Every pattern in this lesson is the reverse of one from Polynomials. Multiplying \((x+3)(x-3)\) gives \(x^{2}-9\); factoring runs that exact process backward, starting from \(x^{2}-9\) and recovering \((x+3)(x-3)\). The special product patterns you already know for expanding are exactly the patterns worth recognizing for factoring too.
What Is Factoring?
Factoring means rewriting a polynomial as a product of simpler polynomials, called factors. A polynomial is factored completely when none of its factors can be broken down any further — no common factor left to pull out, and no recognizable pattern left hiding inside a factor.
Factoring Out the GCF
The first move on every factoring problem, no exceptions: find the greatest common factor (GCF) of every term, and factor it out using the distributive property in reverse.
$$ 6x^{2}+9x = 3x(2x+3) $$
Here the GCF is \(3x\) — the largest expression that divides evenly into both \(6x^{2}\) and \(9x\). Skipping this step doesn’t make later factoring wrong, but it does make the numbers bigger and the patterns harder to spot.
Properties of Factoring
| Pattern | Name | Statement |
|---|---|---|
| GCF | \(ab+ac = a(b+c)\) | |
| Difference of Squares | \(a^{2}-b^{2} = (a+b)(a-b)\) | |
| Perfect Square Trinomial | (sum) | \(a^{2}+2ab+b^{2} = (a+b)^{2}\) |
| Perfect Square Trinomial | (difference) | \(a^{2}-2ab+b^{2} = (a-b)^{2}\) |
| Trinomial, leading coefficient 1 | \(x^{2}+(p+q)x+pq = (x+p)(x+q)\) |
The last row is the key to factoring \(x^{2}+bx+c\): find two numbers that multiply to \(c\) and add to \(b\) — those two numbers are exactly \(p\) and \(q\).
Example 1: Factoring Out the GCF
| Expression | GCF | Factored |
|---|---|---|
| \(6x^{2}+9x\) | \(3x\) | \(3x(2x+3)\) |
| \(12x^{3}-8x^{2}\) | \(4x^{2}\) | \(4x^{2}(3x-2)\) |
| \(20x^{3}+15x^{2}\) | \(5x^{2}\) | \(5x^{2}(4x+3)\) |
| \(14x^{2}-21x\) | \(7x\) | \(7x(2x-3)\) |
Example 2: Factoring Special Patterns
| Expression | Pattern | Factored |
|---|---|---|
| \(x^{2}-25\) | Difference of squares | \((x+5)(x-5)\) |
| \(4x^{2}-9\) | Difference of squares | \((2x+3)(2x-3)\) |
| \(x^{2}+10x+25\) | Perfect square trinomial | \((x+5)^{2}\) |
| \(x^{2}-14x+49\) | Perfect square trinomial | \((x-7)^{2}\) |
Example 3: Factoring x² + bx + c
Find two numbers that multiply to \(c\) and add to \(b\).
| Expression | Two Numbers | Factored |
|---|---|---|
| \(x^{2}+7x+12\) | \(3, 4\) | \((x+3)(x+4)\) |
| \(x^{2}-8x+15\) | \(-3, -5\) | \((x-3)(x-5)\) |
| \(x^{2}+2x-15\) | \(5, -3\) | \((x+5)(x-3)\) |
| \(x^{2}-x-12\) | \(-4, 3\) | \((x-4)(x+3)\) |
Worked Example A: GCF, Then a Pattern
Factor \(2x^{2}-8\) completely.
Step 1 — factor out the GCF:
$$ 2x^{2}-8 = 2(x^{2}-4) $$
Step 2 — recognize the difference of squares inside:
$$ x^{2}-4 = (x+2)(x-2) $$
Final answer: \(2(x+2)(x-2)\). Stopping after Step 1 would be a common, incomplete answer — always check what’s left inside the parentheses for another pattern.
Worked Example B: The ac Method
Factor \(3x^{2}+11x+6\), where the leading coefficient isn’t 1.
Step 1 — multiply \(a\) and \(c\), then find two numbers multiplying to that product and adding to \(b\):
$$ a \cdot c = 3 \cdot 6 = 18, \qquad 9 \cdot 2 = 18, \qquad 9 + 2 = 11 $$
Step 2 — split the middle term using those two numbers:
$$ 3x^{2}+11x+6 = 3x^{2}+9x+2x+6 $$
Step 3 — factor by grouping, pulling a common factor from each pair:
$$ 3x(x+3) + 2(x+3) = (x+3)(3x+2) $$
Check by expanding: \((x+3)(3x+2) = 3x^{2}+2x+9x+6 = 3x^{2}+11x+6\). ✓
Worked Example C: GCF, Then a Trinomial
Factor \(4x^{3}-4x^{2}-24x\) completely.
Step 1 — factor out the GCF:
$$ 4x^{3}-4x^{2}-24x = 4x(x^{2}-x-6) $$
Step 2 — factor the trinomial: find two numbers multiplying to \(-6\) and adding to \(-1\):
$$ -3 \cdot 2 = -6, \qquad -3+2=-1 $$
Step 3 — write the factored trinomial:
$$ x^{2}-x-6 = (x-3)(x+2) $$
Final answer: \(4x(x-3)(x+2)\).
Common Mistakes to Avoid
- Skipping the GCF. Missing a common factor doesn’t make the rest wrong, but it often hides the pattern that would have made the problem simple.
- Sign errors on the two-numbers step. For \(x^{2}-x-6\), the numbers need a product of \(-6\) (one positive, one negative) and a sum of \(-1\) — check both conditions, not just one.
- Assuming \(a^{2}+b^{2}\) factors like \(a^{2}-b^{2}\). It doesn’t; a sum of squares has no real-number factorization.
- Stopping before it’s complete. \(2(x^{2}-4)\) still has a difference of squares inside — always re-check every factor for one more pattern.
- Never checking the answer. Multiplying the factors back out catches sign and arithmetic errors in seconds.
Practice Problems
Work each problem yourself before opening the answer.
Problem 1. Factor \(8x^{2}+12x\).
Show answer
$$ 8x^{2}+12x = 4x(2x+3) $$
Answer: \(4x(2x+3)\)
Problem 2. Factor \(10x^{3}-15x^{2}\).
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$$ 10x^{3}-15x^{2} = 5x^{2}(2x-3) $$
Answer: \(5x^{2}(2x-3)\)
Problem 3. Factor \(x^{2}-36\).
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$$ x^{2}-36 = (x+6)(x-6) $$
Answer: \((x+6)(x-6)\)
Problem 4. Factor \(x^{2}+12x+36\).
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$$ x^{2}+12x+36 = (x+6)^{2} $$
Answer: \((x+6)^{2}\)
Problem 5. Factor \(x^{2}+5x+6\).
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Two numbers multiplying to \(6\) and adding to \(5\): \(2\) and \(3\).
$$ x^{2}+5x+6 = (x+2)(x+3) $$
Answer: \((x+2)(x+3)\)
Problem 6. Factor \(x^{2}-2x-24\).
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Two numbers multiplying to \(-24\) and adding to \(-2\): \(-6\) and \(4\).
$$ x^{2}-2x-24 = (x-6)(x+4) $$
Answer: \((x-6)(x+4)\)
Problem 7. Factor \(2x^{2}+7x+3\) using the ac method.
Show answer
Step 1 — multiply \(a\) and \(c\), find two numbers multiplying to that and adding to \(7\):
$$ a \cdot c = 2 \cdot 3 = 6, \qquad 6 \cdot 1 = 6, \qquad 6+1=7 $$
Step 2 — split and group:
$$ 2x^{2}+6x+x+3 = 2x(x+3) + 1(x+3) = (x+3)(2x+1) $$
Answer: \((x+3)(2x+1)\)
Problem 8. Factor \(3x^{2}-27\) completely.
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Step 1 — factor out the GCF:
$$ 3x^{2}-27 = 3(x^{2}-9) $$
Step 2 — factor the difference of squares:
$$ x^{2}-9 = (x+3)(x-3) $$
Answer: \(3(x+3)(x-3)\)
Problem 9. Factor \(5x^{3}+5x^{2}-30x\) completely.
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Step 1 — factor out the GCF:
$$ 5x^{3}+5x^{2}-30x = 5x(x^{2}+x-6) $$
Step 2 — factor the trinomial (numbers multiplying to \(-6\), adding to \(1\)): \(3\) and \(-2\):
$$ x^{2}+x-6 = (x+3)(x-2) $$
Answer: \(5x(x+3)(x-2)\)
Problem 10. Factor \(9x^{2}-16\).
Show answer
$$ 9x^{2}-16 = (3x+4)(3x-4) $$
Answer: \((3x+4)(3x-4)\)
Quick Reference
| Situation | Move |
|---|---|
| Every term shares a common factor | Factor out the GCF first, always |
| \(a^{2}-b^{2}\) | \((a+b)(a-b)\) |
| \(a^{2}+2ab+b^{2}\) or \(a^{2}-2ab+b^{2}\) | \((a+b)^{2}\) or \((a-b)^{2}\) |
| \(x^{2}+bx+c\) | Find two numbers multiplying to \(c\), adding to \(b\) |
| \(ax^{2}+bx+c\), \(a \neq 1\) | The ac method: split the middle term, factor by grouping |
| \(a^{2}+b^{2}\) | Does not factor over the real numbers |
Once these patterns feel automatic, they carry forward directly into Rational Expressions, where factoring is the tool that simplifies, multiplies, and divides fractions built from polynomials. Build speed with the Factoring Generator, browse the rest of the Algebra lessons as they publish, revisit Polynomials if any pattern above felt unfamiliar, or explore the site’s calculators to check your own arithmetic against a worked result.