Quadratic Equations • Topic 1 of 4

Factoring Quadratics

Factoring rewrites ax² + bx + c as a product of two binomials, and for GRE purposes it is the first thing to try because it is faster than any formula. When a = 1 you need two numbers that multiply to c and add to b: for x² + 5x + 6 you want 2 and 3, giving (x + 2)(x + 3). Watch the signs — a positive c with a negative b means both numbers are negative (x² − 7x + 12 = (x − 3)(x − 4)), and a negative c means the numbers have opposite signs. Once factored, the zero-product rule finishes the job: a product is zero only if one of its factors is zero, so (x − 3)(x − 4) = 0 gives x = 3 or x = 4. When a ≠ 1, either factor out a common number first or split the middle term (find two numbers multiplying to a·c and adding to b). Always move everything to one side so the equation equals zero before you factor — factoring x² + 5x = 6 as written is a classic error.

✅ Solved examples

1. Solve x² + 7x + 12 = 0.
Need two numbers that multiply to 12 and add to 7 → 3 and 4. So (x + 3)(x + 4) = 0, giving x = −3 or x = −4.
2. Solve x² − 5x = 14.
First set to zero: x² − 5x − 14 = 0. Product −14, sum −5 → −7 and +2. (x − 7)(x + 2) = 0, so x = 7 or x = −2.
3. Solve 2x² + 7x + 3 = 0.
a·c = 6; two numbers multiplying to 6 and adding to 7 are 6 and 1. Split: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0. So x = −1/2 or x = −3.
4. Quantitative Comparison. x² − 9x + 20 = 0. Quantity A: the larger root. Quantity B: 4. Which is greater?
Factor to (x − 4)(x − 5) = 0, roots 4 and 5. The larger root is 5. Quantity A (5) > Quantity B (4), so A is greater.

✏️ Practice — try these, take hints as needed

1. Solve x² + 9x + 18 = 0.
Two numbers multiplying to 18, adding to 9.
Try 6 and 3.
(x + 6)(x + 3) = 0.
x = −6 or x = −3
2. Solve x² − 3x − 10 = 0.
Product −10, sum −3.
Opposite signs; 5 and −2... check the sum.
−5 and +2 give sum −3.
x = 5 or x = −2
3. Solve x² = 6x − 9.
Set to zero first: x² − 6x + 9 = 0.
This is a perfect square.
(x − 3)² = 0.
x = 3 (a repeated root)
4. Solve 3x² − 12x = 0.
Do not divide by x — you would lose a root.
Factor out 3x.
3x(x − 4) = 0.
x = 0 or x = 4
5. Solve 2x² − x − 6 = 0.
a·c = −12; find two numbers multiplying to −12, adding to −1.
They are −4 and 3.
Split the middle and group.
x = 2 or x = −3/2

📝 Topic test — 8 questions

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