Definition of a Polynomial

The governing concept is the required form of a polynomial in x. Every exponent of x in a polynomial must be a non-negative integer, while numerical coefficients may be any real numbers. Options A and D clearly satisfy this rule. Option C also satisfies it because √2 is only a constant coefficient multiplying x^3; the radical does not contain x. In option B, 7/(x - 3), the variable x occurs in the denominator. This makes the expression rational rather than polynomial, and it is undefined at x = 3. Hence option B is correct. The presence of a radical constant in option C is not a violation, and a polynomial may contain several terms and constant terms.

In a polynomial, every exponent of the variable must be a non-negative integer. Although x⁴ and the constant 1 satisfy this rule, x^(1/2) has the fractional exponent 1/2, so it violates the definition. Therefore option C is correct. Having a fourth-power term or three terms is allowed, and the constant 1 is a valid degree-zero term.

The governing definition states that every exponent of the variable in a polynomial must be a non-negative integer: 0, 1, 2, 3 and so on. Coefficients may be rational or irrational real numbers, so the presence of √2 as a constant does not disqualify an expression. Options A and B use only allowed integer exponents. Option C has x³ and the constant √2, so it is also a polynomial in x of degree 3. Option D contains x^(5/2); 5/2 is a fractional exponent rather than a non-negative integer, so this expression is not a polynomial in x. Therefore option D is correct. A negative exponent, a variable in a denominator or a radical involving the variable would similarly violate the usual polynomial definition.

The governing concept is the degree of a polynomial: it is the greatest exponent of the variable whose coefficient is non-zero. In p(x) = kx² + 3x + 4, the term kx² normally gives degree 2. To make the polynomial have degree 1, the coefficient of x² must vanish, so k = 0. The expression then becomes p(x) = 3x + 4, whose highest power of x is 1. Therefore option A is correct. Choosing k = 1 would leave an x² term and give degree 2; k = 3 or k = 4 would also leave a non-zero quadratic term. The non-zero coefficient 3 of x ensures that the resulting polynomial is genuinely linear.

A polynomial in x is a finite sum of terms of the form axⁿ, where each exponent n is a non-negative integer. In option A, 1/x⁴ can be rewritten as x⁻⁴. The exponent −4 is negative, so this term violates the defining condition for a polynomial; consequently x⁴ + 1/x⁴ is not a polynomial. Option B has powers 4, 2 and 0, all allowed. Option C uses x⁰ = 1 and has powers 0 and 1, while option D has powers 3 and 0. Thus only option A is invalid. The issue is not merely that a fraction appears: numerical fractional coefficients may be allowed, but a variable in the denominator creates a negative exponent.

To answer, apply two conditions: every exponent must be a non-negative integer, and every power must be even. In option A the exponents are 6, 4, 2 and 0; all are even and all are permitted in a polynomial. The constant term −8 corresponds to x⁰, and zero is even, so it does not violate the condition. Option B contains x⁵, an odd power, although the expression is a polynomial. Option C contains x⁻², a negative exponent, so it is not a polynomial. Option D contains x¹ᐟ², a fractional exponent, so it is not a polynomial. Hence option A is the only expression satisfying both requirements.

The degree of a non-zero polynomial is the greatest exponent of x whose coefficient is non-zero. Here the highest-power term is λx². For the polynomial to retain degree 2, its coefficient λ must not vanish. Thus λ ≠ 0 is necessary. If λ = 0, the x²-term disappears and the expression becomes 4x − 3, which has degree 1; the particular values 4 and −3 are irrelevant to this condition.

The defining test is whether the variable appears only with non-negative integer powers in a finite sum. Option C contains √(x² + 1), in which the variable occurs inside a square root applied to a non-monomial expression. This is not a polynomial term, so √(x² + 1) + x is not a polynomial in x. Option A is a polynomial with powers 2 and 0. Option B is also a polynomial: √2 is a real constant coefficient, not a variable power. Option D has powers 4 and 2 and is plainly a polynomial. Thus option C is correct. The important distinction is that a root of a constant, such as √2, is allowed as a coefficient; a variable-containing radical is a different kind of expression.

Being an integer is not sufficient for an exponent to be permitted in a polynomial. Polynomial exponents must be non-negative integers. In option B, the powers −2 and 3 are both integers, but −2 is negative. Since x⁻² = 1/x², the expression contains the variable in a denominator and therefore is not a polynomial. Option A has powers 2, 1 and 0, all allowed. Option C has powers 4 and 0, also allowed. In option D, x⁰ = 1, so its effective powers are 0 and 1, both valid. Therefore option B is correct. This question distinguishes the broader set of integers, which includes negative numbers, from the narrower set of non-negative integer exponents required by the polynomial definition.

A polynomial in x has only non-negative integer powers of x, and a quadratic polynomial has highest power exactly 2. Option A is 3x²−πx+√7. Its x-powers are 2, 1 and 0, so it meets the polynomial definition and has degree 2. The coefficients π and √7 are real numbers, and irrational coefficients are allowed; they do not change the degree. Hence option A is the required quadratic polynomial. Option B has highest power 3 and is cubic. Option C contains x^(1/2), a fractional power, so it is not a polynomial. Option D contains 1/x=x^−1, a negative power, and is also not a polynomial. Thus the degree and exponent rules identify A unambiguously.

The governing definition permits real coefficients, including irrational numbers such as √5 and rational fractions such as 3/8. It requires only that the powers of x be non-negative integers. In the expression, x⁰ is the constant 1, −√5x⁴ has power 4, and 3x/8 can be written as (3/8)x, whose power is 1. Every term therefore satisfies the polynomial rule, and the greatest power with a non-zero coefficient is 4. The expression is a polynomial of degree 4, so option B is correct. It is not the zero polynomial because its terms do not all vanish. Options A and D incorrectly treat allowable coefficients as forbidden; coefficients need not be integers for an expression to be a polynomial.

In a polynomial in x, every exponent of x must be a non-negative integer: 0, 1, 2, 3, and so on. The terms x⁶ and 2 satisfy this rule, but x^(3/2) has the fractional exponent 3/2. Therefore q(x) is not a polynomial in x. The number of terms and the presence of a constant do not violate the definition.

A polynomial in x may have real coefficients, including irrational numbers such as √2 and fractional numbers such as −4/9. The essential restriction is on the exponents of x: they must be non-negative integers. In the given expression, x⁰ is a constant term, √2x⁵ has exponent 5, and −4x²/9 has exponent 2. Every exponent is therefore allowed, and the coefficient of x⁵ is non-zero. The greatest exponent is 5, so the expression is a polynomial of degree 5. Option B is correct. Options A and D impose an incorrect restriction that coefficients must be integers, while C is wrong because the expression contains non-zero terms.

A polynomial may contain only non-negative integer powers of its variable. The extra condition here is that every non-zero variable term must have an even exponent. In option B, the variable terms are 3x⁴ and −2x²; their exponents 4 and 2 are both even, and the constant 9 is permitted because constants have exponent 0, which is also even. Option A contains x⁵, an odd power. Option C contains x³ and x, both odd powers. Option D contains √x = x^(1/2), which is not a polynomial term because 1/2 is not an integer. Thus B is the only option satisfying both the polynomial rule and the even-power condition.

The definition of a polynomial in x permits only non-negative integer powers of x, such as 0, 1, 2, 3 and so on. Option B contains x^(7/3), whose exponent is fractional rather than an integer. Therefore it does not satisfy the polynomial definition, even though the other terms 2x and −5 are acceptable. Option A has exponents 7, 1 and 0, so it is a polynomial. Option C has exponents 3 and 0, and option D has exponents 1 and 0; both are also valid polynomials. The coefficients may be fractional or irrational without causing a problem, but a fractional exponent of the variable is not allowed. Hence option B is uniquely correct.

To classify the expression, apply three rules: polynomial exponents must be non-negative integers, the degree is the greatest exponent with a non-zero coefficient, and terms are counted after combining like terms. Option A has four distinct terms: x⁶, −2x⁴, x, and −3. Its exponents are 6, 4, 1 and 0, all allowed, and its greatest exponent is 6. Therefore it is a four-term polynomial of degree 6. Option B is not a polynomial because it has the negative exponent −2. Option C has four terms but degree 5. Option D has the fractional exponent 6/5. Thus only option A satisfies both required conditions.

A polynomial requires every exponent of the variable to be a non-negative integer. Merely being an integer is not enough. In option B, the exponent −3 is an integer but it is negative. Since x⁻³ equals 1/x³, the variable occurs in a denominator, which violates the definition of a polynomial in x. Option A has exponents 4, 2 and 0; option C has 3, 1 and 0; and option D has 1 and 0. All these exponents are non-negative integers, so A, C and D are polynomials. The negative exponent is therefore the decisive reason that B is not a polynomial, making option B the unique correct answer.

A polynomial in x is a finite sum of terms of the form axⁿ, where a is a constant and n is a non-negative integer. The term log x is logarithmic, not a power of x with an allowed integer exponent. Consequently, option A is not a polynomial in x. Option B uses powers 2, 1 and 0, so it is a polynomial. In option C, √2 is merely a constant coefficient and can be viewed as √2x⁰; its presence does not cause a problem. Option D uses powers 0 and 1. Thus irrational constants and addition are allowed, but logarithmic dependence on x is not. Therefore option A is the unique correct answer.

The condition x ≠ 0 makes division by x valid. Divide each term in the numerator by x: x⁴/x = x³, x²/x = x, and x/x = 1. Therefore (x⁴ + x² + x)/x simplifies to x³ + x + 1. The resulting expression has exponents 3, 1 and 0, all of which are non-negative integers, so it is a polynomial. Option B incorrectly keeps the first exponent as 4. Option C incorrectly changes x²/x into 1/x, whereas x²/x equals x for x ≠ 0. Option D ignores the given restriction and the valid simplification. Hence option A is correct; the original fraction remains undefined at x = 0, but the obtained polynomial formula is valid on the stated domain.

The governing concept is that every exponent of the variable in a polynomial must be a non-negative integer. In the expression x^2 + x^(-2) + x^0, the powers are 2, -2, and 0. Although 2 and 0 are permitted, -2 is not permitted. In fact, x^(-2) equals 1/x^2, which places the variable in a denominator. A single invalid term is enough to make the entire expression non-polynomial. Therefore option C is correct. Having three terms, or containing x^2 or x^0, does not override the negative exponent. The presence of one valid term cannot make an expression polynomial when another term violates the definition.

In \(5x^{-3}+1\), the exponent of \(x\) is \(-3\), which is negative. In a polynomial, the exponents of variables must be non-negative integers, so this expression is not a polynomial. \(7x^0-4x\) is a polynomial because \(x^0=1\). Exam tip: while identifying polynomials, check that no variable has a negative or fractional exponent, or appears in the denominator.

The degree of a polynomial is the greatest exponent of the variable in any term with a non-zero coefficient. The exponents of the terms here are 7, 3, 1, and 0. The greatest is 7, so the degree is 7. The number 8 is a constant term, not an exponent of x. Exam tip: a non-zero constant has degree 0.

Direct answer: Option A, constant polynomial, is correct. The number 23 contains no variable, so it does not change when x changes. It can be written as 23x^0 because x^0 equals 1 for non-zero x, so the only effective term has exponent 0. A non-zero constant polynomial therefore has degree 0. Option B, linear polynomial, must have degree 1 and normally contains a non-zero x term, which 23 does not. Option C, quadratic polynomial, must have degree 2 and contain a non-zero x^2 term, which is absent. Option D, zero polynomial, is specifically the polynomial whose value is always 0; 23 is non-zero, so it is not the zero polynomial. The word constant means the value stays fixed, not that it must be zero. Memory cue: no variable and a non-zero number means constant polynomial of degree 0; only the number 0 is the zero polynomial.

A linear polynomial has highest power of the variable equal to 1, with a non-zero coefficient of that variable. In \(5t-9\), the highest power of \(t\) is 1, so it is a linear polynomial. \(t^2+1\) is quadratic, while \(12\) is a constant polynomial of degree 0. Exam tip: To find the degree of a polynomial, look for the greatest exponent of its variable.

(0) is the zero polynomial but its degree is not defined. Keep it separate from a non-zero constant polynomial.