Definition of a Polynomial

In (x^3-2x^2+5x-7), the highest power of x is 3, so it is a cubic polynomial. It has four terms: x^3, -2x^2, 5x, and -7. Hence, it is a cubic four-term polynomial. A quadratic polynomial would have highest degree 2. Exam tip: find the degree from the highest exponent of the variable, and count terms separated by + or - signs.

The degree of a polynomial is the greatest power of the variable among terms with non-zero coefficients. Here, the coefficient of 0x^5 is 0, so this term does not determine the degree. Among the remaining terms, 8x^3 has the greatest power; therefore, the degree of p(x) is 3. Choosing 5 would be incorrect because a term with a zero coefficient is ignored. Exam tip: first remove terms whose coefficients are zero before finding the degree.

Option A is correct. In √13x² − 4x + 1, √13 is a constant coefficient multiplying x²; it is not an exponent and x is not inside the radical. The powers of x are 2, 1, and 0, all of which are non-negative integers. Polynomial coefficients may be irrational real numbers, so the fact that √13 is irrational does not disqualify the expression. In option B, √x equals x^(1/2), giving a fractional exponent. In option C, x^(-2) has a negative exponent. In option D, √13/x can be written as √13x^(-1), so x has a negative exponent. Therefore only option A meets the power condition required for a polynomial.

In \(x^4+7x-3\), the powers of \(x\) in \(x^4\), \(7x\), and \(-3\) are \(4\), \(1\), and \(0\), respectively. Since \(-3=-3x^0\), a constant term has power \(0\) of the variable. Option B contains \(x^{-1}\), and option C contains the fractional power \(x^{1/2}\). Exam tip: write a constant as a multiple of \(x^0\) to identify its power quickly.

In option B, \(\frac{2x^2+5}{x}=2x+\frac{5}{x}=2x+5x^{-1}\). The exponent of \(x\) is \(-1\). In a polynomial, the exponent of a variable must be a non-negative integer, so this expression is not a polynomial. In contrast, option D has \(x^0=1\), giving \(4+7x\), which is a polynomial. Exam tip: When a variable occurs in the denominator, rewrite it with a negative exponent to test whether the expression is a polynomial.

In p(x)=4x^2-9x+11, the highest power of x is 2, so its degree is 2 and it is a quadratic polynomial. It has three terms: 4x^2, -9x, and 11; therefore, it is also a trinomial. Hence, the correct identification is a quadratic trinomial. A cubic trinomial would have highest degree 3. Exam tip: first find the highest exponent, then count the terms to classify a polynomial.

Option C, \(2x^2+x-5\), has three terms: \(2x^2\), \(x\), and \(-5\). The powers of the variable are \(2\), \(1\), and \(0\), all of which are non-negative integers, so it is a polynomial. Since it has three terms, it is a trinomial; therefore, it is neither a monomial nor a binomial. Option B is a polynomial but has only two terms. Option D contains \(x^{-1}\), so it is not a polynomial. Exam tip: Before counting terms, check that no variable has a negative or fractional exponent.

The powers of x in the polynomial are 6, 4, 2, and 0; the constant term 4 has power 0. All of these are even integers, so the correct choice is only even powers. A polynomial with only odd powers would contain powers such as 1, 3, or 5. Exam tip: Always count the power of a constant term as 0.

Option B is correct. A polynomial in x can contain only non-negative integer powers of x. The exponent -1/2 is both negative and fractional, so it is not permitted. In fact, x^(-1/2) can be rewritten as 1/√x, which places the variable in a denominator and makes the violation especially clear. The other options satisfy the exponent rule: option A has powers 2, 1, and 0; option C has powers 0 and 1; and option D has powers 3 and 0. All of those are non-negative integers. Therefore B is the only expression that is not a polynomial, and its defining defect is precisely the exponent -1/2.

When \(b\) is treated as a constant coefficient, the powers of \(x\) in \(bx^4-3bx+2\) are \(4,1\), and \(0\). All are non-negative integers, so it is a polynomial in \(x\). Option B has \(\frac{1}{x}=x^{-1}\), and option C has \(x^{-4}\); both contain negative powers. In option D, \(\sqrt{x}=x^{1/2}\), which gives a fractional power. Exam tip: powers of the variable in a polynomial can only be \(0,1,2,\ldots\).

The expression has four terms: x^3, x^2, x, and 1. Its highest power of x is 3, so its degree is 3 and it is a cubic polynomial. Option B has an incorrect number of terms, while option C has an incorrect degree. Exam tip: To find the degree of a polynomial, identify the greatest exponent of the variable among terms with non-zero coefficients.

Polynomial terms use fixed non-negative integer powers of the variable. (\tan x) is not such a term.

A non-zero term in a polynomial has a coefficient that is not zero. Here, \(5x^3\) has coefficient 5 and the constant term \(-12\) has coefficient \(-12\), so both are non-zero terms. The terms \(0x^2\) and \(0x\) have coefficient 0 and do not contribute any term to the polynomial. Therefore, there are 2 non-zero terms. Exam tip: Remove terms with zero coefficients before counting terms.

In (4x^4-3x^2), there is no separate constant term, so the constant term is considered (0). All powers are valid.

(\frac{x^4+x^3}{x^2}=x^2+x). It is a polynomial, so the correct reason is that powers (2) and (1) are valid.

On expanding


y((x+2)^2), we get


y(x^2+4x+4). The highest power of x is 2, so the polynomial has degree 2 and is quadratic. Brackets do not make an expression non-polynomial; they can be removed by expansion. Exam tip: Simplify or expand bracketed expressions before identifying the degree of a polynomial.

The direct answer is option A: 5+x^3-2x is a polynomial, but its terms are not written in descending powers. A polynomial allows only non-negative integer powers of x. Rewrite the terms mentally as 5x^0+x^3-2x^1. The powers are 0, 3, and 1, and all are valid, so the expression is a polynomial. Descending order would be x^3, then x^1, then x^0, giving x^3-2x+5; this is only the standard arrangement, not a condition for being a polynomial. Option A is correct because it satisfies the polynomial rule while being unsorted. Option B is wrong because x^{-1} has a negative exponent. Option C is wrong because sqrt{x}=x^{1/2} has a fractional exponent. Option D is wrong because 1/x=x^{-1}, also a negative exponent. Exam cue: first check whether powers are allowed; arrange powers only afterward.

Substitute a = 2 into the coefficient of x³. We obtain a − 2 = 2 − 2 = 0, so the cubic term becomes 0x³ and disappears. Thus p(x) = 0x³ + 5x − 7 = 5x − 7. The highest power of x with a non-zero coefficient is now 1, so p(x) is a linear polynomial. Option C is correct. It is not cubic because the coefficient of x³ is zero. It is not quadratic because no x² term is present. It is not the zero polynomial because 5x − 7 is not identically zero; for example, at x = 0 its value is -7. The degree must be determined after substitution and simplification, not from the original displayed form.

In the second expression, the powers are (4), (2), and (0), and the coefficients are real. So it is a polynomial of degree (4).

The coefficient of \(0x^5\) is 0, so this term has value 0 and is not considered while finding the degree of a polynomial. In the remaining polynomial, \(-3x^2+8x-1\), the highest power of \(x\) is 2. Therefore, the degree is 2. Option 5 is incorrect because a term with zero coefficient does not determine the degree. Exam tip: Remove all zero-coefficient terms first, then identify the highest exponent of the variable.

A linear polynomial has degree 1, so the coefficient of \(x^2\) must be zero. Here, the coefficient of \(x^2\) is \(k+4\). Thus, \(k+4=0\), which gives \(k=-4\). If \(k=0\), the coefficient of \(x^2\) remains 4, so the polynomial is still quadratic. Exam tip: To reduce the degree of a polynomial, make the coefficient of its highest-degree term zero.

The term (|x|) is not written using non-negative integer powers of (x). Therefore it is not a polynomial.

On substituting m=0, mx^4+(m+2)x^3-5 becomes 0x^4+2x^3-5, or 2x^3-5. The highest power of x in the remaining polynomial is 3, so its degree is 3. Option 4 is not correct because the coefficient of x^4 becomes 0, so that term vanishes. Exam tip: After substitution, remove all zero-coefficient terms before identifying the highest exponent.

In \(5x^2-\pi x+\sqrt{3}\), the powers of \(x\) are \(2\), \(1\), and \(0\), all of which are non-negative integers. Hence, it is a polynomial. Although \(\pi\) and \(\sqrt{3}\) are irrational, they are real constants and may be valid polynomial coefficients. Since the highest power is \(2\), it is a quadratic polynomial. Option B is incorrect because the presence of an \(x\)-term alone does not make a polynomial linear; its highest power determines the degree. Exam tip: find the greatest exponent of the variable to determine the degree.

The coefficient of a term is the number multiplying the variable power. If a term is missing from a polynomial, its coefficient is understood to be zero. This convention allows us to compare all powers systematically, even when some terms are not written explicitly. It is different from confusing an exponent with a coefficient: 5 is an exponent in \\(x^5\\), not its coefficient.

The polynomial \\(4x^5-7x^3+2\\) contains no \\(x^4\\) term and no \\(x\\) term. We may write the missing parts as \\(0x^4\\) and \\(0x\\). Therefore the coefficients of \\(x^4\\) and x, respectively, are 0 and 0. Option B is correct; 4 and -7 are the coefficients of \\(x^5\\) and \\(x^3\\).