Definition of a Polynomial

Correct answer: C. 5

Explanation: The degree of a polynomial is the highest power of the variable having a non-zero coefficient. Here, \(0x^8\) is a zero term, so \(x^8\) is not counted. Among the remaining terms, the highest power is \(5\), in \(-6x^5\). Option \(3\) is a close distractor, but \(2x^3\) has a lower power. Exam tip: remove all terms with zero coefficients before finding the degree.

Correct answer: A. (k=1)

Explanation: For degree (3), the (x^4) coefficient must be zero and the (x^3) coefficient must be non-zero. At (k=1), (k+2=3) is also non-zero.

Correct answer: C. No such (r) exists

Explanation: The direct answer is C, no such value of r exists. For the polynomial to be constant, the coefficients of x³ and x² must both be zero. From r−3=0 we get r=3. From r+1=0 we get r=−1. Since one value of r cannot equal both 3 and −1, the requirements are inconsistent. The constant term 9 is non-zero, so if the variable terms could vanish, the result would indeed be a non-zero constant; however, they cannot vanish together. Option A, r=3, removes x³ but leaves 4x². Option B, r=−1, removes x² but leaves −4x³. Option C is correct because no single r satisfies both equations. Option D, r=1, leaves both variable terms. Exam method: write one equation for every variable-term coefficient and check whether all equations have a common solution.

Correct answer: C. Degree is not defined

Explanation: The degree is defined as the greatest exponent of the variable in a non-zero polynomial. A non-zero constant, such as \(7\), has degree \(0\), because it can be written as \(7x^0\). The zero polynomial is different: every coefficient is zero, so there is no highest power with a non-zero coefficient. Consequently, its degree is not defined in the usual school-level definition.

For \(p(x)=0\), there is no term such as \(ax^n\) with \(a\neq0\) from which a greatest exponent can be selected. Calling its degree \(0\) would incorrectly confuse it with a non-zero constant polynomial. Therefore the mathematically appropriate statement is that the degree of the zero polynomial is not defined. Hence option C is correct.

Correct answer: B. (-4)

Explanation: For degree (4), the (x^5) coefficient must be zero and the (x^4) coefficient must be non-zero. At (m=-4), (m-1=-5) is non-zero.

Correct answer: B. It is not a polynomial because the variable is inside a root

Explanation: The correct answer is B: the expression is not a polynomial in the Grade 9 sense. A polynomial may contain powers such as x^0, x^1, x^2, but it cannot contain the variable inside a square root in this form. The second part is √(x^2+4), where x occurs under the radical, so it is not a polynomial term. Although x^2 itself is a valid quadratic term, one non-polynomial part makes the entire sum non-polynomial. Option A is wrong because the expression is not a quadratic polynomial. Option B is correct. Option C is wrong because x is present, so the expression is not constant. Option D is wrong because it is not a linear polynomial either. The important school-level test is to inspect the form of every term, not just the first term. Memory cue: a variable under a root generally signals “not a polynomial.”

Correct answer: B. It is not a polynomial because it contains x^{-2}

Explanation: Option B is correct. Here, x^0=1, so a zero exponent is perfectly valid. However, x^{-2}=\(\frac{1}{x^2}\), which gives the variable a negative exponent. In a polynomial, variable exponents must be 0 or positive integers; therefore, the given expression is not a polynomial. Option A considers only x^0 and ignores x^{-2}. Exam tip: To identify a polynomial, first check whether any variable has a negative or fractional exponent.

Correct answer: C. (\frac{x^3+1}{x})

Explanation: (\frac{x^3+1}{x}=x^2+\frac{1}{x}). Since (\frac{1}{x}) has a negative power, it is not a polynomial.

Correct answer: A. Always 4

Explanation: The degree of a polynomial is the highest power whose coefficient is non-zero. Here, the coefficient of x^4 is 2, which cannot become zero for any value of b. Hence, the degree of p(x) is always 4. When b=3, only the coefficient of x^2 becomes zero; the x^4 term remains. Exam tip: To find the degree, first check the coefficient of the term with the highest power.

Correct answer: A. \(x^7-4x^5+2x-6\)

Explanation: In option A, the variable terms have powers \(7, 5\), and \(1\), all of which are odd positive integers. The term \(-6\) is a constant, so \(x^7-4x^5+2x-6\) satisfies the condition. Option B includes the even power \(x^6\). Option C has the negative exponent in \(x^{-7}\), while option D has the fractional exponent \(\sqrt{x}=x^{1/2}\); therefore, neither is a polynomial in \(x\). Exam tip: Exponents of a variable in a polynomial must be non-negative integers.

Correct answer: C. Constant polynomial

Explanation: Terms with a zero coefficient contribute nothing to a polynomial and can be removed without changing its value. In the given expression, 0x^4 and 0x^2 are both zero terms. Thus p(x) = 0x^4 + 0x^2 + 9 simplifies to p(x) = 9. Since 9 is not zero and does not depend on x, it is a non-zero constant polynomial.

A constant polynomial has degree 0, but the question asks for its type. It is therefore a constant polynomial, so option C is correct. It is not the zero polynomial because its remaining value is 9, not 0. It is also neither linear nor quadratic, because no x or x^2 term with a non-zero coefficient remains after simplification.

Correct answer: A. (0)

Explanation: For degree (2), the (x^3) coefficient must be zero and the (x^2) coefficient must be non-zero. At (k=0), (k-2=-2) is non-zero.

Correct answer: C. 3

Explanation: Given \(f(x)=x^2-1\), we get \(g(x)=x^3+f(x)=x^3+x^2-1\). The highest power of \(x\) in this polynomial is 3, so the degree of \(g(x)\) is 3. Degree 2 is the degree of \(f(x)\), not of \(g(x)\). Exam tip: Substitute first, simplify the polynomial, and then identify the highest power of \(x\).

Correct answer: B. (\sqrt{7}x^4-3x+2)

Explanation: The direct answer is B, √7x⁴−3x+2. A polynomial may have real coefficients such as integers, fractions, or irrational real numbers, provided the powers of the variable are non-negative integers. In option B, √7 is an irrational but real coefficient, and the powers of x are 4 and 1, while the constant term has power 0. Thus it satisfies both requirements. Option A contains √x=x^(1/2), whose exponent is not a non-negative integer, so it is not a polynomial. Option C contains x^(−2), a negative power, so it is not a polynomial, even though √7 is a valid coefficient. Option D contains 1/x=x^(−1), also a negative power. Option B is therefore the only expression with the required form. Memory cue: irrational coefficients are allowed; fractional or negative exponents of the variable are not.

Correct answer: C. No such \(a\) exists

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

Correct answer: A. (x^n+2), where (n) varies

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

Correct answer: C. 7x^2+5

Explanation: In \(7x^2+5\), the highest power of \(x\) is 2, so it is a quadratic polynomial. It has only two terms, \(7x^2\) and 5; therefore, it is a binomial, not a trinomial. Although \(2x^2-3x+1\) is also quadratic, it has three terms. Exam tip: find the degree from the highest exponent and the type from the number of terms separately.

Correct answer: B. Any real (h)

Explanation: The degree of a polynomial is the highest power of the variable having a non-zero coefficient. Here, the coefficient of (x^3) is 2, which remains non-zero for every real value of (h). Hence, the polynomial always has degree 3. For (h=5), only the (x) term disappears; the (x^3) term does not. Exam tip: To determine degree, first check the coefficient of the highest-power term.

Correct answer: A. It actually simplifies to \(x^3+1\), so it is a polynomial.

Explanation: Factor the numerator: \(x^5+x^2=x^2(x^3+1)\). Hence \(\frac{x^5+x^2}{x^2}=x^3+1\), for \(x\ne0\). The exponents of the variable in \(x^3+1\) are 3 and 0, both non-negative integers, so it is a polynomial. Option B comes from incorrect division by \(x^2\). Exam tip: factor the numerator before cancelling a common factor in an algebraic fraction.

Correct answer: B. It is not a polynomial because (\sin x) is not a polynomial term

Explanation: In a polynomial, terms are powers of the variable with non-negative integer exponents. (\sin x) is not such a term.

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

Explanation: The degree of a polynomial is the highest exponent whose coefficient is non-zero. Here, the coefficient of \(x^2\) is \(1\), and it never becomes zero for any value of \(a\). Hence, \(p(x)\) has degree \(2\) for all real \(a\). Even when \(a=0\), \(p(x)=x^2\), which still has degree \(2\). Exam tip: In a polynomial containing a parameter, first check whether the coefficient of the highest-power term can become zero.

Correct answer: B. \(4+\frac{1}{x^5}\)

Explanation: Option B, \(4+\frac{1}{x^5}\), is not a polynomial because \(\frac{1}{x^5}=x^{-5}\). The exponent of \(x\) is \(-5\), whereas exponents of variables in a polynomial must be non-negative integers: \(0,1,2,\ldots\). In option C, \(x^0=1\), so it is a polynomial. Exam tip: If a variable is in the denominator, rewrite it with a negative exponent and check it.

Correct answer: A. 4 terms, degree 6

Explanation: The terms of the polynomial are x^6, -2x^4, x^2, and -1. Therefore, it has 4 terms. The degree of a polynomial is the greatest exponent of the variable in any term, which is 6 here. Degree 4 is incorrect because the term x^6 is present. Exam tip: Count terms by separating the expression at addition and subtraction signs.

Correct answer: B. \(z^2x^4+3zx^2-8\)

Explanation: In \(z^2x^4+3zx^2-8\), the powers of \(x\) are \(4\), \(2\), and \(0\). All are non-negative integers, so it is a polynomial in \(x\). The expressions \(z^2\) and \(3z\) act as coefficients of powers of \(x\). In option A, \(x\) has exponent \(-2\); in option C, \(x\) is in the denominator, so its exponent is \(-1\); and in option D, the exponent of \(x\) is \(\frac12\). Hence, these are not polynomials in \(x\). Exam tip: when testing for a polynomial in \(x\), check the exponents of \(x\); other letters may be coefficients.

Correct answer: A. (2)

Explanation: For degree (0), the coefficients of both (x^2) and (x) must be zero. From (a-2=0), (a=2), and (7) remains.