Definition of a Polynomial

The correct answer is C: the expression is not a polynomial because it contains x^{-1}. A polynomial is a sum of terms in which the variable has a whole-number exponent 0, 1, 2, 3, and so on; negative exponents are not allowed. Here x^0=1, so that part is valid, and 2 is also a constant. However, x^{-1}=1/x has a negative exponent and places x in the denominator. One invalid term makes the whole expression non-polynomial. Option A is wrong because the presence of x^0 alone is not enough. Option B is wrong because the expression still contains x^{-1}, so it is not constant. Option C correctly identifies the decisive restriction. Option D is wrong because the expression is not a polynomial at all, and it does not consist only of a highest power x^2. Memory cue: denominator or negative exponent means not a polynomial.

(\sqrt{11}) is a real constant and is allowed. All powers of the variable are non-negative integers.

A polynomial need not contain all middle powers. In (x^5+2x^2-3), the present powers (5), (2), (0) are valid.

On simplifying, p(x)=6 because every term containing x has coefficient 0. Since 6 is a non-zero constant polynomial, its degree is 0. “Not defined” applies to the zero polynomial, not to a non-zero constant polynomial. Exam tip: remove terms with zero coefficients before finding the degree.

The direct answer is C, no such value of s exists. A non-zero constant polynomial must have no variable terms, while its constant term must remain non-zero. Here the coefficients of both variable terms must be zero: s−2=0 gives s=2, whereas s+1=0 gives s=−1. One number cannot be both 2 and −1, so the two required conditions cannot hold together. The remaining term is 5, which is non-zero, so the issue is only the incompatible coefficients. Option A, s=2, removes the cubic term but leaves (3)x². Option B, s=−1, removes the x² term but leaves (−3)x³. Option C is correct because no single s removes both terms. Option D, s=0, leaves both variable terms. Memory cue: for a constant polynomial, every variable-term coefficient must be zero.

In \(5x^3-2x+7\), the powers of \(x\) are \(3\), \(1\), and \(0\). All are non-negative integers, so it is a polynomial in one variable. In option B, \(\frac{3}{x}=3x^{-1}\), and option D also has a negative exponent; option C has \(\sqrt{x}=x^{1/2}\), a fractional exponent. Exam tip: every exponent of the variable in a polynomial must be a non-negative integer.

The terms of the polynomial are \(x^4\), \(-2x^3\), \(3x^2\), \(-4x\), and \(5\). Hence, it has 5 terms. The degree of a polynomial is the highest exponent of \(x\), which is 4 in \(x^4\). Therefore, the correct answer is 5 terms and degree 4. Do not omit the constant \(5\), as it is also a term. Exam tip: Count terms by separating them at addition and subtraction signs.

In \(5x^2-11\), the highest power of \(x\) is 2, so it is a quadratic polynomial. It has only two non-zero terms, \(5x^2\) and \(-11\), making it a binomial rather than a trinomial. \(x^2+3x+2\) is quadratic, but it has three terms. Exam tip: find the degree from the highest exponent and count only non-zero terms to identify the type of polynomial.

In \(p(x)=2x^3+(h-4)x^2+7\), the coefficient of \(x^3\) is \(2\). It is non-zero and does not depend on \(h\). Hence the highest-power term \(2x^3\) remains present for every real value of \(h\), so the degree is always \(3\). If \(h=4\), only the \(x^2\) coefficient becomes zero; the \(x^3\) term does not disappear. Exam tip: to determine degree, first check the coefficient of the highest-power term.

In \(3x^2+1\), the term \(3x^2\) contains \(x\), and the polynomial has degree 2. Hence, it is a non-constant polynomial in \(x\). The expressions \(14\), \(0\), and \(-9\) contain no \(x\), so they are constant polynomials. Exam tip: A polynomial with a variable raised to a positive power cannot be a constant polynomial.

To simplify the fraction, divide each term in the numerator by x squared: \\(x^4/x^2=x^2\\) and \\(-1/x^2=-1/x^2\\). Therefore, for x not equal to zero, the expression becomes \\(x^2-1/x^2\\), or equivalently \\(x^2-x^{-2}\\). A polynomial cannot contain a negative power of its variable.

The presence of subtraction is not a problem, because polynomials may have both positive and negative coefficients. The term x to the fourth power is also completely valid, and the constant 1 is valid. The real difficulty is the denominator x squared, which creates the negative exponent. Hence option A gives the correct reason that the simplified expression is not a polynomial.

In a polynomial, powers of x must be non-negative integers; coefficients such as \(\sqrt{2}\) are allowed. Option A has powers 3, 1, and 0. Option B contains \(x^{-1}\), while C has \(x^{1/2}\). Exam tip: if the variable appears in a denominator, the expression is not a polynomial.

The degree of a polynomial is the highest power of \(x\) having a non-zero coefficient. Here, the coefficient of \(x^2\) is always \(1\), and it cannot become zero for any value of \(a\). Hence, \(p(x)\) has degree \(2\) for every real \(a\). The condition \(a\ne0\) is unnecessary because even when \(a=0\), \(p(x)=x^2\) still has degree \(2\). Exam tip: In a polynomial containing a parameter, first check the coefficient of the highest-power term.

In a polynomial, powers must be fixed non-negative integers. A power depending on a changing variable is not valid in the definition.

In \(x^3-5x^2+2\), the highest power of \(x\) is 3, so its degree is 3 and it is a cubic polynomial. It has three non-zero terms: \(x^3\), \(-5x^2\), and \(2\); therefore, it is a trinomial. Hence, it is a cubic trinomial. “Cubic binomial” is incorrect because the expression has three terms, not two. Exam tip: find the highest exponent for the degree, then count the non-zero terms separately.

In \(x^5-2x^3+x-9\), the powers of \(x\) are 5, 3, and 1, all odd positive integers, and \(-9\) is the constant term. Therefore, it satisfies both conditions in the question. \(x^4+x^2+1\) is a polynomial, but its variable terms have even powers. \(x^{-3}+x\) has a negative exponent, while \(\sqrt{x}+x^3\) contains \(x^{\frac12}\); hence neither is a polynomial. Exam tip: exponents of variables in a polynomial must be 0 or positive integers.

For a polynomial to have degree 0, it must reduce to a non-zero constant. Since the constant term here is 2, the coefficients of both x-terms must be zero: a-1=0 and a+3=0. These give a=1 and a=-3 respectively, which cannot hold for the same value of a. Therefore, no such a exists. Exam tip: for degree 0, set the coefficients of every variable term equal to zero.

The direct answer is B. A polynomial is a finite expression in which the powers of x are non-negative integers; x must not remain in a denominator after simplification. For option B, divide each term by x: (x^3+x)/x=x^2+1 for x≠0. The result has powers 2 and 0, so it is a polynomial. Option A gives (x²+1)/x=x+1/x, which contains a negative power and is not a polynomial. Option C gives (x+1)/x²=1/x+1/x², also non-polynomial. Option D is already 1/x+2 and still has x in the denominator. The condition x≠0 is important because cancellation is not permitted at x=0. Exam cue: simplify term by term and check whether any negative power remains.

In (x^3-4x), there is no separate constant term, so the constant term is considered (0). Its powers are valid.

This is the zero polynomial because all its coefficients are zero. The degree of a polynomial is the greatest exponent having a non-zero coefficient; the zero polynomial has no such exponent. Hence, its degree is not defined. Degree 0 belongs only to a non-zero constant polynomial, such as 5. Exam tip: Do not confuse the zero polynomial with a non-zero constant polynomial.

Since \(q(x)=x^3+p(x)\) and \(p(x)=2x^2-3x+5\), we get \(q(x)=x^3+2x^2-3x+5\). The highest power of \(x\) in this polynomial is 3, so the degree of \(q(x)\) is 3. Option 5 is incorrect because degrees are not added when polynomials are added; the highest power is considered. Exam tip: simplify the polynomial first, then identify the highest power with a non-zero coefficient.

In the polynomial x^2-4, the highest exponent is 2, so its degree is quadratic. It has two terms, x^2 and -4, so it is a binomial. Therefore, it is a quadratic binomial. A quadratic trinomial would have three terms. Exam tip: first count the terms, then identify the highest exponent.

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

For exactly quadratic, the coefficient of (x^6) must be zero and the coefficient of (x^2) is non-zero. From (a-5=0), (a=5).

(\frac{2}{x}=2x^{-1}), where the variable has a negative power. One invalid term prevents the whole expression from being a polynomial.