Definition of a Polynomial

On substituting \(a=0\), the term \(ax^2\) becomes \(0\). Thus, \(p(x)=2x^3+5\). The highest power of \(x\) with a non-zero coefficient is 3, so the degree remains 3. Option A is incorrect because the \(x^3\) term is still present even though the \(x^2\) term disappears. Exam tip: remove all terms with zero coefficients before finding the degree.

A polynomial in x is formed from terms in which the power of x is a non-negative integer: 0, 1, 2, 3, and so on. Any other symbol, such as y, may be treated as a fixed coefficient when the expression is considered only in x. In option A, y^2 is therefore a coefficient of x^3, and the expression has x-powers 3, 1, and 0. The highest power is 3, so it is a polynomial in x of degree 3. Option B contains y/x, which introduces x^(-1); option C explicitly contains a negative power; and option D contains sqrt(x), equivalent to x^(1/2). Negative and fractional powers of x are not allowed. Hence option A is the only correct choice.

In \(x^6+2x^4-x^2+9\), the powers of \(x\) are \(6,4,2\), and the constant term \(9\) has power \(0\). All of these are even, non-negative integers, so it is a polynomial with only even powers. Option B contains odd powers, while C has a negative power and D has a fractional power because \(\sqrt{x}=x^{1/2}\); hence they are not the required polynomial. Exam tip: in a polynomial, the exponent of a variable must be a non-negative integer.

The parts of a polynomial separated by plus or minus signs are called terms. Here the terms are \(x^4\), \(2x^3\), \(x^2\), \(x\), and \(1\). Therefore, there are 5 terms. \(x^4\) and \(2x^3\) cannot be combined because they have different exponents. Exam tip: Count every positive or negative part as a separate term.

In x^2-3x+2, the highest power of x is 2, so it is a quadratic polynomial. It has three terms: x^2, -3x, and 2; hence, it is a trinomial. Therefore, it is a quadratic trinomial polynomial. A linear polynomial has highest degree 1. Exam tip: find the degree from the greatest exponent of the variable, and count terms separated by + or - signs.

The defining condition for a polynomial in x is that, after simplification, every power of x must be a non-negative integer. In option C, division by x gives (x^2 + 1)/x = x + 1/x = x + x^(-1). The exponent -1 is negative, so the expression is not a polynomial in x. Options A and B have only permitted powers: 2 and 0 in A, and 3 and 1 in B. In option D, x^0 equals 1, so the expression becomes 7 + 4x, which is a polynomial of degree 1. A numerical fraction in a coefficient would be acceptable; the problem here is specifically that the variable x occurs in the denominator. Thus C is correct.

A polynomial in x may have real coefficients, including the irrational coefficient √5, but every exponent of x must be a non-negative integer. Option C has powers 3, 1, and 0, so it is a polynomial and its greatest power is 3. Option A has x in the denominator, option B has degree 2, and option D has a fractional exponent.

For a polynomial in x, every exponent of x must be a non-negative integer. Option C contains x^(-1/3), whose exponent is both negative and fractional. It therefore fails the defining rule and is not a polynomial. It can also be interpreted as a reciprocal involving a fractional root, which confirms the same issue. Option A has powers 5, 1, and 0, all allowed. Option B has powers 2, 1, and 0; pi is simply a real coefficient and does not cause any problem. In option D, x^0 equals 1, so the expression simplifies to 11 - 5x, a valid linear polynomial. Therefore the variable power in option C is the reason it is excluded.

Under the standard algebraic definition, a polynomial in x is a finite sum of terms of the form ax^n, where n is a non-negative integer. The terms 2x^2 and 1 satisfy this condition, but |x| does not have that form. Absolute value is defined piecewise: |x| = x for x greater than or equal to 0 and |x| = -x for x less than 0. Consequently, the whole expression has different formulas on different intervals and is not one polynomial in x. The number of terms, the presence of a constant, or the presence of x^2 cannot by itself make an expression a polynomial. Therefore option B identifies the precise reason and is correct.

In a polynomial in x, the variable may appear only with non-negative integer powers. In option C, sqrt(x + 1) places x inside a square root; algebraically this involves the fractional power (x + 1)^(1/2), so it does not satisfy the polynomial definition. The other radicals contain numerical constants only. In option A, sqrt(2) is a real coefficient multiplying x; in option B, -sqrt(5) is a constant term; and in option D, sqrt(7) is the coefficient of x. Irrational real coefficients are allowed in polynomials. Thus the presence of a radical is not automatically disqualifying; what matters is whether the variable occurs inside it. Hence option C is correct.

A polynomial is an expression made from constants and variables whose powers are non-negative whole numbers. The variable may be squared, cubed, or raised to another whole-number power, and irrational constants such as \(\sqrt{2}\) may be used as coefficients. However, a variable in the denominator is equivalent to a negative or non-polynomial power, so that expression does not meet the definition of a polynomial.

In option B, \(5/(x+2)\), the variable occurs in the denominator. Its value is not formed by a finite sum of terms with non-negative integer powers of \(x\), so it is not a polynomial. Option A has powers 2 and 1, option C has power 3, and option D uses \(\sqrt{2}\) only as a constant coefficient. Hence option B is correct.

The defining rule for a polynomial in x is that every exponent of x must be a fixed non-negative integer: 0, 1, 2, 3, and so on. The coefficients may be real numbers, but negative and fractional powers of the variable are not allowed. In x^k + 2x - 5, the powers in the last two terms are 1 and 0, so only k needs to be tested. For k = -1 or k = -4, x has a negative exponent and would occur in a denominator. For k = 1/2, the exponent is fractional and gives a square-root type term. When k = 4, the expression becomes x^4 + 2x - 5, with powers 4, 1, and 0. Thus option C is the only valid answer.

A polynomial can have any real coefficients, including irrational real numbers such as √3 and √5. However, every exponent of the variable must be a fixed non-negative integer. Option A has x-powers 3, 1, and 0, so it is a polynomial despite the coefficient √3. Option B has powers 2, 1, and 0 and is also a polynomial. Option C has powers 4 and 0; √5 is merely a real coefficient, so it remains a polynomial. In option D, x has exponent -2, and 6x^(-2) means 6/x^2. A negative exponent or a variable in the denominator violates the polynomial definition. Therefore, option D is correct.

For an expression to be a polynomial in x, every power of x must be a non-negative integer, and x must not occur in a denominator. In option B, divide each term in the numerator by x: (x² + 4)/x = x + 4/x = x + 4x⁻¹. The term 4x⁻¹ has a negative exponent, so the expression is not a polynomial. Thus option B is correct. Option A has powers 2, 1, and 0, making it a quadratic polynomial. Option C has the valid powers 3 and 0, so it is a cubic polynomial. In option D, x⁰ = 1, so the expression becomes 1 + 3x, a linear polynomial. The essential issue is the variable denominator, not the mere appearance of a fraction.

To classify the expression, count its nonzero terms and identify the greatest exponent of x. The terms are 2x⁶, −4x⁴, x², and −9, so there are four terms. Their powers are 6, 4, 2, and 0, making the highest power, and hence the degree, 6. Therefore option B is correct; it is not degree 4, a trinomial, or a constant.

The governing rule is that every exponent of the variable in a polynomial must be a fixed non-negative integer. Option C contains x^(5/2), whose exponent 5/2 is fractional and therefore is not an integer. It may also be written as x²√x, but that equivalent form still contains a fractional power of x, so it does not become an ordinary polynomial. Hence option C is correct. Option A uses exponents 7 and 0, option B uses exponent 0 for its variable term, and option D uses exponents 3 and 1. All of these are non-negative integers, so those expressions are polynomials. The fact that an exponent is positive is not enough; it must specifically be an allowed whole-number exponent.

In a polynomial in x, x may appear as the base raised to a fixed non-negative integer power. Coefficients must be constants with respect to x. Option B contains 3^x, where x is in the exponent and the base 3 is constant. This is an exponential term, not a polynomial term, so the whole expression is not a polynomial in x. Therefore option B is correct. The additional term x² in option B does not remove the invalid exponential term. Option A has powers 3 and 1, option C has x⁰ and a constant, and option D has powers 5 and 1. All their variable exponents are fixed non-negative integers. The distinction is important: x^3 is polynomial, whereas 3^x is exponential.

A polynomial may have any real coefficients, so both the fractional coefficient 2/3 and the irrational constant −√7 are permitted. The exponents of x are 2 and 1, while the constant has exponent 0. The greatest exponent is 2, so option C is correct. It is neither excluded by its coefficients nor the zero polynomial, since its terms are not all zero.

When an expression is described as a polynomial in u, only the powers of u are checked. Any quantity independent of u may be treated as a coefficient, so x acts as a constant coefficient in this question. Option A, xu³ − 4u + 2, has u-powers 3, 1, and 0, all of which are non-negative integers. It is therefore a polynomial in u, even though its coefficient xu³ contains x. Option B has u in the denominator, giving u⁻¹. Option C has the negative exponent −2. Option D has √u = u^(1/2), a fractional exponent. Each of those violates the polynomial condition in u, so option A is the only valid choice.

A polynomial in x may contain x^0, x, x², x³, and other fixed non-negative integer powers. Option B contains x^(1/3), whose exponent is one-third rather than an integer. This term represents the cube root of x, and a fractional power of the variable is not allowed in an ordinary polynomial in x. Thus option B is correct. Option A uses exponents 3, 1, and 0. Option C contains x⁰ = 1, so it simplifies to 3 + 4, a constant polynomial. Option D uses powers 5 and 0, since −2 is a constant term. All the exponents in A, C, and D satisfy the definition. The issue is the fractional exponent, not the fact that the expression includes a root-like operation.

To classify the expression, check two features independently: its degree and its number of terms. A quadratic polynomial has degree 2, meaning the greatest exponent of x is 2. A binomial has exactly two non-zero terms after simplification. Option A, x² − 16, has two terms and greatest exponent 2, so it is a quadratic binomial polynomial. Option B also has degree 2 but has three terms, making it a trinomial. Option C has greatest exponent 3, so it is cubic, not quadratic. Option D contains x⁻², a negative exponent, and therefore is not a polynomial at all. Thus only option A satisfies both required conditions. The constant −16 counts as one term, with implicit power x⁰.

A non-zero constant polynomial contains no variable-dependent terms and has a non-zero constant value. In f(x) = cx² + 4x + 1, the value of c controls only the coefficient of x². Setting c = 0 removes cx², so removing the x² term is possible. Keeping the constant term 1 is also possible and ensures that the remaining value is non-zero. However, the coefficient of 4x is fixed at 4 and is independent of c. No selection of c can change 4 into 0, so the term 4x cannot be eliminated using the given parameter. Therefore making the 4x term zero is impossible, and option C is correct. The question tests which coefficient is actually controlled by c.

Having two terms means an expression may be a binomial, but it does not automatically make the expression a polynomial. The polynomial condition must also be checked: every exponent of x must be a non-negative integer. Option C contains x⁻¹, which is equal to 1/x. Thus x has a negative exponent and occurs in a denominator, so the expression is not a polynomial even though it has two terms. Option A has valid exponents 5 and 0, option B has exponents 2 and 0, and option D has exponents 1 and 0. These are all binomial polynomials. Therefore option C is correct. This demonstrates that the number of terms and the validity of the exponents are separate classification tests.

The governing concept is the definition and degree of a polynomial. In a polynomial in x, every exponent of x must be a non-negative integer, such as 0, 1, 2 or 3. The degree is the greatest exponent whose coefficient is not zero. Option A is 2x³ − √5x² + 1; its exponents are 3, 2 and 0, so it satisfies the polynomial definition and has degree 3. The coefficient √5 is an allowed real coefficient and does not cause a problem. Thus option A is cubic. Option B contains x^−1 because 1/x=x^−1, so it is not a polynomial. Option C has the fractional exponent 3/2, which is not allowed, while option D has highest exponent 1 and is therefore linear, not cubic.

The governing concept is a polynomial in one variable when another symbol is treated as a fixed constant. For an expression to be a polynomial in x, each power of x must be a non-negative integer. The coefficients may contain y because y is being held constant. In option A, yx² + 3x − 4 has x-powers 2, 1 and 0, so it is a polynomial in x of degree 2; y simply forms part of the coefficient of x². Option B contains x^−1, since 1/x=x^−1. Option C contains the fractional power x^(1/2), and option D again contains a negative power of x. Negative and fractional exponents violate the polynomial rule. Therefore option A is the only correct answer.