Methods of Factoring Polynomials- Techniques and Examples
What Factoring Polynomials Actually Is
Factoring polynomials means rewriting a polynomial as a product of simpler polynomials. That's it. You're breaking down something complex into pieces that multiply back together.
This isn't busywork. Factoring shows up everywhere—in solving equations, simplifying expressions, and finding roots. If you're working with quadratics or higher-degree polynomials, you need these techniques in your toolkit.
Greatest Common Factor (GCF)
This is the first thing you should always check. Find the largest factor that divides every term.
How to find GCF
- Factor each coefficient into primes
- Identify common variables across all terms
- Multiply the common factors together
Examples
Example 1: Factor 12x³ + 18x²
GCF of 12 and 18 is 6. GCF of x³ and x² is x². So the GCF is 6x².
12x³ + 18x² = 6x²(2x + 3)
Example 2: Factor 5a²b³ - 15ab² + 10a³b
GCF is 5ab².
5a²b³ - 15ab² + 10a³b = 5ab²(a²b - 3 + 2a²)
Always factor out the GCF first. It's not optional—it's the foundation everything else builds on.
Factoring by Grouping
Use this when you have four terms with no obvious GCF across all of them. The trick is grouping terms that share common factors.
Process
- Group the first two terms together, last two terms together
- Factor out the GCF from each group
- If the binomials match, factor them out
Example
Factor 3x³ + 6x² + 2x + 4
Group: (3x³ + 6x²) + (2x + 4)
Factor each: 3x²(x + 2) + 2(x + 2)
Factor out (x + 2): (x + 2)(3x² + 2)
Sometimes you need to rearrange terms first. If grouping doesn't work immediately, try a different grouping arrangement.
Factoring Trinomials
Trinomials have three terms, usually in the form ax² + bx + c. The approach depends on whether a = 1 or a ≠ 1.
When a = 1 (x² + bx + c)
Find two numbers that multiply to c and add to b.
Example: Factor x² + 7x + 12
Find two numbers that multiply to 12 and add to 7. Those numbers are 3 and 4.
x² + 7x + 12 = (x + 3)(x + 4)
Example: Factor x² - 5x + 6
Need numbers that multiply to 6 and add to -5. That's -2 and -3.
x² - 5x + 6 = (x - 2)(x - 3)
When a ≠ 1 (AC Method)
For ax² + bx + c where a ≠ 1, use the AC method:
- Multiply a and c (the "AC" product)
- Find two numbers that multiply to AC and add to b
- Split the middle term using those numbers
- Factor by grouping
Example: Factor 2x² + 7x + 3
AC = 2 × 3 = 6
Find numbers that multiply to 6 and add to 7: 6 and 1
2x² + 7x + 3 = 2x² + 6x + x + 3
Group: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3)
= (x + 3)(2x + 1)
Box Method (Alternative)
Some students prefer the box method. Draw a 2×2 grid. Put ax² in the top-left, c in the bottom-right. Fill the remaining corners so the diagonals multiply correctly, then read off the factored form.
It works. It's visual. Use it if the AC method gives you trouble.
Difference of Squares
Recognize this pattern: a² - b² = (a + b)(a - b)
The key is identifying perfect squares within your expression.
Examples
Example 1: Factor x² - 16
x² is a perfect square (x)². 16 is 4².
x² - 16 = (x + 4)(x - 4)
Example 2: Factor 49a² - 25b²
49a² = (7a)². 25b² = (5b)².
49a² - 25b² = (7a + 5b)(7a - 5b)
Example 3: Factor x⁴ - 16
This is nested squares: (x²)² - 4². Apply the formula twice.
x⁴ - 16 = (x² + 4)(x² - 4)
= (x² + 4)(x + 2)(x - 2)
Remember: you cannot factor a sum of squares over the real numbers. a² + b² stays as is. Don't try to force it.
Perfect Square Trinomials
These are trinomials that come from squaring a binomial. Recognize them quickly:
- a² + 2ab + b² = (a + b)²
- a² - 2ab + b² = (a - b)²
Check the middle term. Is it exactly twice the product of the square roots of the first and last terms? If yes, you have a perfect square trinomial.
Example
Factor 9x² + 12x + 4
Check: √(9x²) = 3x, √4 = 2. Twice their product: 2(3x)(2) = 12x. Matches the middle term.
9x² + 12x + 4 = (3x + 2)²
Sum and Difference of Cubes
Less common but you need them for higher-degree polynomials.
Sum of cubes: a³ + b³ = (a + b)(a² - ab + b²)
Difference of cubes: a³ - b³ = (a - b)(a² + ab + b²)
Example
Factor 8x³ + 27
8x³ = (2x)³. 27 = 3³.
8x³ + 27 = (2x + 3)((2x)² - (2x)(3) + 3²)
= (2x + 3)(4x² - 6x + 9)
The quadratic factor (a² - ab + b² or a² + ab + b²) won't factor further over the real numbers. That's your final answer.
Method Comparison
| Method | Pattern/Form | When to Use |
|---|---|---|
| GCF | Common factor in all terms | Always check this first |
| Grouping | Four terms, no overall GCF | Four-term expressions |
| Trinomials (a=1) | x² + bx + c | Two numbers to c, sum to b |
| AC Method | ax² + bx + c, a ≠ 1 | Harder trinomials |
| Difference of Squares | a² - b² | Two perfect squares subtracted |
| Perfect Square | a² ± 2ab + b² | Middle term is 2√(first × last) |
| Cubes | a³ ± b³ | Two perfect cubes added or subtracted |
Getting Started: A Practical Approach
Don't try to guess which method to use. Work through this checklist in order:
- Check for GCF first. Factor it out completely before doing anything else.
- Count the terms. Two terms? Check for squares or cubes. Three terms? Look for perfect square or use trinomial methods. Four terms? Try grouping.
- Look for recognizable patterns. Difference of squares, cubes, perfect squares—these have distinct forms.
- Count negative signs. They tell you whether binomial factors contain plus or minus.
- Multiply your answer back to verify it gives you the original polynomial.
Practice problem: Factor 2x³y + 8xy - 4x²y²
Step 1: GCF is 2xy.
2xy(x² + 4 - 2xy)
Step 2: Rearrange the trinomial inside: x² - 2xy + 4. Can this factor?
Check discriminant: (-2y)² - 4(1)(4) = 4y² - 16. This is negative unless y² > 4, so it doesn't factor nicely over integers. Your answer is 2xy(x² - 2xy + 4).
Common Mistakes to Avoid
- Forgetting to check for GCF before trying other methods
- Factoring a sum of squares (can't be done)
- Getting the signs wrong in binomial factors
- Not verifying your answer by multiplying back
- Over-factoring: stopping when the quadratic won't factor further