Factoring Polynomials — Complete Guide
Factoring a polynomial means rewriting it as a product of simpler expressions. Just as 12 = 3 × 4, the polynomial x² − 5x + 6 = (x−2)(x−3). Factoring is the reverse of expanding (multiplying out), and it is essential for solving polynomial equations, simplifying rational expressions, and finding roots.
This guide covers every standard factoring method in the order you should try them: always start by pulling out the GCF, then look for special patterns (difference of squares, perfect square trinomials, sum/difference of cubes), and finally apply trial-and-error or the ac method for general trinomials.
Factoring Strategy — Always Follow This Order
Before choosing a method, ask these questions in order:
1. Is there a Greatest Common Factor (GCF)? If yes, factor it out first. This simplifies every subsequent step.
2. How many terms does the remaining expression have?
— 2 terms: check for difference of squares (a²−b²), sum of cubes (a³+b³), or difference of cubes (a³−b³).
— 3 terms (trinomial): check for perfect square (a²±2ab+b²), then try trial-and-error or the ac method.
— 4+ terms: try factoring by grouping.
3. Can the result be factored further? Factor completely — every polynomial factor should be irreducible over the integers.
Step 1: Greatest Common Factor (GCF)
The GCF of a polynomial is the largest monomial that divides every term. Always factor out the GCF first.
Finding the GCF: Take the lowest power of each common variable, and the GCF of the numerical coefficients.
Example 1: 6x³ + 9x² − 3x. Numerical GCF of 6, 9, 3 = 3. Lowest power of x = x¹. GCF = 3x. Factor: 3x(2x² + 3x − 1).
Example 2: 4x²y³ − 8xy² + 12x²y. GCF = 4xy². Result: 4xy²(xy − 2 + 3x).
Example 3 — Negative leading coefficient: −2x² + 8x − 6. Factor out −2: −2(x² − 4x + 3) = −2(x−1)(x−3). Factoring out a negative makes the leading coefficient of the remaining polynomial positive.
Special Pattern: Difference of Squares — a² − b² = (a+b)(a−b)
The difference of squares pattern appears whenever a binomial is the subtraction of two perfect squares. It never applies to a sum (a² + b² is irreducible over the reals).
Recognition: Two terms, both perfect squares, connected by subtraction.
Examples:
x² − 9 = (x+3)(x−3). Check: (x+3)(x−3) = x² − 3x + 3x − 9 = x² − 9 ✓
4x² − 25 = (2x)² − 5² = (2x+5)(2x−5).
(x+1)² − 16 = ((x+1)+4)((x+1)−4) = (x+5)(x−3). The "a" can be any expression.
x⁴ − 81 = (x²)² − 9² = (x²+9)(x²−9). Factor again: (x²+9)(x+3)(x−3). Always check if further factoring is possible.
Trap: a² + b² does NOT factor over the reals. x² + 4 is irreducible.
Special Pattern: Perfect Square Trinomials — a²±2ab+b²
A perfect square trinomial results from squaring a binomial. Recognising the pattern saves time compared to the general trinomial methods.
Pattern A: a² + 2ab + b² = (a+b)²
Pattern B: a² − 2ab + b² = (a−b)²
Recognition test: The first and last terms are perfect squares, and the middle term is ±2 times the product of their square roots.
Examples:
x² + 6x + 9: √(x²) = x, √9 = 3, 2·x·3 = 6x ✓ → (x+3)².
4x² − 12x + 9: √(4x²) = 2x, √9 = 3, 2·2x·3 = 12x ✓ → (2x−3)².
25x² + 20xy + 4y² = (5x+2y)². Check: 2·5x·2y = 20xy ✓.
General Trinomials — Trial-and-Error and the ac Method
Case 1: Leading coefficient = 1 (x² + bx + c). Find two integers whose product is c and sum is b. Write them as (x + r)(x + s).
Example: x² − 7x + 12. Need two numbers with product 12 and sum −7. Pairs: (−3)(−4) = 12, (−3)+(−4) = −7 ✓. Factor: (x−3)(x−4).
Case 2: Leading coefficient ≠ 1 (ax² + bx + c) — ac method.
Step 1: Compute ac. Step 2: Find two numbers m and n where mn = ac and m + n = b. Step 3: Rewrite bx as mx + nx. Step 4: Factor by grouping.
Example: 6x² + 11x + 3. ac = 18. Find m, n: mn=18, m+n=11 → m=2, n=9. Rewrite: 6x² + 2x + 9x + 3. Group: 2x(3x+1) + 3(3x+1) = (2x+3)(3x+1).
Verify: Always expand your factors to confirm. (2x+3)(3x+1) = 6x²+2x+9x+3 = 6x²+11x+3 ✓.
Factoring by Grouping — 4-Term Polynomials
Grouping applies to 4-term polynomials. Split into two pairs, factor the GCF from each pair, then factor out the common binomial factor.
Example 1: x³ + 3x² + 2x + 6. Group: (x³ + 3x²) + (2x + 6). Factor each: x²(x+3) + 2(x+3). Common factor (x+3): (x+3)(x²+2).
Example 2 — Rearranging may help: 2x³ − x² + 4x − 2. Try: (2x³ + 4x) + (−x² − 2) = 2x(x²+2) − 1(x²+2) = (x²+2)(2x−1).
When grouping fails: If no common binomial factor appears after the first grouping, try rearranging the terms or using a different pairing. If no pairing works, the polynomial may be irreducible or require the ac method after multiplying by a constant.
Sum and Difference of Cubes
Two special factoring formulas handle cubic binomials:
Sum of cubes: a³ + b³ = (a + b)(a² − ab + b²)
Difference of cubes: a³ − b³ = (a − b)(a² + ab + b²)
Memory trick: "SOAP" — Same sign, Opposite sign, Always Positive for the trinomial factor.
Examples:
x³ + 8 = x³ + 2³ = (x+2)(x²−2x+4). The trinomial x²−2x+4 is irreducible over the reals (D = 4−16 = −12 < 0).
27x³ − 125 = (3x)³ − 5³ = (3x−5)(9x²+15x+25).
x⁶ + 1 = (x²)³ + 1³ = (x²+1)(x⁴−x²+1).
Trap — there is no "sum of squares" formula: a² + b² is irreducible over the reals. Only difference of squares and both cube formulas exist.
Factoring Completely — Always Check the Result
"Factoring completely" means every factor is irreducible over the integers — no further factoring is possible.
Checklist for each factor: (1) Is it a monomial? Done. (2) Is it a binomial? Check for difference/sum of squares or cubes. (3) Is it a trinomial with D < 0 and integer coefficients? Irreducible. (4) Does it have 4 terms? Try grouping again.
Multi-step example: 2x⁴ − 32. GCF first: 2(x⁴ − 16). Difference of squares: 2(x²+4)(x²−4). Factor again: 2(x²+4)(x+2)(x−2). Check each factor: x²+4 is irreducible (sum of squares). Done: 2(x²+4)(x+2)(x−2).
Verification strategy: Expand your final answer and compare to the original polynomial. If they match, the factoring is correct.