Definition of a Polynomial

Correct answer: A. The exponent of \(x\) is \(-1\), which is not allowed in a polynomial.

Explanation: Since \(\frac{-3}{x}=-3x^{-1}\), the exponent of \(x\) is \(-1\). A polynomial permits only \(0,1,2,\dots\) as variable exponents; negative coefficients and constants are allowed. Exam tip: check each exponent.

Correct answer: B. No, because in \(\frac{2}{x}\), the variable is in the denominator

Explanation: In a polynomial, the exponent of a variable must be 0 or a positive integer. Here, \(\frac{2}{x}=2x^{-1}\), so the exponent of \(x\) is \(-1\). Therefore, \(5x^2-\frac{2}{x}\) is not a polynomial. Although \(5x^2\) is a polynomial term, one negative exponent makes the complete expression non-polynomial. Exam tip: If a variable occurs in the denominator, rewrite it with a negative exponent and check it.

Correct answer: A. \(\frac{3}{x}\)

Explanation: \(\frac{3}{x}=3x^{-1}\) has only one algebraic term, but the exponent of \(x\) is \(-1\). A polynomial cannot have a negative exponent of a variable, so it is not a polynomial. In contrast, \(3x\), \(x^2\), and \(5\) have exponents \(1\), \(2\), and \(0\), respectively, so they are polynomials. Exam tip: Rewrite a variable in the denominator using a negative exponent before checking whether an expression is a polynomial.

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

Explanation: For every value of \(x\), \(0x=0\). Hence, \(3x^2+0x-7=3x^2-7\), so option A is correct. Option B incorrectly adds an \(x\)-term, although its coefficient is 0. Exam tip: Remove any term whose coefficient is 0 when simplifying a polynomial.

Correct answer: B. \(4x^2+x+6\)

Explanation: In \(4x^2+x+6\), there is no \(x^3\) term, so the coefficient of \(x^3\) is \(0\). In contrast, the coefficient of \(x^3\) is \(1\) in \(x^3+2\) and \(7x^3+1\), and \(2\) in \(2x^3-x\). Exam tip: If a term of a particular power is missing, its coefficient is taken as \(0\).

Correct answer: C. Because it contains the negative exponent \(x^{-2}\) of the variable

Explanation: In a polynomial, the exponent of a variable must be 0 or a positive integer. Here, \(x^{-2}=\frac{1}{x^2}\) has a negative exponent, so the expression is not a polynomial. The exponent 3 in \(4x^3\) and the constant term 2 are allowed, and having three terms is not a problem. Exam tip: check for negative, fractional, or radical exponents of variables.

Correct answer: A. Yes, because all powers of x are non-negative integers.

Explanation: Yes, (5x^3-2x^2+x-1) is a polynomial because the powers of x are 3, 2, 1, and 0, all of which are non-negative integers. The constant term -1 can be viewed as -1x^0. Having four terms or a term containing x^3 does not prevent an expression from being a polynomial. Exam tip: in a polynomial, the exponent of a variable must be 0 or a positive integer.

Correct answer: B. -4

Explanation: The linear term containing x is -4x. Since -4x = (-4)\(x\), the coefficient of x is -4. Here, 7 is the coefficient of x², while 10 is the constant term. Exam tip: To find a variable’s coefficient, identify the number multiplying that variable in its term.

Correct answer: D. 7

Explanation: The highest exponent of the variable x is 7, occurring in the term 2x^7. Therefore, the degree of the polynomial is 7. Although -5x^4 has exponent 4, it is not the highest exponent. Exam tip: To find the degree, look for the greatest exponent of the variable in the simplified polynomial.

Correct answer: C. 8

Explanation: The degree of a polynomial is the greatest exponent of the variable with a non-zero coefficient. Here, the term 4x^8 has exponent 8, while the constant term -3 has exponent 0. Therefore, the degree of the polynomial is 8. The numbers 11, 4, and 3 are not exponents of x in the polynomial. Exam tip: To find degree, look for the highest power of the variable, not the coefficient.

Correct answer: A. Yes, because \(-\sqrt{2}\) is a real coefficient

Explanation: In \(-\sqrt{2}x+9\), the power of \(x\) is \(1\), and \(9\) is a constant term. Coefficients of a polynomial may be any real numbers, so \(-\sqrt{2}\) is allowed. The square root is in the coefficient, not on \(x\); hence the expression is a polynomial. Exam tip: check that powers of the variable are only \(0,1,2,\ldots\).

Correct answer: B. Because the variable is in the denominator

Explanation: A polynomial in \\(x\\) can contain only non-negative whole-number powers of \\(x\\). In the expression \\(\\frac{x^2+1}{x}\\), division by \\(x\\) gives \\(\\frac{x^2}{x}+\\frac{1}{x}=x+x^{-1}\\), wherever the expression is defined. The term \\(x^{-1}\\) has a negative exponent, so the expression is not a polynomial in \\(x\\).

Therefore, option B is correct: the variable occurs in the denominator. The presence of \\(x^2\\), the constant 1, or addition is not a problem by itself. For example, \\(x^2+1\\) is a polynomial. The important test is whether every variable power is a non-negative integer and no variable remains in a denominator after simplification.

Correct answer: C. x⁶ + x³ + x⁰

Explanation: A polynomial in x has terms of the form axⁿ, where n must be a non-negative integer. In option C, the exponents are 6, 3, and 0, so every power is valid; moreover, x⁰ equals 1. Thus x⁶ + x³ + x⁰ is a polynomial, and option C is correct. Option A contains the negative exponent −3, which is equivalent to placing x³ in a denominator. Option B contains the fractional exponent 1/4, which is not an integer. Option D contains 1/x = x⁻¹, another negative power. Therefore the other three expressions violate the exponent rule, while option C satisfies it completely.

Correct answer: C. 4

Explanation: A constant term is a term that contains no variable. In 10x^2-7x+4, both 10x^2 and -7x contain x, whereas 4 has no variable. Therefore, 4 is the constant term. Exam tip: Identify the term with no variable to find the constant term.

Correct answer: C. नहीं, क्योंकि \(x^{-1}\) में \(x\) की घात ऋणात्मक है

Explanation: In a polynomial, the powers of the variable must be 0 or positive integers. Here, \(2x^{-1}=\frac{2}{x}\), so the power of \(x\) is \(-1\). Therefore, the expression is not a polynomial. Having three terms or real coefficients alone does not make an expression a polynomial. Exam tip: Check first for negative or fractional powers, or a variable in the denominator.

Correct answer: A. 3x^5

Explanation: The leading term of a polynomial is the term with the greatest exponent of the variable. In the given polynomial, the exponents of x are 5, 4, 2, and 0. Since 5 is the greatest exponent, 3x^5 is the leading term. Although -2x^4 has a negative coefficient, its exponent is only 4, so it is not the leading term. Exam tip: Write a polynomial in descending order of exponents; the first term is the leading term.

Correct answer: A. -9

Explanation: In the polynomial \(-9x^6+x-8\), the term with the highest power is \(-9x^6\), since the power of \(x\) is 6. The numerical coefficient of this leading term is \(-9\), so the correct answer is \(-9\). The number \(6\) is the degree of the polynomial, not its coefficient. Exam tip: first identify the term with the highest exponent, then write its numerical coefficient.

Correct answer: A. \(x^4-16\)

Explanation: \(x^4-16\) has exactly two terms, \(x^4\) and \(-16\), so it is a binomial. Its exponent of \(x\) is 4, a non-negative integer; hence it is a polynomial. \(x^2+x+1\) is a trinomial, while \(\frac{1}{x}+2\) contains \(x^{-1}\), so it is not a polynomial. Exam tip: a binomial has exactly two terms, and variable exponents in a polynomial must be non-negative integers.

Correct answer: B. (2x^2-3x+9)

Explanation: A trinomial is an algebraic expression with exactly three unlike terms. For it to be a polynomial in \(x\), the powers of \(x\) must be non-negative integers. In option B, \(2x^2-3x+9\), the three terms are \(2x^2\), \(-3x\), and \(+9\). Their powers are 2, 1, and 0, so all are allowed in a polynomial.

Option A has only two terms, so it is a binomial. Option C has only one term, so it is a monomial. Option D contains \(\sqrt{x}=x^{1/2}\), which has a fractional power and therefore is not a polynomial in x. Thus option B is the only expression that is both a polynomial in x and a trinomial. The supplied answer and explanation are correct.

Correct answer: B. Yes, because (\pi) is a real number and the powers are valid

Explanation: A polynomial in \\(x\\) may have real-number coefficients, and \\(\\pi\\) is a real number. The expression \\(\\pi x^2-5x+1\\) has powers \\(2\\), \\(1\\), and \\(0\\), all of which are non-negative whole numbers. Its coefficients are \\(\\pi\\), -5, and 1, so it satisfies the definition of a polynomial in \\(x\\).

Thus option B is correct. The symbol \\(\\pi\\) is not an obstacle; irrational real numbers may be coefficients. Also, having three terms or containing \\(x^2\\) is completely allowed. A non-polynomial example would contain a negative or fractional power of \\(x\\), or place \\(x\\) in a denominator. Neither issue occurs here.

Correct answer: B. Because (\frac{1}{\sqrt{x}}) has variable power (-\frac{1}{2})

Explanation: Direct answer: Option B. A polynomial in x can have only non-negative integer powers of x: 0, 1, 2, 3 and so on. In the expression x^2 + 1/sqrt(x), the first term x^2 is acceptable. But 1/sqrt(x) can be rewritten as 1/x^(1/2) = x^(-1/2). The power -1/2 is negative and fractional, so it is not allowed in a polynomial. Option A is wrong because x^2 is a perfectly valid polynomial term. Option B is correct because it identifies the exact problem: the variable power is -1/2. Option C is wrong because addition is allowed in polynomials; many polynomial terms can be added. Option D is wrong because having two terms is also allowed; a polynomial may have one, two or many terms. Step by step: rewrite the root as a power; move the denominator to the numerator using a negative exponent; inspect -1/2; reject the expression as a polynomial. Memory cue: denominator or root containing the variable often signals an invalid negative or fractional power.

Correct answer: A. \(8x^3-6\)

Explanation: Since \(0x^2=0\), the term \(0x^2\) contributes nothing to the polynomial and is removed. Therefore, the simplified form is \(8x^3-6\). Option B incorrectly treats the zero coefficient as 1. Exam tip: Remove any term whose coefficient is 0 before simplifying a polynomial.

Correct answer: D. 0

Explanation: The polynomial 2x^4-3x+5 can be written as 2x^4+0x^2-3x+5. Its x^2 term is 0x^2, so the coefficient of x^2 is 0. Here, 2, -3, and 5 are associated with the x^4 term, the x term, and the constant term respectively. Exam tip: If a term of a given power is missing, its coefficient is 0.

Correct answer: B. x⁴ − 3x² + 2x − 7

Explanation: The governing concept is the standard form of a polynomial. In standard form, like terms are combined and the remaining terms are arranged in descending powers of the variable. Option B is x⁴ − 3x² + 2x − 7, whose exponents occur in the order 4, 2, 1, and 0. It is therefore already written in standard form, so option B is correct. Option A has powers 0, 3, and 1; option C has powers 0, 1, and 2; and option D has powers 1, 5, and 0. Each of those expressions is a polynomial, but each must be rearranged to place the highest power first. Missing powers, such as x³ in option B, are completely acceptable.

Correct answer: A. Yes, because x/3 has x to the power 1

Explanation: A polynomial in x may have real-number coefficients, including fractions. Rewrite x/3 as (1/3)x; its exponent is the whole non-negative integer 1. The other powers are 2 and 0 for x² and −4, so every term satisfies the definition and option A is correct. A fractional coefficient, a square term, or three terms does not disqualify a polynomial.