ASVAB Mathematics Knowledge — Algebra and Equations

Factoring a trinomial x² + bx + c: find two numbers that multiply to c and add to b. Here: multiply to 12, add to 7. Try pairs: 1 × 12 = 12, but 1 + 12 = 13 ✗; 2 × 6 = 12, but 2 + 6 = 8 ✗; 3 × 4 = 12, and 3 + 4 = 7 ✓. So x² + 7x + 12 = (x + 3)(x + 4). Verify by FOIL (First, Outer, Inner, Last): (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 ✓. Signs guide factoring: (1) c positive, b positive — both factors positive; (2) c positive, b negative — both factors negative; (3) c negative — factors have opposite signs (the larger absolute value matches the sign of b). Example with negative signs: x² − 5x + 6 = (x − 2)(x − 3); x² + x − 6 = (x + 3)(x − 2). Factoring is the reverse of FOIL. Used to solve quadratic equations: set the factored form = 0 and use zero-product property.

Difference of squares pattern: a² − b² = (a + b)(a − b). Here x² − 16 = x² − 4², so a = x and b = 4. Factored: (x + 4)(x − 4). Verify by FOIL: (x + 4)(x − 4) = x² − 4x + 4x − 16 = x² − 16 ✓. The middle terms cancel because of the +/− pattern. Key pattern recognition: a² − b² (difference) factors easily; a² + b² (sum) does NOT factor over real numbers (factors using complex numbers as (a + bi)(a − bi)). Common applications: (1) x² − 25 = (x + 5)(x − 5); (2) 4x² − 9 = (2x + 3)(2x − 3); (3) x⁴ − 16 = (x² + 4)(x² − 4) = (x² + 4)(x + 2)(x − 2) — factor repeatedly. Perfect square trinomial patterns to also recognize: a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)². Example: x² + 6x + 9 = (x + 3)².

Solve quadratic equations by factoring when possible. Step 1: factor — find numbers that multiply to 6 and add to −5: that's −2 and −3 (since (−2)(−3) = 6 and −2 + −3 = −5). So x² − 5x + 6 = (x − 2)(x − 3). Step 2: set the factored form equal to zero: (x − 2)(x − 3) = 0. Step 3: zero-product property — if a product is zero, at least one factor is zero. So x − 2 = 0 (giving x = 2) or x − 3 = 0 (giving x = 3). Solutions: x = 2 or x = 3. Verify: 2² − 5(2) + 6 = 4 − 10 + 6 = 0 ✓; 3² − 5(3) + 6 = 9 − 15 + 6 = 0 ✓. A quadratic equation can have 0, 1, or 2 real solutions. If factoring doesn't work, use the quadratic formula: x = (−b ± √(b² − 4ac)) / (2a) for ax² + bx + c = 0. The discriminant b² − 4ac: positive → 2 solutions; zero → 1 solution; negative → no real solutions.

Two-step linear equation. Add 9 to both sides: 4y − 9 + 9 = 11 + 9, giving 4y = 20. Divide both sides by 4: 4y ÷ 4 = 20 ÷ 4, giving y = 5. Verify: 4(5) − 9 = 20 − 9 = 11 ✓. The algorithm: undo operations in reverse order of operations. The original expression performs (1) multiply variable by 4, then (2) subtract 9. To solve, reverse: first add 9 (undo subtraction), then divide by 4 (undo multiplication). For more complex equations with variables on both sides: first move all variables to one side (by adding/subtracting), then constants to the other, then solve. Example: 3x + 5 = 2x − 8 → x + 5 = −8 → x = −13. Variables in equations represent unknown quantities; solving identifies the value(s) that satisfy the equation.

To solve a linear equation, isolate x. Start with 3x − 7 = 14. Add 7 to both sides: 3x = 21. Then divide both sides by 3: x = 7. The strategy is to undo operations in reverse order — first undo the subtraction (add 7), then undo the multiplication (divide by 3). You can check by substituting back: 3(7) − 7 = 21 − 7 = 14. ✓ Linear equations are a staple of Mathematics Knowledge, and the reliable method is to perform the same operation on both sides until x stands alone.

To factor x² + 5x + 6, find two numbers that multiply to the constant (6) and add to the middle coefficient (5). Those numbers are 2 and 3, since 2 × 3 = 6 and 2 + 3 = 5. So the factors are (x + 2)(x + 3). You can verify by expanding: (x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6. ✓ The other options either multiply to the wrong constant or add to the wrong middle term. The product-and-sum method is the standard approach for factoring simple quadratics.

First distribute or divide. Dividing both sides by 2: x + 3 = 8. Then subtract 3: x = 5. Alternatively, distribute first: 2x + 6 = 16, then 2x = 10, so x = 5. Either order works. Check: 2(5 + 3) = 2(8) = 16. ✓ Equations with parentheses can be solved by distributing the multiplier or by dividing both sides by it first; dividing first is often quicker when the right side is divisible. The value of x is 5.

Solve an inequality much like an equation: subtract 5 from both sides of x + 5 > 12 to get x > 7. The solution is all values greater than 7. Note that the inequality sign stays the same when adding or subtracting; it only flips when you multiply or divide both sides by a negative number (which doesn't happen here). So the answer is x > 7. Inequalities are common on Mathematics Knowledge, and the steps mirror solving equations, with that one rule about flipping the sign for negative multiplication or division.

First distribute the 3 across the parentheses: 3 × 2x = 6x and 3 × 4 = 12, giving 6x + 12. Then subtract 2x: 6x + 12 − 2x. Combine like terms (6x − 2x = 4x): the result is 4x + 12. The key steps are distributing first, then combining like terms (the x-terms together, the constants separately). A common error is forgetting to distribute the 3 to both terms inside the parentheses. The simplified expression is 4x + 12.

Negative exponents indicate reciprocals: a^(−n) = 1/a^n. So x⁻² = 1/x². With x = 4: x⁻² = 1/4² = 1/16. The negative exponent does NOT make the result negative — it makes it a reciprocal. Negative exponent rules: (1) a^(−n) = 1/a^n; (2) (a/b)^(−n) = (b/a)^n (flip the fraction, change sign); (3) 1/a^(−n) = a^n. Examples: 2⁻³ = 1/2³ = 1/8; (1/3)⁻² = 3² = 9; 10⁻⁴ = 0.0001. Zero exponent: a⁰ = 1 for any non-zero a. 0⁰ is undefined (or 1 by convention in some contexts). Fractional exponents: a^(1/n) = ⁿ√a; a^(m/n) = ⁿ√(a^m) = (ⁿ√a)^m. Example: 8^(2/3) = (³√8)² = 2² = 4. These rules combine: (1/2)⁻³ = 2³ = 8.