Divide First, Then Substitute — This Quartic Factors Into Two Pairs | Math Olympiad

Polynomial division, radical substitution, olympiad algebra — the equation hides a clean factorization that only appears after dividing the numerator by x+7 before substituting anything. (2x²+3x+21)/(7x) = √(x+7). Set y = √(x+7), so x = y²−7. Substituting directly into the numerator gives a messy degree-4 expression in y — unless you divide first. Polynomial division gives 2x²+3x+21 = (x+7)(2x−11)+98. Now (x+7) is exactly y², and 2x−11 rewrites as 2y²−25 after substituting x = y²−7. The numerator collapses to 2y⁴−25y²+98 in two clean steps. Cross multiply with the denominator 7(y²−7) and the equation becomes 2y⁴−7y³−25y²+49y+98 = 0. This quartic factors as (2y²−y−14)(y²−3y−7) = 0 — two independent quadratics, each solvable by the quadratic formula. Case 1 gives y = (1+√113)/4 after rejecting the negative root since y = √(x+7) ≥ 0. Case 2 gives y = (3+√37)/2 by the same rejection. Square each and subtract 7 to recover x: x = (1+√113)/8 and x = (9+3√37)/2. Both verify numerically — plug either value back in and both sides agree to three decimal places. The polynomial division step before substitution is the entire trick. Substitute first without dividing and you get a quartic with no visible factorization. Divide first and the structure x = y²−7 maps perfectly onto both pieces of the division. This division-before-substitution technique works for any rational equation where the numerator is a polynomial in x and the substitution variable equals a linear shift of x. Divide first to expose the shifted variable, then substitute — the factorization becomes visible that would otherwise stay hidden. 🔔 Daily olympiad problems — algebra, number theory, geometry, combinatorics. #OlympiadMath #PolynomialDivision #RadicalSubstitution #AlgebraTrick #CompetitionMath #AMCPrep #AIМЕPrep #MathOlympiad #QuarticFactoring #SubstitutionMethod #TwoSolutions #OlympiadAlgebra