Factoring — common factor extraction, grouping, using identities: Worked examples

This page is a workout. The parent note taught the three tricks (pull out a common factor, group, use an identity). Here we hit every kind of factoring problem you can meet — one at a time, each labelled with which "cell" of the scenario map it fills.

Before each solution there is a Forecast line. Cover the answer, guess what technique wins, then check yourself. Guessing first is how the pattern-recognition muscle grows.

The scenario matrix

Every factoring problem you meet is one (or a mix) of these cells. The last column names the example that covers it.

#	Case class	What makes it tricky	Covered by
A	Common factor, positive coeffs	spotting the GCF including variable powers	Ex 1
B	Common factor with a negative leading term	pulling out a minus sign flips signs	Ex 2
C	Grouping, 4 terms, works first try	matching binomials	Ex 3
D	Grouping that needs rearranging	naive pairing fails	Ex 4
E	Difference of squares (a special identity)	recognising two squares	Ex 5
F	Perfect-square trinomial vs a fake one	the middle term test	Ex 6
G	Sum / difference of cubes (sign trap)	which sign goes where	Ex 7
H	Nested — factor, then factor again	knowing when you're done	Ex 8
I	Degenerate / prime inputs (won't factor over reals)	knowing when to stop	Ex 9
J	Word problem (real-world) + exam twist	translating words to a polynomial	Ex 10

We also fold in the zero / limiting cases as we go: what happens when a coefficient is $0$ , or when the two "squares" are equal, or when a variable is missing.

Let's first fix one picture in your mind, because factoring is really turning a sum into a rectangle.

Figure — Factoring — common factor extraction, grouping, using identities

Two words we use constantly:

GCF (Greatest Common Factor) — the biggest chunk every term shares, counting both numbers and letters.
prime / irreducible — a polynomial that cannot be broken into smaller polynomial factors (over the number system we're allowed to use).

Cell A — Common factor, positive coefficients

Find the number part of the GCF. $g cd (12, 18, 30) = 6$ . Why this step? The GCF's number is the biggest integer dividing all coefficients; anything bigger would fail on at least one term.
Find the variable part. The powers of $x$ are $4, 3, 2$ . The lowest is $2$ , so $x^{2}$ is shared by all. Why this step? You can only pull out a power that every term actually contains. The smallest power is the limiting one.
GCF $= 6 x^{2}$ . Divide each term by it. $\frac{12 x ^{4}}{6 x ^{2}} = 2 x^{2}, \frac{18 x ^{3}}{6 x ^{2}} = 3 x, \frac{30 x ^{2}}{6 x ^{2}} = 5.$
Write as a product. $12 x^{4} + 18 x^{3} + 30 x^{2} = 6 x^{2} (2 x^{2} + 3 x + 5) .$ Why stop here? The bracket $2 x^{2} + 3 x + 5$ has no common factor and (as we'll see later) no real roots — it's prime.

Verify: expand back: $6 x^{2} \cdot 2 x^{2} = 12 x^{4}$ , $6 x^{2} \cdot 3 x = 18 x^{3}$ , $6 x^{2} \cdot 5 = 30 x^{2}$ . ✓

Cell B — Negative leading term

GCF of magnitudes: $g cd (4, 6, 2) = 2$ , lowest power of $x$ is $x^{1}$ . So the size of the GCF is $2 x$ .
Choose the sign. Because the leading term $- 4 x^{3}$ is negative, pull out $- 2 x$ so the leading term inside becomes positive (this is the exam-standard, tidy form). Why this step? A leading $+$ inside makes later identity-matching far easier and avoids sign slips.
Divide each term by $- 2 x$ (dividing a negative by a negative flips the sign): $\frac{- 4 x ^{3}}{- 2 x} = 2 x^{2}, \frac{6 x ^{2}}{- 2 x} = - 3 x, \frac{- 2 x}{- 2 x} = 1.$
Assemble: $- 4 x^{3} + 6 x^{2} - 2 x = - 2 x (2 x^{2} - 3 x + 1) .$
Bonus — the inside still factors. $2 x^{2} - 3 x + 1 = (2 x - 1) (x - 1)$ , so fully: $- 2 x (2 x - 1) (x - 1) .$

Verify: $- 2 x (2 x - 1) (x - 1)$ . First $(2 x - 1) (x - 1) = 2 x^{2} - 3 x + 1$ , then $- 2 x$ times that $= - 4 x^{3} + 6 x^{2} - 2 x$ . ✓

Cell C — Grouping, works first try

Grouping means: split into two pairs, factor each pair, and hope a shared bracket appears. Watch the picture — the two pairs must land on the same binomial.

Split into pairs: $(x^{3} + 5 x^{2}) + (3 x + 15)$ . Why this pairing? The first two share $x^{2}$ ; the last two share $3$ . Look for pairs with an obvious common factor.
Factor each pair: $x^{2} (x + 5) + 3 (x + 5) .$ Why this step? We want the leftover brackets to match. Here both are $(x + 5)$ — success.
Extract the shared bracket $(x + 5)$ as a common factor: $(x + 5) (x^{2} + 3) .$ Why this works: $(x + 5)$ is now a common factor of the two terms $x^{2} (x + 5)$ and $3 (x + 5)$ — exactly the "pull out a common factor" move, but the common thing is a whole binomial.

Verify: $(x + 5) (x^{2} + 3) = x^{3} + 3 x + 5 x^{2} + 15 = x^{3} + 5 x^{2} + 3 x + 15$ . ✓

Cell D — Grouping that needs rearranging

Naive pairing $(6 x y + 4) + (8 x + 3 y)$ : factor each — $2 (3 x y + 2)$ and $(8 x + 3 y)$ . No shared bracket. Dead end. Why show the failure? Because in an exam your first grouping often fails; the skill is recognising it and rearranging, not panicking.
Rearrange so matchable terms sit together. Put the two terms containing $x$ next to each other: $6 x y + 8 x + 3 y + 4.$ Why this step? Group $6 x y$ with $8 x$ (both have $x$ ) and $3 y$ with $4$ .
Pair and factor: $(6 x y + 8 x) + (3 y + 4) = 2 x (3 y + 4) + 1 \cdot (3 y + 4) .$ Why the " $1$ "? $3 y + 4$ has no common factor other than $1$ ; writing the $1$ makes the shared bracket visible.
Extract $(3 y + 4)$ : $(3 y + 4) (2 x + 1) .$

Verify: $(3 y + 4) (2 x + 1) = 6 x y + 3 y + 8 x + 4$ , same set of terms as the original. ✓

Cell E — Difference of squares

Now we bring in identities — memorised factorings. The first is the difference of squares. See why the middle cancels:

Write each part as a square. $25 x^{2} = (5 x)^{2}$ and $49 = 7^{2}$ . So $a = 5 x$ , $b = 7$ . Why this step? The identity only applies once you can name the two squares.
Apply the identity: $25 x^{2} - 49 = (5 x - 7) (5 x + 7) .$

Zero / limiting note: if the problem were $25 x^{2} - 0 = 25 x^{2}$ , the " $b$ " is $0$ , the identity gives $(5 x) (5 x) = 25 x^{2}$ — degenerate but consistent (it just becomes a repeated factor).

Verify: $(5 x - 7) (5 x + 7) = 25 x^{2} + 35 x - 35 x - 49 = 25 x^{2} - 49$ . ✓

Cell F — Real perfect square vs a fake one

Name the outer squares. $9 x^{2} = (3 x)^{2}$ and $25 = 5^{2}$ , so try $a = 3 x$ , $b = 5$ .
Run the middle-term test. A perfect square needs the middle term to be exactly $2 ab$ : here $2 ab = 2 (3 x) (5) = 30 x$ . The polynomial has $- 30 x$ — matches $- 2 ab$ . Why this test? Two squares alone are not enough; the identity is only valid when the middle term is precisely $\pm 2 ab$ .
Conclude: $9 x^{2} - 30 x + 25 = (3 x - 5)^{2} .$
Now the fake: $9 x^{2} + 25$ has no middle term, so $2 ab$ is missing. It is not a perfect square, and (being a sum of squares) it's prime over the reals.

Verify: $(3 x - 5)^{2} = 9 x^{2} - 30 x + 25$ . ✓ And $(3 x + 5)^{2} = 9 x^{2} + 30 x + 25 \neq = 9 x^{2} + 25$ . ✓

Cell G — Sum / difference of cubes (the sign trap)

Name the cubes. $27 x^{3} = (3 x)^{3}$ , $64 = 4^{3}$ . So $a = 3 x$ , $b = 4$ .
Sign 1 (Same): original is a minus, so the binomial is $(3 x - 4)$ .
Sign 2 (Opposite): the middle term of the trinomial takes the opposite sign → plus $ab$ .
Sign 3 (Always Positive): last term is $+ b^{2}$ . Why SOAP? It removes the #1 exam error — mixing up which sign lives where.
Assemble: $27 x^{3} - 64 = (3 x - 4) ((3 x)^{2} + (3 x) (4) + 4^{2}) = (3 x - 4) (9 x^{2} + 12 x + 16) .$

Verify: $(3 x - 4) (9 x^{2} + 12 x + 16)$ expands to $27 x^{3} - 64$ . ✓ (checked in VERIFY)

Cell H — Nested: factor, then factor again

First difference of squares: $x^{4} = (x^{2})^{2}$ , $81 = 9^{2}$ , so $x^{4} - 81 = (x^{2} - 9) (x^{2} + 9) .$
Check each factor again. $x^{2} - 9 = (x - 3) (x + 3)$ — another difference of squares. But $x^{2} + 9$ is a sum of squares → prime over reals. Why re-check? "Factor completely" means keep going until every factor is prime. Stopping too early loses roots.
Full factorisation over the reals: $x^{4} - 81 = (x - 3) (x + 3) (x^{2} + 9) .$

Recall

Over the complex numbers (Complex Numbers), can $x^{2} + 9$ go further? Yes — $x^{2} + 9 = (x - 3 i) (x + 3 i)$ , giving four linear factors, matching the Fundamental Theorem of Algebra (a degree-4 polynomial has exactly 4 roots).

Verify: $(x - 3) (x + 3) (x^{2} + 9) = (x^{2} - 9) (x^{2} + 9) = x^{4} - 81$ . ✓

Cell I — Degenerate / prime inputs (know when to stop)

$x^{2} + 4$ : a sum of two squares, no middle term. No real difference-of-squares or perfect-square form fits. It is prime over the reals. Why prime? Its roots solve $x^{2} = - 4$ , which has no real solution — so no real linear factors exist (see Zero Product Property).
$x^{2} + x + 1$ : try the discriminant from the Quadratic Formula. The discriminant is $b^{2} - 4 a c = 1^{2} - 4 (1) (1) = 1 - 4 = - 3 < 0$ . Why the discriminant? A negative discriminant means no real roots, hence no real linear factors — the trinomial is prime over the reals.
Conclusion: both are already fully factored over $R$ . Recognising "prime" is a correct answer, not a failure.

Verify: discriminant of $x^{2} + x + 1$ is $- 3$ (negative), and $x^{2} + 4$ has no real root since $x^{2} = - 4$ is impossible. ✓

Cell J — Word problem + exam twist

Factor the area. No common factor. It's a trinomial $2 x^{2} + 7 x + 3$ ; we need two factors whose product is $2 x^{2} + 7 x + 3$ and one of which is $(x + 3)$ . Factor by the "split the middle" version of grouping: find two numbers multiplying to $2 \times 3 = 6$ and adding to $7$ : those are $6$ and $1$ . $2 x^{2} + 7 x + 3 = 2 x^{2} + 6 x + 1 x + 3 = 2 x (x + 3) + 1 (x + 3) = (x + 3) (2 x + 1) .$ Why split the middle? It converts the trinomial into a 4-term grouping problem (Cell C machinery) so we can extract the known bracket.
Read off the length. Area $=$ width $\times$ length $= (x + 3) (2 x + 1)$ , and width $= (x + 3)$ , so $length = (2 x + 1) m .$ Why divide by factoring instead of long division? Because it factored exactly, division is instant — no Polynomial Long Division needed.
Exam twist: width $= 5$ . Set $x + 3 = 5 \Rightarrow x = 2$ .
Then length $= 2 (2) + 1 = 5$ m and area $= 5 \times 5 = 25$ m². Sanity/units: metres × metres = square metres. ✓ All lengths positive, so $x = 2$ is physically valid.

Verify: $(x + 3) (2 x + 1) = 2 x^{2} + 7 x + 3$ . At $x = 2$ : width $= 5$ , length $= 5$ , area $= 25 = 2 (4) + 7 (2) + 3 = 8 + 14 + 3$ . ✓

One-screen recap

Recall

Which cell? Match the trigger to the technique. Every term shares a chunk ::: pull out the GCF (Cells A, B) — mind the sign if the leading term is negative. Four terms, no global factor ::: grouping; rearrange if the first pairing fails (Cells C, D). Two squares, a minus between ::: difference of squares $(a - b) (a + b)$ (Cell E). Three terms, middle $= \pm 2 ab$ ::: perfect square $(a \pm b)^{2}$ (Cell F). Two cubes ::: SOAP — Same, Opposite, Always Positive (Cell G). Factored once ::: check each factor again; stop only at primes (Cell H). Sum of squares / negative discriminant ::: prime over the reals — a valid final answer (Cell I). Word problem ::: translate to a polynomial, factor, divide exactly (Cell J).

Related tools worth a click: Distributive Property (the engine behind every extraction), Quadratic Formula and Rational Root Theorem (for trinomials that resist tricks), Completing the Square (when nothing factors nicely).