Degree of a Polynomial

For a multivariable polynomial, the total degree of a term is the sum of the exponents of all variables in that term, and the total degree of the polynomial is the largest such sum among its nonzero terms. For x^7y^4z^3, the sum is 7+4+3=14. For 5x^3y^2z^10, the numerical coefficient 5 is ignored when calculating degree, and the exponent sum is 3+2+10=15. The constant term −14 has degree 0. The greatest term-degree is therefore 15, so the polynomial has total degree 15. Option A considers only the first term, option C uses an incomplete exponent sum, and option D considers only the exponent of x. Hence option B is correct.

A polynomial has degree 0 when every positive-power term vanishes and the remaining constant is nonzero. Here the coefficient of x^15 is u^2−196=(u−14)(u+14), and the coefficient of x^8 is u−14. To eliminate x^8, we must take u=14. Substituting this value into the first coefficient gives 14^2−196=196−196=0, so x^15 also disappears. The expression then reduces to the constant 13, which is nonzero and therefore has degree 0. Although u=−14 makes the x^15 coefficient zero, it leaves the x^8 coefficient −28, so the degree is not 0. The values 0 and 196 do not eliminate both variable terms. Thus option A is correct.

The governing concept is that the degree of a polynomial is the greatest exponent of a term with a nonzero coefficient. In 2x^n + 7x^8 − 1, the coefficient of x^n is 2, so the term cannot disappear. Since n > 8, the exponent n is greater than the exponent 8, and the degree is consequently n. We only need to identify which option is strictly greater than 8. The value 11 satisfies the condition. Values 6 and 4 are less than 8, while 8 is equal to 8 and does not satisfy the strict inequality n > 8. Therefore n = 11 is the only possible value, making option C correct. The constant term −1 has degree 0 and does not affect the comparison of the leading exponents.

The governing concept is that the degree of a nonzero polynomial is the largest exponent whose coefficient is nonzero. Here the coefficient of x^12 is b−3. Because b ≠ 3, we know b−3 ≠ 0, so the x^12 term definitely remains in the polynomial. The other terms have exponents 5 and 0, which are both smaller than 12. Consequently, the greatest surviving exponent is 12 and option B is correct. Option A ignores the nonzero x^12 term. Option C is the excluded value of the parameter, not the degree; if b were 3, the x^12 term would vanish, but that case is explicitly ruled out. Option D has no relevance to the highest exponent.

The governing concept is that the degree of a nonzero polynomial is the greatest exponent of the variable whose coefficient is nonzero. The term 0x^15 does not contribute, even though it displays the largest exponent, because its coefficient is zero and the entire term equals zero. Since t ≠ 0, the term tx^13 is present with a nonzero coefficient. The other nonzero terms are 6x^4 and −8, having degrees 4 and 0. Therefore the greatest exponent attached to a nonzero term is 13, so option B is correct. Option A incorrectly counts the zero term, while options C and D overlook the surviving thirteenth-degree term.

A polynomial is linear only when its degree is exactly 1. To determine the degree of this product, expand the factors: (3x − 2)(x + 5) = 3x² + 15x − 2x − 10 = 3x² + 13x − 10. The highest power present is x², and its coefficient 3 is nonzero, so the product has degree 2. It is therefore a quadratic polynomial rather than a linear one, making option B correct. Option A incorrectly assumes that multiplying two linear factors keeps the degree at 1; in fact, their leading terms produce 3x². Options C and D are also incorrect because the result contains variable terms and is neither a constant nor the zero polynomial. No cancellation removes the x² term.

The governing concept is the degree of a polynomial. A polynomial is linear only when its highest power of x is 1. Expand the product: (4x+1)(x−6)=4x²−24x+x−6=4x²−23x−6. The leading term is 4x², whose coefficient is nonzero, so the degree is 2. Therefore the product is quadratic rather than linear, and option D is correct. Option A would be appropriate only if the highest surviving power were x¹. The expression is not a constant because it contains x² and x terms, and it is not the zero polynomial because its leading coefficient 4 is nonzero. Thus B and C are also impossible. The expansion provides a direct and complete verification.

The governing concept is the degree of a polynomial. A linear polynomial has degree exactly 1. Multiplying two nonconstant linear factors usually produces a quadratic expression. Expanding here gives (5x−2)(x+7)=5x²+35x−2x−14=5x²+33x−14. The coefficient of x² is 5, which is nonzero, so the highest power of x is 2. The product is therefore a quadratic polynomial, not a linear polynomial, and option C is correct. Option A states the wrong degree. Options B and D are also incorrect because the result contains variable terms and is neither a constant nor the zero polynomial. No cancellation removes the x² term.