.

In a polynomial, the exponents of variables must be zero or positive integers. Therefore, an expression containing a negative exponent, such as \(x^{-1}\), is not a polynomial because it places the variable in the denominator. A zero exponent gives a constant term, and real coefficients are allowed. Exam tip: while identifying a polynomial, check that no variable has a negative or fractional exponent.

The governing concept is the classification of polynomials by degree. A constant polynomial contains no variable term and has degree zero, while a non-constant polynomial contains a variable with a positive whole-number power. Option C, 7x + 3, uses powers 1 and 0, so it is a valid polynomial and has degree 1; it is therefore not constant. Options A, B and D are numbers without x. Each is a constant polynomial, including the zero polynomial, whose degree is treated separately in many textbooks but which is still constant. Hence option C is the only suitable answer.

(\sqrt{x^2}) is not treated as a polynomial term in the usual class (9) definition. If the variable is under a root, it is not considered a polynomial.

The degree of a polynomial is the highest power of its variable with a non-zero coefficient. Here, x has power 1 in 7x, while -4 is a constant term. Therefore, p(x)=7x-4 has degree 1 and is a linear polynomial. Degree 0 applies only to non-zero constant polynomials. Exam tip: identify the type of a polynomial by finding the highest exponent of the variable.