Definition of a Polynomial

When \(a=0\), the term \(ax^5\) becomes 0. Thus, \(p(x)=3x^2-7\). The highest power of \(x\) in the remaining polynomial is 2, so its degree is 2. Degree 5 would apply only if the coefficient of \(x^5\) were non-zero. Exam tip: remove all terms with zero coefficients before finding the degree.

The degree of a polynomial is the greatest exponent among terms with non-zero coefficients. Here, the coefficient of 0x^7 is 0, so this term does not contribute to the degree. Among the remaining terms, -5x^4 has the greatest exponent, 4. Hence, the correct answer is 4. Option 7 is a close distractor, but a term with zero coefficient is not counted. Exam tip: First ignore all terms whose coefficients are zero before finding the degree.

In a polynomial, the exponents of the variable must be non-negative integers. For \(n=-2\), the expression contains \(4x^{-2}=\frac{4}{x^2}\), so the variable occurs in the denominator; hence it is not a polynomial. In contrast, when \(n=0\), \(4x^0=4\) is a constant term, so the expression remains a polynomial. Exam tip: a negative or fractional exponent means the expression is not a polynomial.

13 is a non-zero constant polynomial because it has no term involving t. Every non-zero constant polynomial has degree 0. The zero polynomial 0 is usually treated as having an undefined degree, so it is not the correct choice. Both t+13 and t^0+t = 1+t have degree 1. Exam tip: the degree of a non-zero constant polynomial is always 0.

In class (9), when the variable is under a root, the power is not treated as a valid integer power. So it is not considered a polynomial.

\(x^4+5x^2-6\) is a polynomial because the powers of \(x\) present are \(4, 2\), and \(0\), all of which are non-negative integers. It has no \(x^3\) or \(x\) term; a polynomial does not need to include every intermediate power. Option B contains an \(x^3\) term, option C contains an \(x\) term, and option D is not a polynomial because it has a negative exponent. Exam tip: in a polynomial, every exponent of the variable must be a non-negative integer.

The zero polynomial is (0), but its degree is not defined. Keep zero and non-zero constant polynomials separate.

On substituting \(a=2\), the coefficient of \(x^4\) becomes \(a-2=0\). Hence, the \(x^4\) term vanishes and the polynomial becomes \(5x^2-3\). The highest power of \(x\) in this polynomial is 2, so its degree is 2. Option 4 is incorrect because a term with zero coefficient is not counted. Exam tip: Always simplify a polynomial and remove zero-coefficient terms before finding its degree.

In (3x^5-2x^2+7), all powers are valid and the highest power is (5). The other options have invalid powers.

In option A, the powers of \(x\) are \(3\), \(1\), and \(0\). All are non-negative integers, so \(ax^3+bx+9\) is a polynomial in \(x\). Option B contains the negative exponent \(x^{-1}\). In option C, \(\frac{a}{x^2}=ax^{-2}\), which also has a negative exponent. Option D contains \(\sqrt{x}=x^{\tfrac{1}{2}}\), a fractional exponent. Exam tip: every exponent of the variable in a polynomial must be one of \(0,1,2,3,\ldots\).

The degree of a polynomial is determined by the highest-power term whose coefficient is non-zero. Here, if k has any non-zero value, the term kx³ remains and the polynomial has degree 3. To make the degree 2, the x³ term must vanish; hence k=0. Exam tip: In parameter-based polynomials, check the coefficient of the highest-power term first.

In \(x^5-2x^3+7x-4\), the powers of \(x\) are \(5,3\), and \(1\), all of which are odd positive integers, and \(-4\) is a constant term. Therefore, it satisfies the given condition. Option B is also a polynomial, but it contains the even powers \(4\) and \(2\). Option C has a negative exponent, while option D has \(\sqrt{x}=x^{1/2}\), a fractional exponent, so neither is a polynomial. Exam tip: variable exponents in a polynomial must be non-negative integers.

The direct answer is C, degree 0. Degree is found after removing terms whose coefficients are zero. In the given expression, 0x^4=0 and 0x^2=0, so it becomes 6. The remaining polynomial is the non-zero constant 6, whose degree is 0. Option A, 4, is wrong because the x^4 term has coefficient zero and therefore is not actually present. Option B, 2, is wrong for the same reason: the x^2 term also disappears. Option C is correct because the highest power of x in the simplified non-zero polynomial 6 is 0. Option D, not defined, is wrong because the expression is not the zero polynomial; it equals 6. The degree of the zero polynomial is treated specially, but that rule is irrelevant here. Do not identify degree by looking at the largest exponent before checking coefficients. Memory cue: first delete zero-coefficient terms, then find the highest remaining exponent.

The main condition of a polynomial is about powers of the variable. Coefficients may be real, but powers must also be valid.

The degree of a polynomial is determined by the non-zero term having the greatest power of the variable. The given polynomial has \(x^8\), \(x^5\), \(x^2\), and 1; the highest power is 8 in \(x^8\). Hence, \(x^8\) determines the degree. Although \(x^5\) is the closest distractor, its power is only 5. Exam tip: compare the powers of the variable in all terms; a non-zero constant has degree 0.

The direct answer is C. A polynomial in x may use addition, multiplication, and non-negative whole powers of x, but the variable cannot remain in a denominator. Check each option. Option A becomes x(x+2)+3=x^2+2x+3, which is a polynomial because all powers are non-negative. Option B becomes 2(x^2-1)-x=2x^2-x-2, also a polynomial. Option C is x+1/(x+1). The fraction contains x in its denominator, and it cannot be cancelled with anything in the expression. Its denominator is zero at x=-1, and it is not a polynomial term; therefore C is correct. Option D becomes 5(x-3)+x^2=5x-15+x^2=x^2+5x-15, a polynomial. The number of terms is not the deciding test: a polynomial may have one, two, or many terms. Option C is not rejected merely because it looks complicated; it fails because of the variable denominator. Memory cue: expand brackets freely, but a variable under a fraction bar is a warning sign.

For degree (2), the coefficient of (x^3) must be (0) and the coefficient of (x^2) non-zero. This happens at (m=-1).

A linear polynomial has degree 1, so the coefficient of \(x^2\) must be zero. Here, the coefficient of \(x^2\) is \(a-3\). Thus, \(a-3=0\) gives \(a=3\). For values such as \(a=0\) or \(a=4\), the \(x^2\) coefficient is non-zero, so the polynomial remains quadratic. Exam tip: To reduce the degree of a polynomial, set the coefficient of its highest-power term to zero.

The degree of a polynomial is the greatest exponent of the variable among its terms with non-zero coefficients. Here the exponents are 4, 2, and 0; since \(x^0=1\), \(3x^0=3\) is a constant term. Therefore, the greatest exponent is 4. Option 2 is only the exponent of \(-2x^2\), not the degree of the entire polynomial. Exam tip: Treat \(x^0\) as 1 first, then identify the highest exponent.

Every term has coefficient 0, so \(p(x)=0\) for every value of \(x\). Therefore, it is the zero polynomial. Although \(0x^3\) appears in the expression, it is not a cubic polynomial because the coefficient of \(x^3\) is zero. Exam tip: remove terms with zero coefficients before identifying the degree; the degree of the zero polynomial is generally considered undefined.

In (x^3-5x+6), the highest power is (3) and the (x^2) term is absent. A missing term does not make a polynomial invalid.

In a polynomial, every exponent of the variable must be a non-negative integer. Here, \(\frac{1}{x^2}=x^{-2}\), so \(x\) has exponent \(-2\). Since a negative exponent occurs, the expression is not a polynomial. Both \(x^4\) and \(x\) are valid polynomial terms, and an expression may have three terms. Exam tip: When a variable appears in a denominator, rewrite it using a negative exponent before checking whether it is a polynomial.

For degree (3), the coefficient of (x^5) must be (0). From (k+2=0), (k=-2), and the (x^3) coefficient remains non-zero.

A polynomial in \(x\) may have powers such as 0, 1, 2, 3, and so on, but not negative or fractional powers. The question adds another condition: every non-zero power of the variable must be even. In option A, \(x^6-3x^4+2x^2-5\), the powers are 6, 4, 2, and 0. The non-zero powers 6, 4, and 2 are all even, and the constant term has power 0.

Option B contains odd powers 5 and 3, so it fails the even-power condition. Option C contains the negative power \(x^{-2}\), so it is not a polynomial. Option D contains the fractional power \(x^{1/2}\), so it is also not a polynomial. Therefore option A is the only expression satisfying both requirements. The supplied answer and explanation are correct.

The terms of 2x^3+5x^2+x+7 are 2x^3, 5x^2, x, and 7, so it has four terms. The highest exponent of the variable x is 3; therefore, the degree of the polynomial is 3. The number 7 is a constant term, not the degree. Exam tip: Count terms separated by + or − signs, then identify the greatest exponent of the variable.