Lesson 2 of 12 · Algebra Foundations
Factoring
Factoring is pattern recognition under time pressure. The identities themselves are simple; the skill is spotting which one applies — including when it's hidden behind a substitution, an extra common factor, or an unfamiliar degree.
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Learning objectives
- Apply the standard identities — difference of squares, sum/difference of cubes, perfect square trinomials — without deriving them each time.
- Factor four-term polynomials by grouping.
- Recognize a "quadratic in disguise" (an expression that's quadratic in x², x³, or another sub-expression) and substitute to reveal it.
- Know the Sophie Germain identity and recognize when a fourth-degree expression is a candidate for it.
The core idea
Every factoring problem starts the same way, regardless of degree: pull out the greatest common factor first. Skipping this step is the single most common reason a factorization looks "stuck" — after removing a GCF, what's left is almost always simpler and matches a standard pattern. From there, the order of attack is: check for a difference of squares or sum/difference of cubes, check whether the expression is a quadratic in some sub-expression, and only then consider grouping or a named identity.
Difference of squares
a² − b² = (a − b)(a + b)
The most common factoring pattern on the exam by a wide margin. Watch for it disguised as a difference of even powers: x⁴ − 16 = (x²)² − 4² factors the same way.
Sum and difference of cubes
a³ − b³ = (a − b)(a² + ab + b²), a³ + b³ = (a + b)(a² − ab + b²)
Unlike a² + b², a sum of cubes does factor. The quadratic factor does not factor further over the integers in general — don't try to force it.
Perfect square trinomial
a² + 2ab + b² = (a + b)², a² − 2ab + b² = (a − b)²
Recognizable by the middle term being exactly twice the product of the square roots of the outer terms.
Sophie Germain identity
a⁴ + 4b⁴ = (a² + 2b² − 2ab)(a² + 2b² + 2ab)
A specific, memorable case that lets a sum of fourth powers factor even though it isn't a difference of squares. Signal: a fourth-degree binomial of the form (something)⁴ + 4·(something)⁴.
Worked example 1 — GCF, then difference of squares
Problem
Factor 2x³ − 8x completely.
Key insight
There's a common factor of 2x before any identity applies.
Solution
2x³ − 8x = 2x(x² − 4) = 2x(x − 2)(x + 2).
Takeaway
Always remove the GCF first — it's the step most often skipped, and skipping it hides the pattern underneath.
Worked example 2 — Difference of squares, twice
Problem
Factor x⁴ − 1 completely over the integers.
Key insight
x⁴ − 1 = (x²)² − 1² is a difference of squares — and one of the resulting factors is too.
Solution
x⁴ − 1 = (x² − 1)(x² + 1) = (x − 1)(x + 1)(x² + 1). The factor x² + 1 has no real roots, so it stops here.
Takeaway
After factoring once, check whether either resulting factor can be factored again before declaring the expression fully factored.
Worked example 3 — Factoring by grouping
Problem
Factor x³ − 3x² − 4x + 12 completely.
Key insight
Four terms with no single common factor is the classic signal for grouping: pair the terms and factor each pair.
Solution
Group as (x³−3x²)+(−4x+12) = x²(x−3) − 4(x−3) = (x−3)(x²−4). The second factor is a difference of squares, so the complete factorization is (x − 3)(x − 2)(x + 2).
Takeaway
Grouping produces an intermediate factorization — always check whether any resulting factor can be broken down further.
Worked example 4 — Quadratic in disguise
Problem
Solve x⁴ − 5x² + 4 = 0 for all real x.
Key insight
The expression is quadratic in x² — substitute y = x² to see the familiar shape.
Solution
With y = x²: y² − 5y + 4 = 0, which factors as (y − 1)(y − 4) = 0, so y = 1 or y = 4. Substituting back: x² = 1 gives x = ±1, and x² = 4 gives x = ±2. All four values: x = −2, −1, 1, 2.
Takeaway
When you see only even powers of x (or any repeated sub-expression), a substitution can turn an unfamiliar degree into an ordinary quadratic.
Strategy notes
GCF first, always
Before checking any identity, ask whether every term shares a common factor. This single habit prevents most "I don't see how to factor this" moments.
Sum of squares doesn't factor — usually
a² + b² has no factorization over the reals in general. If a problem seems to need one, check whether it's actually a disguised Sophie Germain form (a⁴ + 4b⁴) rather than a plain sum of squares.
Common mistakes
Stopping after the first factoring step
A factorization isn't complete just because it "looks factored." Check every resulting factor for further structure, as in Example 2.
Sign errors when grouping
When the second pair in a grouping starts with a negative term, distributing that negative sign incorrectly is the most common error in this technique. Write the sign explicitly rather than doing it mentally.
Forgetting to substitute back
After solving for y = x² (or another substitution), the problem usually asks for x, not y. Forgetting the back-substitution — and forgetting that x² = k has two real solutions when k > 0 — is a common source of lost points.
Practice
20 questions across four difficulty tiers. Most ask you to factor completely — type your answer in any equivalent form (any order of factors, any equivalent grouping) and it's checked for algebraic equivalence, not exact text. A few ask you to solve for x — separate multiple values with commas. Up to three tries per question before the solution is shown.