Factoring polynomials means breaking an expression into simpler factors that multiply back to the original. It is one of the highest-leverage skills in algebra: it makes solving equations, simplifying expressions, and graphing functions dramatically easier. This guide covers the five methods you need, with worked examples for each.
Why Factoring Matters
Many hard-looking problems collapse once a polynomial is factored. Solving x² + 5x + 6 = 0 is tedious by trial and error, but writing it as (x + 2)(x + 3) = 0 reveals instantly that x = −2 or x = −3. Factoring turns a puzzle into a reading exercise.
Method 1: Factor Out the GCF
The first step with any polynomial is checking whether all terms share a common factor.
Example: 6x² + 12x
Both terms share a factor of 6x:
6x² + 12x = 6x(x + 2)
Verify: 6x·x + 6x·2 = 6x² + 12x ✓
Always look for the GCF first. Factoring out a common factor before anything else makes every other method easier — and skipping this step is the most common beginner error.
Method 2: Factoring Trinomials (x² + bx + c)
When the leading coefficient is 1, find two numbers that multiply to c and add to b.
Example: x² + 7x + 12
You need two numbers that multiply to 12 and add to 7: 3 and 4.
x² + 7x + 12 = (x + 3)(x + 4)
Example with negatives: x² − x − 6
You need two numbers that multiply to −6 and add to −1: −3 and 2.
x² − x − 6 = (x − 3)(x + 2)
Verify: (x − 3)(x + 2) = x² + 2x − 3x − 6 = x² − x − 6 ✓
Method 3: Factoring by Grouping
For polynomials with four terms, pair them up and factor each pair.
Example: x³ + 3x² + 2x + 6
- Group: (x³ + 3x²) + (2x + 6)
- Factor each: x²(x + 3) + 2(x + 3)
- The binomial (x + 3) is common: (x + 3)(x² + 2)
Method 4: Difference of Squares
Anything of the form a² − b² factors into (a − b)(a + b).
Example: x² − 16
Since 16 = 4², this is x² − 4²:
x² − 16 = (x − 4)(x + 4)
Method 5: Perfect Square Trinomials
Recognize the patterns a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)².
Example: x² + 6x + 9
Here a = x, b = 3, and 2ab = 6x matches the middle term:
x² + 6x + 9 = (x + 3)²
Bonus: Trinomials When a ≠ 1
When the leading coefficient is not 1, the “ac method” does the work: multiply a and c, split the middle term using two numbers with that product, then group.
Example: 2x² + 7x + 3
- Multiply a and c: 2 × 3 = 6. Two numbers that multiply to 6 and add to 7: 6 and 1.
- Rewrite the middle term: 2x² + 6x + x + 3
- Group and factor: 2x(x + 3) + 1(x + 3)
- Result: (2x + 1)(x + 3)
Verify: 2x·x + 2x·3 + 1·x + 1·3 = 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓
Checking Your Work
After factoring, multiply the factors back out to confirm you get the original polynomial. Checking each step by hand takes time with larger expressions, so many students use Calc Solver to verify factored answers instantly and catch mistakes before moving on.
Multiplying back is not optional practice — it is the fastest error detector in factoring. Five seconds of FOIL saves five minutes of rework.
Practice Problems (with answers)
Factor each of these:
- x² + 9x + 20
- x² − 4x − 21
- 4x² − 25
- x² + 10x + 25
- 3x² + 11x + 6
Answers: 1) (x + 4)(x + 5); 2) (x − 7)(x + 3); 3) (2x − 5)(2x + 5); 4) (x + 5)²; 5) ac = 18, numbers 9 and 2 → 3x² + 9x + 2x + 6 = 3x(x + 3) + 2(x + 3) = (3x + 2)(x + 3).
Working through a mix of trinomials, grouping problems, and special patterns makes the process feel natural. When a problem gets too complex to verify by hand, running it through Calc Solver is a quick way to confirm the correct factors before continuing.