Definition of a Polynomial

In (x^x), the exponent is not a fixed integer. In a polynomial, the variable power must be a fixed non-negative integer.

On substituting a=5, the coefficient of (a-5)x^2 becomes 0, so the x^2 term disappears. Thus, p(x)=2x^4+3x-1. The highest exponent of x with a non-zero coefficient is 4, so the degree remains 4. Option A is incorrect because the x^4 term is still present. Exam tip: remove terms with zero coefficients before identifying the highest power.

\(\sqrt{3}\) is a real constant, so its presence does not prevent the expression from being a polynomial. In \(x^2+2x+1+\sqrt{3}\), the highest power of \(x\) is \(2\), and its coefficient, \(1\), is non-zero. Therefore, it is a polynomial of degree 2. Option A is incorrect because polynomial coefficients may be irrational real numbers. Exam tip: determine the degree from the highest non-zero exponent of the variable.

The degree of a polynomial is the greatest exponent of the variable having a non-zero coefficient. In \(x^4+x^2\), the greatest power of \(x\) is \(4\), so its degree is \(4\), not \(2\). In option C, \(9\) is a non-zero constant polynomial, whose degree is \(0\). Exam tip: simplify the polynomial first, then identify the highest power of \(x\).

In \(\pi x^2-3x+1\), the powers of \(x\) are \(2,1,0\), all non-negative integers. Since \(\pi\) is a real number, it may be a coefficient of a polynomial. In B, the exponent of \(x\) is \(-2\); in C, \(\sqrt{x}=x^{\tfrac12}\); and in D, \(\frac{1}{x}=x^{-1}\). Hence, none of these is a polynomial. Exam tip: in a polynomial, variable exponents can only be \(0,1,2,\ldots\).

The direct answer is B: the degree of the zero polynomial is not defined. The zero polynomial is the expression that is identically zero, written as 0. It is still a polynomial, so option A is false. The degree of a nonzero polynomial is the greatest exponent of the variable having a nonzero coefficient. For example, the degree of \\(5x^2+1\\) is 2, and the degree of the nonzero constant \\(7\\) is 0. In the zero polynomial there is no nonzero term and therefore no greatest exponent to select. For this special reason, its degree is left undefined rather than called 0. Option B is correct. Option A is wrong because 0 is a valid zero polynomial. Option C is wrong because there is no hidden variable in the expression 0. Option D is wrong because the degree is not 1; it is undefined. The distinction is important: every nonzero constant has degree 0, but the zero polynomial is treated separately. This convention avoids pretending that a missing highest nonzero term exists. Memory cue: nonzero constant means degree 0; zero polynomial means degree undefined.

In a polynomial, every exponent of the variable must be a non-negative integer, that is, 0 or a positive integer. In \(7x^3-4x+1\), the exponents of \(x\) are 3, 1, and 0, so it is a polynomial. However, \(x^{-2}+5\) contains the exponent \(-2\), so it is not a polynomial. Having real coefficients alone is not sufficient. Exam tip: check all variable exponents before classifying an expression as a polynomial.

Combining like terms gives \(4x^2+2x^2=6x^2\) and \(3x-3x=0\). Thus, the simplified expression is \(6x^2+1\). The highest power of \(x\) in it is 2, so the degree of the polynomial is 2. It is not 1 because the linear terms cancel, while the \(x^2\) term remains. Exam tip: simplify by combining like terms first, then identify the highest exponent of the variable left in the polynomial.

In the zero polynomial, all coefficients must be (0). The coefficients of (3x) and (2) are not zero, so no such (k) is possible.

For exactly quadratic, the coefficient of (x^4) must be zero and the coefficient of (x^2) must remain non-zero. From (k-3=0), we get (k=3).

A quadratic polynomial has highest power 2. Therefore, the coefficient of \(x^5\) must be zero: \(a-4=0\). Hence, \(a=4\). Then \(p(x)=3x^2-7\), which has degree 2 because the coefficient of \(x^2\) is the non-zero number 3. If \(a=0\), the \(x^5\) term remains, so the polynomial is not quadratic. Exam tip: in parameter-based questions, first make the coefficient of the highest unwanted power zero.

(\frac{4}{x}=4x^{-1}), where the power of (x) is negative. Having real coefficients alone does not guarantee a polynomial.

The degree of a polynomial is the highest exponent of a term with a non-zero coefficient. Here, \(0x^6\) equals zero, so it does not contribute to the degree. Among the remaining terms, \(5x^4\) has the greatest exponent; therefore, the degree of \(p(x)\) is \(4\). Choosing \(6\) is incorrect because its coefficient is \(0\). Exam tip: remove all zero-coefficient terms before identifying the highest power.

For degree (2), the coefficient of (x^3) must be zero and the coefficient of (x^2) must remain non-zero. From (k+2=0), (k=-2).

In a polynomial, the exponents of the variable can only be non-negative integers. Here, \(\sqrt{x}=x^{1/2}\), and \(1/2\) is not an integer. Therefore, \(x^2+\sqrt{x}+1\) is not a polynomial. Although the expression contains \(x^2\), it cannot be called a quadratic polynomial because the complete expression must first be a polynomial. Exam tip: if a variable appears under a radical or with a fractional exponent, first test whether the expression is a polynomial.

A cubic polynomial has highest exponent 3, and every exponent of the variable must be a non-negative integer. In \(4y^3-2y+8\), the highest exponent is 3, so it is exactly cubic. \(y^4+y^3+1\) is quartic, while \(y^{-3}+2\) is not a polynomial because it has a negative exponent. Exam tip: before identifying the degree, check that no variable has a negative or fractional exponent.

Substituting b=0 makes bx^4=0. Thus, p(x)=3x^3-2x+1. The greatest exponent of x with a non-zero coefficient is 3, so the degree is 3. Option 4 is incorrect because the coefficient of x^4 becomes zero. Exam tip: remove all terms with zero coefficients before finding the degree.

Treating \(m\) as a constant, the powers of \(x\) in \(m^2x^3-5mx+6\) are 3, 1, and 0. All are non-negative integers, so it is a polynomial in \(x\). Option A has \(x^{-2}\), option B has \(\sqrt{x}=x^{1/2}\), and option C has \(\frac{m}{x}=mx^{-1}\); these powers are not permitted in a polynomial. Exam tip: For a polynomial in a given variable, every exponent of that variable must be a non-negative integer.

For a polynomial to be constant, the coefficients of both \(x^2\) and \(x\) must be zero. Thus, \(r-1=0\) requires \(r=1\), while \(r+2=0\) requires \(r=-2\). The same value of \(r\) cannot be both 1 and −2, so no such \(r\) exists. In particular, \(r=1\) removes only the \(x^2\) term; the \(x\) term remains. Exam tip: for a constant polynomial, set the coefficient of every variable term equal to zero.

In the first option, all powers are non-negative integers and the highest power is (6). The other options are not polynomials.

For degree (2), the coefficient of (x^3) must be zero. From (c+5=0), (c=-5), and the (x^2) coefficient (-7) remains non-zero.

Since \(\frac{x}{2}=\frac{1}{2}x\), the variable \(x\) is not in the denominator; \(\frac{1}{2}\) is only a coefficient. The terms are \(x^2\), \(\frac{1}{2}x\), and 9, whose exponents of \(x\) are 2, 1, and 0 respectively. All are non-negative integers, and the greatest exponent is 2, so it is a polynomial of degree 2. Option A wrongly treats the variable as being in the denominator, while option C looks at only one term instead of the whole polynomial. Exam tip: find the degree of a polynomial by identifying the greatest exponent of the variable among all its terms.

The direct answer is C: \\(a=3\\) or \\(a=-3\\). A linear polynomial has degree 1, so the coefficient of every power higher than 1 must be zero, while the coefficient of x must be nonzero. Here \\(p(x)=(a^2-9)x^4+2x+1\\). The coefficient of \\(x^4\\) is \\(a^2-9\\). Set it equal to zero: \\(a^2-9=0\\). Hence \\(a^2=9\\), so \\(a=3\\) or \\(a=-3\\). For either value, the expression becomes \\(0x^4+2x+1=2x+1\\), which has degree 1 because the coefficient of x is 2, not zero. Therefore option C is correct. Option A, \\(a=0\\), gives coefficient \\(-9\\), leaving an \\(x^4\\) term, so the polynomial is degree 4. Option B, \\(a=9\\), gives coefficient 72, again leaving the fourth-degree term. Option D is wrong because two valid values have been found. Do not set \\(a\\) itself to zero; set the coefficient of the unwanted highest-power term to zero. Memory cue: for a polynomial to become linear, remove all powers 2 and above, then confirm the x coefficient remains nonzero.

With respect to (u), (\frac{1}{x}) is like a constant, so it is a polynomial in (u). With respect to (x), (\frac{1}{x}) has power (-1).

The degree of a polynomial is the highest power of the variable having a non-zero coefficient. Here, the coefficient of \(0x^5\) is zero, so this term does not contribute to the degree. Among the remaining terms, \(7x^3\) has the highest power, 3; therefore, the polynomial has degree 3. The degree-5 option is incorrect because the coefficient of \(x^5\) is zero. Exam tip: Ignore all terms with zero coefficients before finding the degree.