Definition of a Polynomial

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.

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.

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.

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.

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.

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.”

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.

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

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.

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.

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.

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.

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 \(x\).

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.

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.

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

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.

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.

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.

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

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.

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.

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.

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.

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