Definition of a Polynomial

Correct answer: C. It is not a polynomial because it has (x^{-1})

Explanation: 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.

Correct answer: D. (3x^3-2x+\sqrt{11})

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

Correct answer: A. (x^5+2x^2-3)

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

Correct answer: D. 0

Explanation: 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.

Correct answer: C. No such (s) exists

Explanation: 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.

Correct answer: A. \(5x^3-2x+7\)

Explanation: 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.

Correct answer: B. 5 terms, degree 4

Explanation: 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.

Correct answer: C. \(5x^2-11\)

Explanation: 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.

Correct answer: B. Any real value of \(h\)

Explanation: 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.

Correct answer: C. \(3x^2+1\)

Explanation: 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.

Correct answer: A. Because it becomes \(x^2-\frac{1}{x^2}\)

Explanation: 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.

Correct answer: A. \(\sqrt{2}x^3-5x+1\)

Explanation: 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.

Correct answer: C. For all real values of \(a\)

Explanation: 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.

Correct answer: A. (x^n+1), where (n) is a changing variable

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

Correct answer: B. Cubic trinomial

Explanation: 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.

Correct answer: A. \(x^5-2x^3+x-9\)

Explanation: 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.

Correct answer: C. No such a exists

Explanation: 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.

Correct answer: B. (\frac{x^3+x}{x})

Explanation: 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.

Correct answer: A. (x^3-4x)

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

Correct answer: C. Its degree is not defined

Explanation: 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.

Correct answer: B. 3

Explanation: 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.

Correct answer: D. Quadratic binomial

Explanation: 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.

Correct answer: A. (2)

Explanation: 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).

Correct answer: A. (5)

Explanation: 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).

Correct answer: B. (x^4+\frac{2}{x}-1)

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