Definition of a Polynomial

Direct answer: Option C, the zero polynomial whose degree is not defined, is correct. First simplify each term: 0 times x^9 is 0, 0 times x^4 is 0, and the final constant 0 remains 0. Thus p(x) is the zero polynomial, not a polynomial with a surviving x^9 or x^4 term. The degree is the greatest exponent of x having a non-zero coefficient. Here every coefficient is zero, so there is no greatest non-zero-power term; in the school convention its degree is not defined. Option A is wrong because the x^9 coefficient is zero, so degree 9 cannot be used. Option B is wrong for the same reason: the x^4 coefficient is also zero. Option D is wrong because the expression equals zero, not the non-zero constant 1. Option C correctly separates the polynomial itself from its degree. Memory cue: for the zero polynomial, do not read the largest written exponent; look for the largest non-zero coefficient, and there is none.

A polynomial is identified mainly by the powers of the variable. The powers must be fixed non-negative integers.

In \(x^8-5x^6+2x^2+4\), the powers of \(x\) are \(8,6,2\), and \(0\) (for the constant term \(4\)). They are all non-negative integers and even, so it is a polynomial containing only even powers. \(x^7+x^2+1\) is a polynomial, but \(7\) is an odd power. \(x^{-2}+x^4\) and \(x^{1/2}+x^2\) are not polynomials because they contain a negative and a fractional exponent, respectively. Exam tip: powers of the variable in a polynomial must be non-negative integers.

For degree (4), the coefficient of (x^5) must be zero. From (k^2-1=0), (k=\pm1), and (2x^4) remains non-zero.

\(x^2-5x+6\) has three terms: \(x^2\), \(-5x\), and \(6\). Its highest exponent is 2, so it is a quadratic polynomial; because it has three terms, it is a trinomial. \(x^3-5x+6\) is also a trinomial, but its degree is 3, so it is cubic. Exam tip: identify degree from the highest exponent and count terms separated by \(+\) or \(-\).

The direct answer is B. A polynomial may contain addition, constants and terms such as x² or x/2, but the variable cannot have a negative exponent. In option B, (x²+4)/x²=1+4/x²=1+4x^(−2), for x≠0. The term x^(−2) shows that x is effectively in the denominator, so the expression is not a polynomial. Option A, x²+x/2, is a polynomial: x/2=(1/2)x and its exponent is 1. Option B is correct because its denominator contains x² and produces a negative power. Option C, 4x²+1, is plainly a polynomial with non-negative integer powers. Option D, x⁰+3x=1+3x, is also a polynomial; x⁰ is allowed and equals 1. A common mistake is to think every fraction is non-polynomial; a fraction such as x/2 is allowed because the denominator is only a constant.

The degree of a polynomial is the greatest exponent of the variable whose coefficient is non-zero. Here, the coefficient of (x^4) is (1), which cannot become zero for any values of (c), (d), and (e). Hence, the degree is definitely (4). Setting (c), (d), or (e) to zero affects only lower-degree terms, not the (x^4) term. Exam tip: always check the coefficient of the highest-power term before deciding the degree.

For a constant polynomial, all variable-term coefficients must be zero. From (a-1=0), (a=1), and (5) remains.

For degree (5), the (x^6) coefficient must be zero and the (x^5) coefficient must be non-zero. At (a=-2), (a-4=-6) remains non-zero.

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

For degree (0), both variable-term coefficients must be zero. Both (k-3=0) and (2k-6=0) give (k=3).

For degree (3), the (x^4) coefficient must be zero and the (x^3) coefficient must be non-zero. At (a=2), (a^2-4=0) and (a+2=4).