Polynomials in one variable

The governing idea is the factor theorem and polynomial division. Since (x − 1) is a factor, divide p(x) by (x − 1), or use synthetic division with 1. The coefficients 1, −4, −7, 10 give a quotient with coefficients 1, −3, −10 and remainder 0. Therefore p(x) = (x − 1)(x² − 3x − 10). Multiplication confirms this: x³ − 3x² − 10x − x² + 3x + 10 = x³ − 4x² − 7x + 10. Thus option A is correct. Option B has the wrong sign for the middle term, option C does not reproduce the cubic, and option D is only part of the original coefficient pattern, not a quadratic quotient.

For a quadratic polynomial ax² + bx + c with zeroes α and β, Vieta’s relation gives αβ = c/a. In this polynomial, a = 1 and c = 16, so αβ = 16, independently of the value of b. If two nonzero numbers are reciprocals of each other, they have the form t and 1/t, and their product is exactly 1. Since the required product would have to be both 16 and 1, the condition is impossible for every value of b. The coefficient b controls the sum α + β = −b, but it cannot alter the fixed product. Thus option A is correct. Options B, C, and D merely assign particular values to b and do not remove the contradiction in the product of the roots.

For ax² + bx + c, if the zeroes are α and β, then α + β = −b/a and αβ = c/a. In this polynomial, a = 2, b = −7, and c = 5. Hence the sum is −(−7)/2 = 7/2, while the product is 5/2. The required difference is sum minus product: 7/2 − 5/2 = 2/2 = 1. Therefore option A is correct. Option B would result from an arithmetic error, option C confuses the reciprocal coefficient relationship, and option D does not follow from either standard relation. The calculation uses the coefficient formulas directly, so solving for the individual zeroes is unnecessary.