| Primary source | OpenStax College Algebra 2e, Section 1.5: "Factoring Polynomials" |
| Direct link | https://openstax.org/books/college-algebra-2e/pages/1-5-factoring-polynomials |
College Algebra 2e Section 1.5 begins on page 73 of the PDF.
Factoring is multiplication in reverse.
You already know how to multiply: $( x + 3 )( x - 2 ) = x^2 + x - 6$. Factoring starts at the end and asks: "which two (or more) things, when multiplied, produce this expression?"
Every factoring technique in this lesson is a pattern you recognize because you have already seen the multiplication. There are six core patterns. Learn to recognize them on sight, and factoring becomes pattern-matching rather than guessing.
One practical reason factoring matters right now: it is how you simplify rational expressions (fractions with polynomials). Without it, you cannot cancel common factors. In MATH161 calculus, you will use factoring to resolve the $\frac{0}{0}$ indeterminate form that appears in every limit calculation.
| 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 trial or 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 GCF from each |
Strategy. Always check for a GCF first. Then count terms: two terms suggests difference of squares or cubes; three terms suggests a trinomial; four terms suggests grouping.
The GCF is the largest factor that divides every term in the expression. Factoring it out is always the first step.
How to find the GCF:
Example 1. Factor $12x^3 - 18x^2 + 6x$.
\[ 12x^3 - 18x^2 + 6x = 6x(2x^2 - 3x + 1) \]
Check by expanding: $6x(2x^2) - 6x(3x) + 6x(1) = 12x^3 - 18x^2 + 6x$. Correct.
Always factor out the GCF first. Students who skip this step create unnecessarily hard trinomials. A trinomial with a large leading coefficient often factors easily after a GCF is removed.
Example 2. Factor $4x^2y - 8xy^2 + 12xy$.
GCF $= 4xy$.
\[ 4x^2y - 8xy^2 + 12xy = 4xy(x - 2y + 3) \]
A trinomial of the form $x^2 + bx + c$ factors as $(x + p)(x + q)$ where: \[ p + q = b \quad \text{and} \quad p \cdot q = c \]
How to find $p$ and $q$: List factor pairs of $c$ and check which pair sums to $b$.
Example 3. Factor $x^2 + 7x + 12$.
Need two numbers that multiply to $12$ and add to $7$.
| 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: $(x+3)(x+4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$. Correct.
Example 4. Factor $x^2 - 5x - 14$.
Need two numbers that multiply to $-14$ and add to $-5$. Since the product is negative, one number is positive and one is negative.
| Factor pair of $-14$ | Sum |
|---|---|
| $1 \times (-14)$ | $-13$ |
| $(-1) \times 14$ | $13$ |
| $2 \times (-7)$ | $-5$ $\checkmark$ |
\[ x^2 - 5x - 14 = (x + 2)(x - 7) \]
Sign rules for the factor pairs:
- Both factors positive: both constant terms are positive.
- Both factors negative: constant terms are both negative, but their product is positive ($c > 0$) and sum is negative ($b < 0$).
- One factor positive, one negative: $c < 0$ (negative product means opposite signs).
For $ax^2 + bx + c$ with $a \neq 1$:
Example 5. Factor $6x^2 + 11x - 10$.
Check: $(2x+5)(3x-2) = 6x^2 - 4x + 15x - 10 = 6x^2 + 11x - 10$. Correct.
Common Error -- Wrong Signs in Grouping. When the middle split creates a negative leading term in the second group (like $-4x - 10$ above), factor out a negative: $-2(2x + 5)$. If you factor out a positive $2$, you get $2(2x - 5)$, which changes the sign of the constant and breaks the factorization.
\[ a^2 - b^2 = (a + b)(a - b) \]
This pattern appears only when both terms are perfect squares AND there is a minus sign between them.
Example 6. Factor $x^2 - 49$.
$x^2 = (x)^2$ and $49 = (7)^2$.
\[ x^2 - 49 = (x + 7)(x - 7) \]
Example 7. Factor $25x^2 - 16y^4$.
$25x^2 = (5x)^2$ and $16y^4 = (4y^2)^2$.
\[ 25x^2 - 16y^4 = (5x + 4y^2)(5x - 4y^2) \]
Example 8. Factor $x^4 - 81$.
$x^4 = (x^2)^2$ and $81 = (9)^2$.
\[ x^4 - 81 = (x^2 + 9)(x^2 - 9) = (x^2 + 9)(x + 3)(x - 3) \]
Note: $(x^2 + 9)$ is a sum of squares and does not factor over the real numbers.
Sum of squares $a^2 + b^2$ does not factor. Students sometimes write $x^2 + 9 = (x+3)^2$ or $(x+3)(x-3)$. Neither is correct. Check: $(x+3)^2 = x^2 + 6x + 9$, which has a middle term. The sum of two squares has no real factorization.
\[ a^2 + 2ab + b^2 = (a + b)^2 \] \[ a^2 - 2ab + b^2 = (a - b)^2 \]
How to recognize: The first and last terms are perfect squares. The middle term is exactly twice the product of their square roots.
Example 9. Factor $x^2 + 10x + 25$.
\[ x^2 + 10x + 25 = (x + 5)^2 \]
Example 10. Factor $4x^2 - 12x + 9$.
\[ 4x^2 - 12x + 9 = (2x - 3)^2 \]
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \] \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
A mnemonic for the signs: SOAP -- Same, Opposite, Always Positive.
Example 11. Factor $x^3 + 8$.
$x^3 = (x)^3$, $8 = (2)^3$.
\[ x^3 + 8 = (x + 2)(x^2 - 2x + 4) \]
Check: $(x+2)(x^2 - 2x + 4) = x^3 - 2x^2 + 4x + 2x^2 - 4x + 8 = x^3 + 8$. Correct.
Example 12. Factor $27x^3 - 64y^3$.
$27x^3 = (3x)^3$, $64y^3 = (4y)^3$.
\[ 27x^3 - 64y^3 = (3x - 4y)(9x^2 + 12xy + 16y^2) \]
The trinomial in sum/difference of cubes does not factor further. Students sometimes try to factor $x^2 - 2x + 4$. Its discriminant is $(-2)^2 - 4(1)(4) = 4 - 16 = -12 < 0$, so it has no real roots and does not factor over the reals.
Used for polynomials with four or more terms. Group terms in pairs that share a common factor.
Example 13. Factor $x^3 + 2x^2 + 3x + 6$.
Group as $(x^3 + 2x^2) + (3x + 6)$.
Factor GCF from each group: \[ x^2(x + 2) + 3(x + 2) \]
Factor out the common binomial $(x + 2)$: \[ (x + 2)(x^2 + 3) \]
Check: $(x+2)(x^2+3) = x^3 + 3x + 2x^2 + 6 = x^3 + 2x^2 + 3x + 6$. Correct.
Example 14. Factor $2x^3 - 3x^2 - 4x + 6$.
This grouping is less obvious. Try grouping the first two and last two:
$(2x^3 - 3x^2) + (-4x + 6)$
$= x^2(2x - 3) - 2(2x - 3)$
$= (2x - 3)(x^2 - 2)$
Grouping does not always work on the first attempt. If one grouping produces no common binomial, try a different pairing. Sometimes you need to rearrange the terms first.
When you see a polynomial to factor, ask these questions in order:
Example 15. Factor $2x^3 - 8x$ completely.
"Completely" means factor until no factor can be factored further. Stopping at $2x(x^2-4)$ is incomplete because $x^2 - 4$ still factors.
| Error | Example | Correction |
|---|---|---|
| Factoring a sum of squares | $x^2 + 4 = (x+2)(x-2)$ | $x^2 + 4$ does not factor over reals |
| Sign error in cubes | $a^3 - b^3 = (a-b)(a^2-ab+b^2)$ | Middle sign must be opposite: $(a-b)(a^2+ab+b^2)$ |
| Incomplete factoring | $2x^3 - 8x = 2x(x^2-4)$ | Must continue: $2x(x+2)(x-2)$ |
| Dropping GCF | $6x^2+9x \to (2x+3)(3x)$ | Factor out $3x$ first: $3x(2x+3)$ |
| Wrong check of perfect square | $x^2+8x+16 \to (x+4)(x+4)$ forgotten as option | Check: $2(x)(4)=8x$. Yes, perfect square: $(x+4)^2$ |
Problem 1. Factor $5x^3 - 15x^2 + 10x$.
GCF $= 5x$.
$5x(x^2 - 3x + 2)$
Factor the trinomial: need $p + q = -3$, $pq = 2$. Try $-1$ and $-2$: $(-1)+(-2)=-3$, $(-1)(-2)=2$. $\checkmark$
$5x(x-1)(x-2)$
Problem 2. Factor $x^2 - 8x + 15$.
Need $p + q = -8$, $pq = 15$. Both negative: try $-3$ and $-5$: $(-3)(-5)=15$, $-3+(-5)=-8$. $\checkmark$
$(x-3)(x-5)$
Problem 3. Factor $4x^2 - 25$.
Difference of squares: $4x^2 = (2x)^2$, $25 = (5)^2$.
$(2x+5)(2x-5)$
Problem 4. Factor $x^3 - 27$.
Difference of cubes: $x^3 = (x)^3$, $27 = (3)^3$.
$(x-3)(x^2 + 3x + 9)$
Check signs using SOAP: Same ($-$), Opposite ($+3x$), Always Positive ($+9$). $\checkmark$
Problem 5. Factor $2x^2 - x - 15$ using the AC method.
$a=2$, $c=-15$, so $ac=-30$. Need $mn=-30$, $m+n=-1$.
Try $-6$ and $5$: $(-6)(5)=-30$, $-6+5=-1$. $\checkmark$
Rewrite: $2x^2 - 6x + 5x - 15$.
Group: $(2x^2-6x) + (5x-15) = 2x(x-3) + 5(x-3) = (x-3)(2x+5)$.
Check: $(x-3)(2x+5) = 2x^2 + 5x - 6x - 15 = 2x^2 - x - 15$. $\checkmark$
Problem 6. Factor $3x^4 - 3$ completely.
GCF $= 3$: $3(x^4 - 1)$.
$x^4 - 1$ is difference of squares: $3(x^2+1)(x^2-1)$.
$x^2 - 1$ is difference of squares again: $3(x^2+1)(x+1)(x-1)$.
$x^2 + 1$ does not factor over the reals.
Final: $3(x^2+1)(x+1)(x-1)$.
Problem 7. Factor $x^3 - 2x^2 - x + 2$.
Group: $(x^3 - 2x^2) + (-x + 2)$.
$= x^2(x-2) - 1(x-2)$
$= (x-2)(x^2-1)$
$= (x-2)(x+1)(x-1)$
Problem 8. Factor $x^4 - 5x^2 + 4$.
Treat as a trinomial in $u = x^2$: $u^2 - 5u + 4$.
Need $p+q=-5$, $pq=4$: try $-1$ and $-4$. $\checkmark$
$(u-1)(u-4) = (x^2-1)(x^2-4)$.
Both are differences of squares:
$(x+1)(x-1)(x+2)(x-2)$.
Problem 9. For which integer values of $k$ does $x^2 + kx + 12$ factor over the integers?
We need integers $p$, $q$ with $pq = 12$. The factor pairs of $12$ (including negatives) are:
$(1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12, 1), (-1, -12), (-2, -6), (-3, -4), (-4, -3), (-6, -2), (-12, -1)$.
The possible values of $k = p + q$ are: $13, 8, 7, -7, -8, -13$.
So $k \in \{-13, -8, -7, 7, 8, 13\}$.
Factoring is reverse multiplication. Every factoring pattern corresponds to a multiplication you already know:
| You know this multiplication... | ...so you can reverse it to this factoring |
|---|---|
| $(x+p)(x+q) = x^2 + (p+q)x + pq$ | $x^2 + bx + c = (x+p)(x+q)$ |
| $(a+b)(a-b) = a^2 - b^2$ | $a^2 - b^2 = (a+b)(a-b)$ |
| $(a+b)^2 = a^2 + 2ab + b^2$ | $a^2 + 2ab + b^2 = (a+b)^2$ |
| $(a+b)(a^2-ab+b^2) = a^3+b^3$ | $a^3 + b^3 = (a+b)(a^2-ab+b^2)$ |
The skill of recognizing a pattern comes from practice. When you see $x^2 - 9$, your first thought should be "two terms, both perfect squares, minus sign between them -- that is $a^2 - b^2$." That recognition becomes fast with enough examples.
For students who find math intimidating: You do not need to memorize the cube formulas on day one. Concentrate on GCF, trinomials, and difference of squares. Those three patterns cover the majority of factoring problems you will encounter. The cube and grouping patterns can be learned when you need them.
For students interested in proof: The Fundamental Theorem of Algebra guarantees that every polynomial of degree $n$ factors completely over the complex numbers into $n$ linear factors. Over the real numbers, irreducible quadratics (negative discriminant) remain. Factoring techniques are the constructive half of this theorem.
For students interested in careers: Factoring is the algebraic equivalent of prime factorization. In computer science, polynomial factorization over finite fields is a core operation in error-correcting codes (used in every digital communication system) and in certain cryptographic protocols.
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