Factoring Techniques

Why This Matters

Factoring is multiplication in reverse. Instead of expanding $(x + 3)(x + 2)$ into $x^2 + 5x + 6$, you're starting with $x^2 + 5x + 6$ and breaking it back into $(x + 3)(x + 2)$. It's one of the most tested skills in algebra because it connects directly to solving equations, simplifying rational expressions, and working with functions later on.

The key to factoring isn't just memorizing formulas. It's learning to recognize which technique fits the expression in front of you. That pattern recognition is what separates quick, confident solutions from frustrating dead-ends.

---

Extracting Common Factors

Before you try anything else, always check for factors shared by every term. This simplifies the expression and often reveals a hidden pattern underneath.

Greatest Common Factor (GCF)

The GCF is the largest factor (number and/or variable) that divides evenly into every term of the polynomial. Always pull it out first.

For example, in $12x^3 + 8x^2 - 4x$, every term is divisible by $4x$:

$12x^3 + 8x^2 - 4x = 4x(3x^2 + 2x - 1)$

To check your work, distribute the GCF back through the parentheses. You should get the original expression.

Factoring by Grouping

Grouping works best for four-term polynomials. If you see four terms, this should be your first instinct.

1. Group the terms into pairs that share a common factor
2. Factor each pair separately
3. Look for a common binomial factor that emerges, and factor it out

For example, with $x^3 + 2x^2 + 3x + 6$:

• Group: $(x^3 + 2x^2) + (3x + 6)$
• Factor each group: $x^2(x + 2) + 3(x + 2)$
• Extract the common binomial: $(x + 2)(x^2 + 3)$

---

Recognizing Special Patterns

These formulas let you factor instantly once you recognize the structure. Memorize the forms — they show up constantly.

Difference of Squares

$a^2 - b^2 = (a + b)(a - b)$

This only works for the subtraction of two perfect squares. Both terms must be perfect squares, so look for coefficients like 1, 4, 9, 16, 25 and even exponents on variables.

For example: $9x^2 - 49 = (3x + 7)(3x - 7)$

One common trap: $a^2 + b^2$ (a sum of squares) cannot be factored over the real numbers. Don't try to force it.

Sum and Difference of Cubes

• Sum of cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
• Difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Notice the sign pattern. The binomial factor uses the same sign as the original expression. The trinomial factor flips the sign on the $ab$ term and keeps the last term always positive. The mnemonic SOAP captures this: Same, Opposite, Always Positive.

For example: $8x^3 - 27 = (2x)^3 - 3^3 = (2x - 3)(4x^2 + 6x + 9)$

The trinomial factor $(a^2 + ab + b^2)$ or $(a^2 - ab + b^2)$ does not factor further over the reals.

Perfect Square Trinomials

$a^2 + 2ab + b^2 = (a + b)^2$ $a^2 - 2ab + b^2 = (a - b)^2$

The test: check whether the middle term equals exactly $2ab$, where $a$ and $b$ are the square roots of the first and last terms.

For example: $x^2 + 10x + 25 = (x + 5)^2$ because $2(x)(5) = 10x$ matches the middle term.

Spotting this pattern is faster than using general trinomial methods, which saves real time on exams.

---

Factoring Trinomials

When special patterns don't apply, use systematic methods to factor expressions of the form $ax^2 + bx + c$.

Trinomials with Leading Coefficient 1

When $a = 1$, you need two numbers $m$ and $n$ that multiply to $c$ and add to $b$. Then:

$x^2 + bx + c = (x + m)(x + n)$

For example, to factor $x^2 + 7x + 12$: you need two numbers that multiply to 12 and add to 7. That's 3 and 4, so $x^2 + 7x + 12 = (x + 3)(x + 4)$.

Watch your signs carefully:

• If $c$ is positive, both numbers have the same sign (both positive if $b$ is positive, both negative if $b$ is negative)
• If $c$ is negative, the two numbers have opposite signs

Trinomials with Leading Coefficient ≠ 1

When $a \neq 1$, the AC method gives you a systematic approach:

1. Multiply $a \cdot c$
2. Find two numbers that multiply to $ac$ and add to $b$
3. Split the middle term $bx$ into two terms using those numbers
4. Factor by grouping the resulting four-term expression

For example, to factor $2x^2 + 7x + 3$:

1. $a \cdot c = 2 \cdot 3 = 6$
2. Two numbers that multiply to 6 and add to 7: that's 1 and 6
3. Rewrite: $2x^2 + x + 6x + 3$
4. Group and factor: $x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)$

You can also use trial and error with factor pairs of $a$ and $c$, but the AC method is more reliable, especially under pressure.

---

Advanced Strategies

When basic techniques don't immediately apply, these methods help you see hidden structure.

Factoring by Substitution

Some expressions look complicated but are actually quadratic in disguise. You can replace a complex piece with a single variable to reveal the pattern.

For example, $x^4 - 5x^2 + 4$ looks tough, but let $u = x^2$:

1. Substitute: $u^2 - 5u + 4$
2. Factor the simpler expression: $(u - 1)(u - 4)$
3. Substitute back: $(x^2 - 1)(x^2 - 4)$
4. Factor further (both are differences of squares): $(x + 1)(x - 1)(x + 2)(x - 2)$

Any time you see a polynomial that looks quadratic but has higher-degree terms (like $x^4$ and $x^2$, or $x^6$ and $x^3$), try substitution.

Complete Factorization

"Factor completely" means every factor should be either prime (unfactorable) or a monomial. To get there, apply techniques in this order:

1. Pull out the GCF
2. Look for special patterns (difference of squares, cubes, perfect square trinomials)
3. Use trinomial methods (direct factoring or AC method)
4. Check each factor to see if it can be broken down further

That last step is easy to forget. After your first factorization, look at every factor again. For instance, $x^4 - 16$ factors as $(x^2 + 4)(x^2 - 4)$, but $x^2 - 4$ is itself a difference of squares: $(x^2 + 4)(x + 2)(x - 2)$.

---

Quick Reference Table

---

Self-Check Questions

1. What is the first step you should always take before applying any other factoring technique?

2. How can you distinguish between a perfect square trinomial and a general trinomial that requires the AC method?

3. Compare the formulas for sum of cubes and difference of cubes. What stays the same, and what changes?

4. Given the expression $x^4 - 16$, which techniques would you need to apply to factor it completely? (Hint: it requires more than one step.)

5. If you encounter $6x^2 + 11x - 10$, explain why you would use the AC method rather than looking for a special pattern.