Definition of a Polynomial

The degree of a polynomial is the greatest exponent of the variable having a non-zero coefficient. The exponents of the terms are 6, 3, 1, and 0, and the greatest is 6. Therefore, the correct answer is 6. The number 7 is a coefficient, not an exponent, so it is not the degree. Exam tip: For degree, check the exponent of the variable, not the numerical coefficient.

Substituting \(k=2\) gives \((k-2)=0\), so the \(x^5\) term becomes zero. The polynomial left is \(3x^2-1\), whose highest exponent is \(2\). Therefore, its degree is \(2\). Although \(5\) appears in the original expression, that term has a zero coefficient and does not count. Exam tip: substitute the given value and remove zero-coefficient terms before finding the degree.

The term \(0x^7\) is zero, so it does not contribute to the degree of the polynomial. In the remaining expression, \(4x^3-2x+6\), the highest power of \(x\) is \(3\). Therefore, its degree is \(3\). Choosing \(7\) would be incorrect because its coefficient is zero. Exam tip: Remove terms with zero coefficients before finding the degree.

In \(5x^3-\sqrt{2}x+7\), the powers of x are 3, 1, and 0, all non-negative integers. An irrational coefficient such as \(\sqrt{2}\) is allowed. Option B has \(x^{-1}\). Exam tip: check exponents of the variable, not the type of coefficient.

A linear polynomial has degree 1. Therefore, the \(x^3\) term must vanish, so its coefficient must be zero: \(m+1=0\). Hence, \(m=-1\). For this value, the expression becomes \(5x-6\), which is a linear polynomial. If \(m=0\), the expression is \(x^3+5x-6\), which has degree 3. Exam tip: To make a polynomial linear, set the coefficient of its highest-degree term to zero.

No. For real \(x\), \(\sqrt{x^2}=|x|\), not always \(x\). Thus the expression becomes \(4x^2+|x|+1\). Since \(|x|\) cannot be written as a term having a non-negative integral power of \(x\), the expression is not a polynomial in \(x\). Option A is true only when \(x\geq0\). Exam tip: whenever you see \(\sqrt{x^2}\), write \(|x|\) first, not simply \(x\).

A polynomial’s degree is the greatest exponent of the variable having a non-zero coefficient. A non-zero constant, such as 7 or -3, has no variable term, so it is treated as having degree 0. The zero polynomial is different because every coefficient is zero, so it has no greatest non-zero exponent. Therefore, its degree is not defined in the usual school-level convention.

Thus option B is correct: the zero polynomial has an undefined degree, while a non-zero constant polynomial has degree 0. Option A is wrong because degree 0 applies to non-zero constants, not to the zero polynomial. The other options incorrectly reject polynomials or assign degree 1.

On putting \(a=0\), \(ax^4=0\), so the polynomial becomes \(2x^3-x+1\). The highest power of \(x\) in the remaining polynomial is 3; therefore, its degree is 3. Option 4 is incorrect because the \(x^4\) term becomes zero. Exam tip: first remove terms with zero coefficients, then identify the highest remaining exponent.

In \(-5\sqrt{x}=-5x^{\frac{1}{2}}\), the exponent of \(x\) is \(\frac{1}{2}\). In a polynomial, variable exponents must be non-negative integers, so this is not a polynomial even though it appears to have just one term. In contrast, \(9x^2\), \(7x\), and \(11\) have exponents 2, 1, and 0 respectively, so they are polynomials. Exam tip: Rewrite a square root involving a variable as an exponent and check that exponent.

In the standard form of a polynomial, terms are arranged in descending powers of the variable. In option B, the powers are 5, 2, 1, and 0: \(x^5+3x^2-2x+7\). Hence, it is in standard form. In option A, \(x^5\) appears after lower-degree terms, so the terms are not in descending order. Exam tip: Check that powers go from the highest power down to the constant term.

The terms of the polynomial are 2x^6, -5x^2, and 4. There is no term containing x^5, so the coefficient of x^5 is 0. Note that -5 is the coefficient of x^2, not x^5. Exam tip: If a power is missing in a polynomial, its coefficient is always 0.

The coefficients of \(0x^4\) and \(0x^3\) are zero, so these terms do not count in the standard form of the polynomial. Thus, \(p(x)=5x-2\), whose highest power of \(x\) is 1; therefore, it is a linear polynomial. It is not a zero polynomial because the coefficients of \(5x\) and \(-2\) are non-zero. Exam tip: Remove terms with zero coefficients before finding the degree of a polynomial.

\(\frac{x^3+1}{x}=\frac{x^3}{x}+\frac{1}{x}=x^2+\frac{1}{x}=x^2+x^{-1}\). In a polynomial, every exponent of the variable must be a non-negative integer. Here, \(x^{-1}\) has exponent \(-1\), so the expression is not a polynomial. Having \(x^3\), a binomial numerator, or a constant term does not prevent an expression from being a polynomial. Exam tip: simplify the expression first, then check the exponents of the variable.

Option C is incorrect. \(5\) is a non-zero constant polynomial, so its degree is \(0\), not \(5\). The degree of a polynomial is the greatest exponent of its variable; a constant has no variable term, so its degree is \(0\). Exam tip: while finding degree, look at the highest power of the variable, not at the coefficient.

In a polynomial, the exponents of a variable must be non-negative integers such as \(0,1,2,\ldots\). Here, \(x^{-1}=\frac{1}{x}\), so \(-5x^{-1}\) cannot be a polynomial term. Having real coefficients is not sufficient; the exponents must also be non-negative integers. Exam tip: before identifying an expression as a polynomial, check every exponent of the variable.

The leading term of a polynomial is the term with the greatest exponent of the variable. In the given polynomial, the exponents of \(3x^4\), \(-2x^2\), \(x\), and \(-5\) are 4, 2, 1, and 0 respectively. Hence, \(3x^4\) is the leading term. Although \(-2x^2\) has the next highest power, its exponent is only 2. Exam tip: identify the term with the greatest exponent to find the leading term.

For the polynomial to have degree 2, the coefficient of the highest-power term \(x^4\) must be zero. Thus, \(2d-6=0\), so \(2d=6\) and \(d=3\). If \(d=2\) or \(d=4\), the coefficient of \(x^4\) is not zero, so the degree remains 4. Exam tip: In questions where the degree is reduced, first set the coefficient of the highest power equal to zero.

The direct answer is B, with an important domain qualification. In the original expression, x/x is defined only when x≠0, and for those values it equals 1. Thus x^2+x/x=x^2+1 for x≠0. However, the original expression is not defined at x=0, whereas the polynomial x^2+1 is defined there; so they are not exactly the same function on all real numbers. Option A is wrong because the original variable denominator prevents calling the displayed form a polynomial without qualification. Option B is correct because it states both the simplification and the restriction. Option C is wrong: the expression is not identically zero. Option D is wrong: x^2+1 has degree 2, not 1. Memory cue: never cancel x/x without recording x≠0; simplification can change the domain.

A non-zero constant polynomial contains only a fixed number and no variable with a positive power. Such a polynomial has degree 0. The expression 0 is the zero polynomial, whose degree is not defined in the standard convention. The expression 13x has degree 1 because the highest power of x is 1, and x^2 - 13 has degree 2 because its highest power is 2.

Option B is -13, a non-zero constant, so its degree is 0. It is important not to confuse the number 0 with a non-zero constant: the former is the zero polynomial, while the latter has a defined degree of 0. Hence the selected answer follows directly from comparing the highest powers and the special rule for the zero polynomial.

Since \(x^0=1\), the expression becomes \(x^4+x^2+1\). The degree of a polynomial is the highest exponent of \(x\) in any term, which is \(4\). Therefore, the correct answer is \(4\). \(2\) is the exponent of one term, not the highest exponent. Exam tip: Simplify terms such as \(x^0\) first, then identify the greatest exponent.

The terms of the polynomial are separated by plus or minus signs: \(2x^3\), \(-4x^2\), \(6x\), and \(-8\). Their coefficients are \(2,-4,6,-8\), all of which are non-zero. Hence, there are \(4\) non-zero terms. The number \(6\) is the coefficient of \(6x\), not the number of terms. Exam tip: Count every positive or negative expression separated by signs as one term.

\(5\) is a constant polynomial in \(x\), because it can be written as \(5x^0\). The exponent of \(x\) is \(0\), which is a whole number. \(7y^2+3\) has \(y\) as its variable, so it is not treated as a polynomial in \(x\). In \(\frac{1}{x}\), the exponent of \(x\) is \(-1\), and in \(\sqrt{x}\), it is \(\frac{1}{2}\); therefore, neither is a polynomial in \(x\). Exam tip: exponents of the variable in a polynomial can only be \(0,1,2,\ldots\).

\(2x^2-\pi x+\sqrt{11}\) is a polynomial because its coefficients \(2\), \(-\pi\), and \(\sqrt{11}\) are real numbers, and the powers of \(x\) are \(2\), \(1\), and \(0\), all non-negative integers. \(\sqrt{x}=x^{\frac12}\) has a fractional exponent, while \(\frac{1}{x}=x^{-1}\) and \(x^{-\pi}\) have negative exponents, so they are not polynomials. Exam tip: variable exponents in a polynomial can only be \(0,1,2,\ldots\); coefficients may be any real numbers, including \(\pi\) and \(\sqrt{11}\).

In a polynomial, the exponents of the variable must be non-negative integers, such as 0, 1, 2, or 3. Here, \(\sqrt{x}=x^{\frac{1}{2}}\), so the exponent of \(x\) is \(\frac{1}{2}\), which is not an integer. Therefore, \(q(x)\) is not a polynomial. The exponent 2 in \(x^2\) is valid, and the constant term 1 is also allowed; the number of terms does not decide whether an expression is a polynomial. Exam tip: a root of a variable or a fractional exponent signals a non-polynomial.

The direct answer is D, never. For an expression to be constant in x, every term containing x must disappear. Here the coefficient of x^2 is n-4, so we need n-4=0, giving n=4. The coefficient of x is n+1, so we also need n+1=0, giving n=-1. One value of n cannot be both 4 and -1. The constant term 6 is already independent of x. Option A, n=4, removes the x^2 term, but the x coefficient becomes 5, so the expression still has an x term. Option B, n=-1, removes the x term, but the x^2 coefficient becomes -5, so an x^2 term remains. Option C says n=4 and n=-1 together; that is impossible for one parameter, so it cannot occur. Option D is correct because no single value satisfies both necessary equations. The reasoning is not that constants are impossible; rather, this particular expression cannot become constant for one common value of n. Exam cue: set every variable-term coefficient equal to zero, then check whether the resulting conditions are compatible.