Degree of a Polynomial

In \(x^2+4\), the powers of \(x\) are 2 and 0, which are non-negative integers; therefore, it is a polynomial. Its highest power is 2, so it is a quadratic polynomial, not a linear polynomial. \(5x-2\) is linear because its degree is 1. \(\frac{1}{x}+1\) and \(\sqrt{x}+2\) are not polynomials because they contain powers \(-1\) and \(\frac{1}{2}\), respectively. Exam tip: the exponent of a variable in a polynomial must be 0 or a positive integer.

In \(5x^4-x+2\), the highest power of \(x\) is \(4\). Therefore, it is a polynomial of degree \(4\), not degree \(2\). \(x^2+x+1\) is also a polynomial, but its degree is \(2\). \(\sqrt{x}+1\) has \(x\) raised to \(\frac{1}{2}\), and \(\frac{1}{x}+x^2\) has \(x\) raised to \(-1\), so neither is a polynomial. Exam tip: exponents of variables in a polynomial must be non-negative integers.

The governing concept is the degree of a polynomial. The degree is the greatest exponent of the variable among all non-zero terms. In p(t) = 2t⁴ − 7t² + 3, the powers of t are 4, 2, and 0, corresponding to 2t⁴, −7t², and 3. The term 2t⁴ has the greatest exponent, so it determines the polynomial’s degree, which is 4. Therefore option C is correct. The constant 3 has degree 0, while 7t² has degree 2, so neither determines the overall degree. Option D lists two terms together rather than identifying the single highest-power term requested by the question.

In \(x^3+2x\), the powers of \(x\) are \(3\) and \(1\), both non-negative integers. The greatest power is \(3\), so it is a polynomial of degree \(3\). In option B, the expression is a polynomial, but its degree is \(2\), not \(1\). Exam tip: To find the degree, identify the greatest power of the variable only after checking that no negative or fractional powers occur.

In \(x^4+x\), the powers of \(x\) are \(4\) and \(1\), both of which are non-negative integers. Hence, it is a polynomial in \(x\), and its highest power, or degree, is \(4\). The expressions \(x^{-4}\), \(\sqrt{x}=x^{1/2}\), and \(\frac{1}{x}=x^{-1}\) contain negative or fractional powers, so they are not polynomials in \(x\). Exam tip: in a polynomial, powers of the variable must be \(0,1,2,\ldots\).

\(x^2+6\) is a polynomial because the exponents of \(x\) are non-negative integers, namely \(2\) and \(0\). Its highest exponent is \(2\), so its degree is \(2\), not \(1\). The closest option, \(4x-7\), has degree \(1\). \(\frac{1}{x}+3\) and \(\sqrt{x}+2\) are not polynomials because they contain \(x^{-1}\) and \(x^{1/2}\), respectively. Exam tip: in a polynomial, every variable exponent must be a non-negative integer.

In \(2x^6-x^3+8\), the powers of \(x\) are \(6\), \(3\), and \(0\), all of which are non-negative integers. The highest exponent is \(6\), so it is a polynomial of degree \(6\). In option A, \(x^{6/2}=x^3\), so its degree is \(3\); options B and D contain negative powers of \(x\), so they are not polynomials. Exam tip: before finding a polynomial’s degree, check that no variable has a negative or fractional exponent.

The degree of a non-zero polynomial is the greatest exponent of x whose coefficient is non-zero. Substitute b = 0 into the expression: ax^4 + bx^6 + 3 becomes ax^4 + 0x^6 + 3, or simply ax^4 + 3. Since a is not equal to zero, the x^4 term remains and has a non-zero coefficient. The x^6 term disappears because its coefficient b is zero. Therefore the greatest remaining power is 4, so option B is correct. Option C incorrectly uses the original largest exponent even though its term vanishes. Option D wrongly adds exponents, and option A would apply only to a non-zero constant polynomial, which this is not.

To determine the degree correctly, first substitute the given value of the parameter r. When r = 0, the coefficient of x^3 is r^2 - r = 0^2 - 0 = 0, so the entire cubic term vanishes. The coefficient of x^2 is r + 2 = 0 + 2 = 2. The expression therefore becomes 2x^2 + 1. Its greatest power of x with a non-zero coefficient is 2, so its degree is 2 and option B is correct. Option C ignores the zero coefficient of x^3. Option D incorrectly adds powers, while option A would describe a non-zero constant, not the remaining quadratic expression.

The governing concept is that the degree must be found after substituting the given value and simplifying; a term whose coefficient becomes zero must be removed. Put n=1 into the expression. Then n²−1=1²−1=0, so the coefficient of x⁴ is zero and the x⁴ term disappears. Also, n+2=1+2=3, so the remaining expression is 3x²−6. The highest power of x with a non-zero coefficient is 2. Therefore the degree is 2 and option C is correct. Option A incorrectly counts the original x⁴ term even though its coefficient vanishes. Option B is not the highest surviving power, and option D would apply only if the result were a non-zero constant, which it is not.

The degree of a non-zero polynomial is the greatest exponent of x with a non-zero coefficient. The leading coefficient is the coefficient of the term having that greatest exponent. Option B, −5x⁴+x²−7, has highest exponent 4, so its degree is 4, and the coefficient of x⁴ is −5, which is negative. It therefore satisfies both conditions and is correct. Option A also has degree 4, but its leading coefficient is +3, so it fails the sign requirement. Option C has a negative leading coefficient, but its highest exponent is 3, giving degree 3 rather than 4. Option D is linear with degree 1. Thus option B is the only unambiguous answer. The revised options are polynomials themselves, avoiding any confusion caused by quotient notation.

The governing concept is degree after substitution and simplification. Substitute n=−2 into each coefficient. First, n²−4=(−2)²−4=4−4=0, so the coefficient of x⁶ is zero and the x⁶ term vanishes. Next, n+5=−2+5=3, so the cubic term becomes 3x³. The expression therefore reduces to 3x³−7. The greatest exponent with a non-zero coefficient is 3, so the degree is 3 and option B is correct. Option A counts the original x⁶ term without checking that its coefficient becomes zero. Options C and D overlook the surviving cubic term; the constant −7 does not determine the degree while the non-zero x³ term remains.

The governing concept is that the degree of a polynomial is the greatest exponent of the variable whose coefficient is non-zero after all given conditions are applied. Here, a = 0, so ax⁶ = 0 and the sixth-degree term disappears completely. Since b ≠ 0, bx⁴ remains a non-zero term. The terms x and −1 have powers 1 and 0, both smaller than 4. Thus the simplified polynomial is bx⁴ + x − 1, and its highest surviving power is 4. Therefore option B is correct. Option A counts the term that vanished, option C ignores the surviving fourth-degree term, and option D ignores all variable terms.

The governing rule is that a polynomial’s degree is found only after substituting the given values and removing terms whose coefficients become zero. Put z = −2. Then z + 2 = −2 + 2 = 0, so (z + 2)x⁵ becomes 0 and the fifth-degree term disappears. Also, z² − 4 = (−2)² − 4 = 4 − 4 = 0, so the cubic term disappears as well. The expression is reduced to 9x − 1. Its highest power with a non-zero coefficient is x¹, so the degree is 1. Therefore option C is correct. Options A and B count terms that vanish after substitution, while D overlooks the surviving term 9x.

The degree of a polynomial is determined by the largest exponent whose coefficient is non-zero. In p(x), the candidate highest-degree term is λx⁴. Hence its coefficient must satisfy λ ≠ 0 for the degree to be 4. If λ = 0, the fourth-degree term vanishes and the remaining polynomial −2x² + 3 has degree 2. The values −2 and 3 are coefficients of lower-degree terms and do not determine this condition.

In \(6x^4-x+3\), the powers of \(x\) are \(4\), \(1\), and \(0\). All are non-negative integers, so it is a polynomial. Its highest power is \(4\), therefore it is not a polynomial of degree \(2\). \(x^2+5x+6\) is a degree-\(2\) polynomial. \(\sqrt{x}+2\) and \(\frac{1}{x}+x^2\) are not polynomials because they contain powers \(\frac{1}{2}\) and \(-1\), respectively. Exam tip: in a polynomial, variable exponents must be non-negative integers, and the greatest exponent gives the degree.

In \(2x^5-x^2+3\), the powers of \(x\) are \(5\), \(2\), and \(0\). All are non-negative integers, so it is a polynomial. Its highest power is \(5\), hence its degree is \(5\). \(x^{5/2}+2\) has a fractional exponent, whereas \(x^{-5}+4\) and \(\frac{1}{x^5}+1\) have negative powers of \(x\); therefore, they are not polynomials. Exam tip: In a polynomial, an exponent of a variable can only be \(0,1,2,\ldots\).

The degree of a polynomial is the highest exponent of its variable with a non-zero coefficient. The exponents of the terms in the given polynomial are 4, 2, 1, and 0. The greatest exponent is 4, so its degree is 4. Option 2 is only the exponent of the term -5x^2, not of the whole polynomial. Exam tip: A constant term has degree 0.

The degree of a polynomial is the greatest exponent among terms with non-zero coefficients. Here, the coefficient of \(0x^7\) is zero, so it does not count as a term of the polynomial. Among the remaining terms, \(8x^5\) has the greatest exponent, 5. Therefore, the degree is 5. Option 7 is a close distractor, but a term with a zero coefficient cannot determine the degree. Exam tip: First ignore all terms whose coefficients are zero before finding the highest power.

Here, 4x^3 and -4x^3 cancel each other, so the expression becomes 9x^2-5. The highest power of x in this polynomial is 2; therefore, its degree is 2. Option 3 is incorrect because the x^3 terms become zero after simplification. Exam tip: Always simplify like terms before finding the degree.

Substituting \(a=3\) gives \(a-3=0\). Hence, the \(x^6\) term vanishes and the polynomial becomes \(p(x)=2x^2-1\). The highest power of \(x\) is \(2\), so its degree is \(2\). Option 6 is incorrect because a term with zero coefficient is not part of the polynomial. Exam tip: Substitute any given parameter values and remove zero-coefficient terms before finding the degree.

The degree of a polynomial is the greatest exponent of the variable having a non-zero coefficient. For the degree to be 1, the coefficient of \(x^4\) must be zero: \(k+5=0\). Hence, \(k=-5\). The polynomial then becomes \(-7x+9\), which has degree 1 because the coefficient of \(x\), \(-7\), is non-zero. Exam tip: In parameter-based polynomials, first set the coefficient of the highest power equal to zero and then check the remaining terms.

For a non-zero polynomial in one variable, the degree is the greatest exponent of the variable attached to a non-zero coefficient. The exponents in −2x^8 + 5x^6 − x^2 + 4 are 8, 6, 2, and 0; the constant 4 may be viewed as 4x^0. The largest exponent is 8, and its coefficient −2 is non-zero. Thus the leading term −2x^8 determines the degree, so option A is correct. The negative sign changes the coefficient but has no effect on the exponent or degree. The terms 5x^6 and −x^2 have smaller exponents, while the constant 4 has degree zero. None of those terms can determine the polynomial’s degree.

The degree of a polynomial is the greatest exponent of the variable in any term with a non-zero coefficient. In the given polynomial, the exponents are 2, 9, 5, and 0; the greatest is 9 in the term \(x^9\). Therefore, the degree is 9. The term \(6x^5\) has exponent 5, so it does not determine the degree. Exam tip: the order of terms does not affect the degree of a polynomial.

Terms with coefficient 0 are not counted in the standard form of a polynomial. Thus, \(q(x)=11x-4\). The highest power of \(x\) in this polynomial is 1, so its degree is 1. Although powers 5 and 3 appear in the expression, their coefficients are 0, so they do not determine the degree. Exam tip: Remove all zero-coefficient terms before finding the degree.