Factoring Techniques
Quick Reference
| Field | Value |
|---|---|
| Textbook | OpenStax College Algebra 2e |
| Chapter | Chapter 1: Prerequisites |
| Section | 1.5 Factoring Polynomials |
| Subsection | 1.5 Factoring Polynomials |
| Course | MATH141 |
| Difficulty | Foundational |
| Time | ~40 minutes |
Page numbers pending faculty verification.
Before You Start: Prerequisite Check
📋 Can you do these? (Click to reveal self-test)
Test yourself on these prerequisite skills:
Multiply two binomials: Expand $(x + 2)(x - 5)$.
Check
$x^2 - 3x - 10$. The outer and inner products combine to give the middle term $-3x$.
Combine like exponents: Simplify $x^2 \cdot x^3$ and $\dfrac{x^7}{x^3}$.
Check
$x^2 \cdot x^3 = x^5$ (add exponents) and $\dfrac{x^7}{x^3} = x^4$ (subtract exponents).
Find a numerical GCF: What is the greatest common factor of $12$ and $18$?
Check
$6$. It is the largest integer dividing both $12$ and $18$.
Recognize perfect squares and cubes: Which of $4$, $9$, $16$, $4x^2$ are perfect squares? Which of $8$, $27$, $8x^3$ are perfect cubes?
Check
$4 = 2^2$, $9 = 3^2$, $16 = 4^2$, and $4x^2 = (2x)^2$ are perfect squares. $8 = 2^3$, $27 = 3^3$, and $8x^3 = (2x)^3$ are perfect cubes.
If you struggled:
- Review Algebraic Simplification for distributing and combining terms
- Review Exponent Laws for adding and subtracting exponents on like bases
Try This First
Before any rule, take a moment with one concrete case.
You already know how to multiply: $(x + 3)(x + 4) = x^2 + 7x + 12$, because multiplying the two binomials and combining like terms gives that result. Factoring asks the reverse question.
Try it: Suppose you are handed $x^2 + 7x + 12$ with no hint about where it came from. Find two numbers that multiply to $12$ and add to $7$. Write them down before reading on.
What did you find?
The pair is $3$ and $4$: they multiply to $12$ and add to $7$. So $x^2 + 7x + 12 = (x + 3)(x + 4)$. Factoring is the act of running multiplication backward, turning a sum of terms into a product of factors.
The Idea: Multiplication Run Backward
Multiplication takes factors and produces a single expanded expression. Factoring is a process that takes an expanded expression as its input and gives back the product of factors as its output. The two operations undo each other, so a factoring answer can always be checked by multiplying the factors back out.
Every factoring technique below is a pattern you can recognize because you have already seen the matching multiplication. Learn to recognize the patterns on sight, and factoring becomes pattern matching rather than guessing.
There is a reason to care beyond tidiness. A product equals zero exactly when one of its factors equals zero, so once a polynomial is written as a product, the values that make it zero become readable directly off the factors. That single fact is what makes factoring the gateway to solving quadratic equations, the skill this one unlocks. The same move also simplifies rational expressions, because common factors cancel only after both numerator and denominator are factored.
Prerequisite Hub
Builds on
| Skill | Why it is needed |
|---|---|
algebraic-simplification |
Factoring reverses the distribution and combination steps used to simplify expressions |
exponent-laws |
Pulling out a common factor relies on subtracting exponents on like bases |
Unlocks
| Skill | What it leads to |
|---|---|
quadratic-equations |
A factored quadratic reveals its roots through the zero-product property |
Factoring also feeds later work in rational expressions, polynomial division, and rational functions, where common factors cancel only after both numerator and denominator are factored. Those topics are covered in their own skills once the course reaches them.
No cross-course prerequisites are listed for this skill.
Official Definitions
Greatest common factor (GCF). The greatest common factor (GCF) of two or more terms is the largest expression (number and variables) that divides evenly into all of the terms.
Source: OpenStax College Algebra 2e, Section 1.5 Factoring Polynomials.
Difference of squares. A perfect square difference of the form $a^2 - b^2$ factors as $(a + b)(a - b)$.
Source: OpenStax College Algebra 2e, Section 1.5 Factoring Polynomials.
The GCF is the first tool to reach for on any factoring problem: pull it out before trying anything else, because a smaller leftover expression is easier to handle. The difference of squares is a named pattern worth memorizing, because it appears constantly and factors instantly once recognized.
Quick Reference
| Pattern name | Form to recognize | Factored form |
|---|---|---|
| GCF | $ax + ay + az$ | $a(x + y + z)$ |
| Trinomial (lead 1) | $x^2 + bx + c$ | $(x + p)(x + q)$ where $p + q = b$, $pq = c$ |
| Trinomial (lead $a$) | $ax^2 + bx + c$ | Factor by the AC method |
| Difference of squares | $a^2 - b^2$ | $(a + b)(a - b)$ |
| Perfect square trinomial | $a^2 + 2ab + b^2$ | $(a + b)^2$ |
| Sum of cubes | $a^3 + b^3$ | $(a + b)(a^2 - ab + b^2)$ |
| Difference of cubes | $a^3 - b^3$ | $(a - b)(a^2 + ab + b^2)$ |
| Grouping | Four terms | Group pairs, factor a GCF from each |
Strategy. Always check for a GCF first. Then count terms: two terms suggests a difference of squares or cubes; three terms suggests a trinomial; four terms suggests grouping.
The Core Techniques
Each technique below is one way to write a polynomial as a product. A polynomial may need more than one of them in sequence. Predict the result before each computation, then verify by multiplying back.
1. Greatest Common Factor (GCF)
The GCF is the largest factor that divides every term. Removing it is always the first step.
How to find the GCF:
- Find the GCF of all numerical coefficients.
- For each variable, take the lowest power that appears in every term.
- Multiply these together.
Always factor out the GCF first. Skipping this step creates unnecessarily hard trinomials. A trinomial with a large leading coefficient often factors easily after a GCF is removed.
2. Trinomials with Leading Coefficient 1
A trinomial $x^2 + bx + c$ factors as $(x + p)(x + q)$ where
$$p + q = b \quad \text{and} \quad p \cdot q = c.$$
Find $p$ and $q$ by listing factor pairs of $c$ and checking which pair sums to $b$. The sign rules follow from the product and sum: a negative $c$ forces opposite signs, while a positive $c$ with negative $b$ forces two negatives.
3. Trinomials with Leading Coefficient $a \neq 1$ (AC Method)
For $ax^2 + bx + c$ with $a \neq 1$:
- Compute $ac$.
- Find two numbers $m$ and $n$ with $mn = ac$ and $m + n = b$.
- Rewrite the middle term: $ax^2 + mx + nx + c$.
- Factor by grouping.
4. Difference of Squares
If an expression has the form $a^2 - b^2$, it factors as $(a + b)(a - b)$. Recognize it by two perfect squares separated by a minus sign. A sum of squares $a^2 + b^2$ does not factor over the real numbers, so the minus sign is essential.
5. Perfect Square Trinomials
$$a^2 + 2ab + b^2 = (a + b)^2, \qquad a^2 - 2ab + b^2 = (a - b)^2.$$
Recognize this when the first and last terms are perfect squares and the middle term is exactly twice the product of their square roots.
6. Sum and Difference of Cubes
$$a^3 + b^3 = (a + b)(a^2 - ab + b^2), \qquad a^3 - b^3 = (a - b)(a^2 + ab + b^2).$$
A mnemonic for the signs is SOAP: Same, Opposite, Always Positive. The sign in $(a \pm b)$ is the same as the original, the sign of $ab$ in the trinomial is opposite, and the last sign is always positive.
7. Grouping (Four Terms)
With four terms, split them into two pairs, factor a GCF out of each pair, and look for a shared binomial factor to pull out. If one pairing produces no common binomial, try a different pairing.
8. Strategy: Choosing the Technique
Ask these questions in order:
- Can a GCF come out? Always check first.
- How many terms remain? Two terms suggests a difference of squares or cubes; three terms suggests a trinomial or perfect square; four terms suggests grouping.
- Is any factor itself factorable? Continue until nothing factors further.
- Does the check pass? Multiply the factors back and compare.
Always Check by Multiplying Back
Because factoring and multiplying are inverse processes, every factored answer can be verified. Multiply the proposed factors, combine like terms, and confirm the result matches the original. This check is fast and catches sign errors and arithmetic slips. It is the predict-then-check habit applied to algebra: you predict the factors, then expand to confirm.
Worked Examples
Example 1: Greatest common factor
Factor $12x^3 - 18x^2 + 6x$.
Predict first. All three coefficients are divisible by $6$, and every term contains at least one $x$. Expect a $6x$ to come out front, leaving a trinomial inside.
Numbers: the GCF of $12$, $18$, and $6$ is $6$.
Variables: the smallest power of $x$ present in every term is $x^1$.
GCF: $6x$. Divide each term by it.
$$\frac{12x^3}{6x} = 2x^2, \qquad \frac{18x^2}{6x} = 3x, \qquad \frac{6x}{6x} = 1.$$
$$12x^3 - 18x^2 + 6x = 6x(2x^2 - 3x + 1).$$
Check. Distribute: $6x(2x^2) - 6x(3x) + 6x(1) = 12x^3 - 18x^2 + 6x$. The result matches the original.
Example 2: Trinomial with leading coefficient 1
Factor $x^2 + 7x + 12$.
Predict first. Two positive numbers must multiply to $12$ and add to $7$. List the factor pairs of $12$ and read off the sums.
| Factor pair of $12$ | Sum |
|---|---|
| $1 \times 12$ | $13$ |
| $2 \times 6$ | $8$ |
| $3 \times 4$ | $7$ $\checkmark$ |
$$x^2 + 7x + 12 = (x + 3)(x + 4).$$
Check. Expand: $(x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$. Correct.
Example 3: Trinomial with a negative constant
Factor $x^2 - 5x - 14$.
Predict first. The constant $-14$ is negative, so one number is positive and one is negative. They multiply to $-14$ and add to $-5$.
| Factor pair of $-14$ | Sum |
|---|---|
| $1 \times (-14)$ | $-13$ |
| $(-1) \times 14$ | $13$ |
| $2 \times (-7)$ | $-5$ $\checkmark$ |
$$x^2 - 5x - 14 = (x + 2)(x - 7).$$
Check. Expand: $(x + 2)(x - 7) = x^2 - 7x + 2x - 14 = x^2 - 5x - 14$. Correct.
Example 4: AC method
Factor $6x^2 + 11x - 10$.
Predict first. Here $a = 6$, $b = 11$, and $c = -10$, so $ac = -60$. Two numbers must multiply to $-60$ and add to $11$. A negative product means opposite signs.
Try $15$ and $-4$: $15 \times (-4) = -60$ and $15 + (-4) = 11$.
Rewrite the middle term: $6x^2 + 15x - 4x - 10$.
Group: $(6x^2 + 15x) + (-4x - 10)$.
Factor each group: $3x(2x + 5) - 2(2x + 5)$.
Shared binomial: $(2x + 5)(3x - 2)$.
Check. Expand: $(2x + 5)(3x - 2) = 6x^2 - 4x + 15x - 10 = 6x^2 + 11x - 10$. Correct.
wrong signs in grouping. When a split creates a negative leading term in the second group (like $-4x - 10$ above), factor out a negative: $-2(2x + 5)$. Factoring out a positive $2$ gives $2(2x - 5)$, which changes the sign of the constant and breaks the factorization.
Example 5: Difference of squares
Factor $25x^2 - 16y^4$.
Predict first. Both terms are perfect squares with a minus sign between them, so the difference of squares pattern should apply.
Write each piece as a square: $25x^2 = (5x)^2$ and $16y^4 = (4y^2)^2$. With $a = 5x$ and $b = 4y^2$,
$$25x^2 - 16y^4 = (5x + 4y^2)(5x - 4y^2).$$
Check. Expand: $(5x + 4y^2)(5x - 4y^2) = 25x^2 - 20xy^2 + 20xy^2 - 16y^4 = 25x^2 - 16y^4$. The middle terms cancel, which is the signature of this pattern.
Example 6: Perfect square trinomial
Factor $4x^2 - 12x + 9$.
Predict first. The first and last terms are perfect squares: $4x^2 = (2x)^2$ and $9 = (3)^2$. If the middle term equals $2ab$ with $a = 2x$ and $b = 3$, the trinomial is a perfect square.
Middle check: $2ab = 2(2x)(3) = 12x$. The minus sign gives $-12x$, which matches.
$$4x^2 - 12x + 9 = (2x - 3)^2.$$
Check. Expand: $(2x - 3)^2 = 4x^2 - 12x + 9$. Correct.
Example 7: Sum of cubes
Factor $x^3 + 8$.
Predict first. Both terms are perfect cubes: $x^3 = (x)^3$ and $8 = (2)^3$. Apply the sum of cubes pattern with $a = x$ and $b = 2$, using SOAP for the signs.
$$x^3 + 8 = (x + 2)(x^2 - 2x + 4).$$
Check. Expand: $(x + 2)(x^2 - 2x + 4) = x^3 - 2x^2 + 4x + 2x^2 - 4x + 8 = x^3 + 8$. Correct.
Example 8: GCF first, then a pattern, factored completely
Factor $2x^3 - 8x$ completely.
Predict first. A GCF of $2x$ is visible. After it comes out, the leftover may match a known pattern, so do not stop at the GCF. The word “completely” is a hint that more than one step is required.
Step 1, GCF: $2x^3 - 8x = 2x(x^2 - 4)$.
Step 2, recognize the pattern: $x^2 - 4 = (x)^2 - (2)^2$ is a difference of squares with $a = x$ and $b = 2$.
$$2x^3 - 8x = 2x(x + 2)(x - 2).$$
Check. Expand: $2x(x + 2)(x - 2) = 2x(x^2 - 4) = 2x^3 - 8x$. Correct. This example shows why the GCF comes first: it exposes a clean pattern that was hidden behind the common factor of $2x$.
Common Misconceptions
a sum of squares factors the same way a difference of squares does. The difference of squares $a^2 - b^2$ factors as $(a + b)(a - b)$, but a sum of squares $a^2 + b^2$ does not factor over the real numbers. Predict-then-check exposes the error directly. Suppose someone claims $x^2 + 9 = (x + 3)(x - 3)$. Multiply the proposed factors: $(x + 3)(x - 3) = x^2 - 9$, not $x^2 + 9$. The sign on the constant is wrong, so the claim fails. The minus sign in the pattern is not optional.
pulling out a common factor is an additive move rather than a multiplicative one. Factoring out a GCF divides each term by the same factor, then writes that factor as a multiplier outside the parentheses. It does not subtract the factor from each term. Predict-then-check on $12x^3 - 18x^2 + 6x$: a tempting but wrong move treats the $6x$ as something subtracted from each term. Multiply the correct answer back instead: $6x(2x^2 - 3x + 1) = 12x^3 - 18x^2 + 6x$. The GCF multiplies the leftover trinomial; it is not subtracted from it.
factoring is finished once a common factor comes out. A GCF is often only the first step. After it is removed, the leftover expression may still factor. In $2x^3 - 8x = 2x(x^2 - 4)$, the leftover $x^2 - 4$ is a difference of squares and factors further into $(x + 2)(x - 2)$. The word “completely” means factoring until no factor can be factored further. Always inspect what remains inside the parentheses before declaring the work done.
the sign in the cube formulas can be guessed. Students sometimes write $a^3 - b^3 = (a - b)(a^2 - ab + b^2)$, copying the outer sign into the trinomial. The SOAP order fixes this: the trinomial sign is opposite the outer sign, so $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Predict-then-check on $x^3 - 8$: expanding $(x - 2)(x^2 + 2x + 4) = x^3 - 8$ confirms the positive middle sign.
Leveled Practice
Factor $5x^3 - 15x^2 + 10x$ completely.
Factor $x^2 - 8x + 15$, then factor $4x^2 - 25$.
Factor $2x^2 - x - 15$ using the AC method.
Factor $3x^4 - 3$ completely.
For which integer values of $k$ does $x^2 + kx + 12$ factor over the integers?
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Factoring a sum of squares | $x^2 + 4 = (x + 2)(x - 2)$ | $x^2 + 4$ does not factor over the reals |
| Sign error in cubes | $a^3 - b^3 = (a - b)(a^2 - ab + b^2)$ | Middle sign is opposite: $(a - b)(a^2 + ab + b^2)$ |
| Incomplete factoring | $2x^3 - 8x = 2x(x^2 - 4)$ | Continue: $2x(x + 2)(x - 2)$ |
| Dropping the GCF | $6x^2 + 9x \to (2x + 3)(3x)$ written without the GCF first | Factor out $3x$ first: $3x(2x + 3)$ |
| Missing a perfect square | $x^2 + 8x + 16 \to$ general trinomial only | Check $2(x)(4) = 8x$: it is $(x + 4)^2$ |
Mastery Checklist
Novice (Level 1-2):
Competent (Level 3-4):
Proficient (Level 5):
Mental Model
Factoring is reverse multiplication. Multiplication is a machine that takes factors in and gives an expanded expression out. Factoring runs that machine backward: it takes the expanded expression as input and returns the factors. Because the two directions undo each other, the answer is always checkable by expanding and comparing.
| You know this multiplication | so you can reverse it to this factoring |
|---|---|
| $(x + p)(x + q) = x^2 + (p + q)x + pq$ | $x^2 + bx + c = (x + p)(x + q)$ |
| $(a + b)(a - b) = a^2 - b^2$ | $a^2 - b^2 = (a + b)(a - b)$ |
| $(a + b)^2 = a^2 + 2ab + b^2$ | $a^2 + 2ab + b^2 = (a + b)^2$ |
| $(a + b)(a^2 - ab + b^2) = a^3 + b^3$ | $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ |
Recognizing a pattern comes from practice, and it is normal for the recognition to feel slow at first. When you see $x^2 - 9$, the goal is for your first thought to be “two terms, both perfect squares, a minus sign between them, so this is $a^2 - b^2$.” That recognition becomes fast with enough examples, and a wrong turn along the way is common and worth understanding rather than a sign you are not a math person.
The payoff arrives in the next skill. A product is zero exactly when one of its factors is zero, so a polynomial written as a product hands you its zeros directly. That is why factoring is the doorway to solving quadratic equations.
Resources
| Resource | Source |
|---|---|
| Textbook section | OpenStax College Algebra 2e, Section 1.5 Factoring Polynomials |
| OpenStax 1.5 Factoring Polynomials | openstax.org/books/college-algebra-2e/pages/1-5-factoring-polynomials |
| OpenStax Chapter 1 Key Concepts | openstax.org/books/college-algebra-2e/pages/1-key-concepts |
Connections
Looking back:
- Algebraic Simplification provides the distribution and combination steps that factoring reverses
- Exponent Laws govern how variable factors combine when a GCF is removed
Looking ahead:
- Quadratic Equations uses factoring with the zero-product property to find roots
- Rational expressions simplify only after numerator and denominator are factored, so common factors can cancel
- In calculus, factoring resolves the $\frac{0}{0}$ indeterminate form: $\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2} = \lim_{x \to 2} (x + 2) = 4$ after canceling the factored $x - 2$
Real-world connections:
- Factoring rewrites an area expressed as a sum into a product of side lengths, recovering the dimensions of a rectangle
- Polynomial factorization over finite fields underlies error-correcting codes used in digital communication
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|---|---|---|
| Linear Equations | Skills Index | Quadratic Equations |
Last updated: 2026-06-16