Algebra: Equations, Inequalities, and Polynomial Structure
What this domain tests
Algebra rewards structure. The exam presents linear equations and inequalities, systems, quadratics, polynomial and rational expressions, absolute value, and literal equations, and it expects you to choose an efficient method rather than default to one procedure. Items often embed algebra inside a context - a break-even point, a mixture, a revenue model - so translating the words into symbols is part of the work.
Core rules
- Balance operations on both sides, and when you multiply or divide an inequality by a negative number, reverse the inequality symbol.
- Systems: substitution suits an equation already solved for a variable; elimination suits matching coefficients. A system with the same slope and different intercepts has no solution; identical lines have infinitely many. Graphically, a solution is an intersection point.
- Quadratics have three routes: factoring when the trinomial is factorable, completing the square to expose the vertex, and the quadratic formula x = (−b ± √(b² − 4ac))/(2a), which always works. The discriminant b² − 4ac tells you the number of real solutions: positive gives two, zero gives one, negative gives none over the reals.
- Zero product property: if ab = 0, then a = 0 or b = 0. This is the justification for setting each factor to zero after factoring - and the reason you must move every term to one side first.
- Polynomial arithmetic follows exponent laws; (x + 3)² expands to x² + 6x + 9, not x² + 9.
- Rational expressions require a common denominator for addition, and their solutions must be checked against values that make any original denominator zero.
Worked example
Solve 3x² − 5x = 2.
Move all terms to one side: 3x² − 5x − 2 = 0. The product of the leading coefficient and constant is 3(−2) = −6; two numbers multiplying to −6 and adding to −5 are −6 and 1. Split the middle term, 3x² − 6x + x − 2 = 0, then group: 3x(x − 2) + 1(x − 2) = 0, so (3x + 1)(x − 2) = 0. Therefore x = −1/3 or x = 2. Checking with the discriminant: 25 + 24 = 49, a perfect square, confirming two rational roots.
Traps that cost marks
- Dividing both sides by x and silently discarding the solution x = 0.
- Distributing a negative sign to only the first term inside parentheses.
- Forgetting to reverse an inequality after dividing by a negative.
- Squaring both sides of a radical equation and accepting extraneous roots without checking.
- Factoring before the equation is set equal to zero.