The four pieces ax, ay, bx, by tile one rectangle. Read its sides — height a + b, width x + y — and the four terms collapse into a single product (a + b)(x + y). That recovery of the sides is exactly what grouping does.
In Lesson 8.4 you learned to factor a three-term quadratic by hunting for a pair of numbers. But what do you do with a polynomial that has four terms and no factor shared by all of them — something like ax + ay + bx + by? You group. Pair the terms up, pull a common piece out of each pair, and a matching bracket falls out that you can then pull out front. We keep the steady color habit of this stage: a teal factor is a side length or a piece pulled outside, an amber term is an expanded piece of the polynomial's area, red flags a sign trap, and the blue frame is the structural outline — the rectangle and its sides.
By the end you will be able to factor four-term polynomials in pairs, recognize when a shared bracket has appeared, pull it out to finish the job, regroup when the first pairing fails, and even handle the slick case where three terms make a perfect square that pairs with a fourth as a difference of squares. The one habit that ties it all together — and the one we will repeat until it is automatic — is to check every answer by multiplying it back out.
8.5.1 Four terms — group them first
Some polynomials wear their common factor on their sleeve. 6x + 9 obviously carries a 3. But ax + ay + bx + by has no single piece in all four terms: the first two share an a, the last two share a b, and a and b are strangers. When nothing pulls out of all four at once, do not give up — group the terms into pairs and look for a factor inside each pair instead.
Think of the four terms as the four pieces of a rectangle, laid out in a 2×2 block. The top row is ax + ay; the bottom row is bx + by. Each row is a strip, and each strip has a side you can read off.
Group the four terms by rows. The top strip has height a and the bottom strip has height b; both span the same width x + y. That shared width is the bracket that will let us finish.
Key idea
When a polynomial has four terms and no common factor in all four, split it into two pairs and factor each pair on its own. Grouping turns one hard problem into two easy common-factor problems.
8.5.2 Factor each group on its own
Take the two pairs one at a time. From the first pair ax + ay, the common factor is a, so it becomes a(x + y). From the second pair bx + by, the common factor is b, so it becomes b(x + y). Line them up:
ax + ay + bx + by = a(x + y) + b(x + y)
Now stop and notice the small miracle: the same bracket(x + y) appears in both pieces. That is not luck — it is the whole point of grouping. If you have chosen your pairs well, the leftover bracket from the first group will match the leftover bracket from the second. A matching bracket is the green light that says "keep going, this is going to work."
If the brackets don't match…
…then either you grouped the wrong pairs (try a different pairing — see Section 8.5.4), or you made an arithmetic slip pulling out a factor. The brackets must come out identical, sign for sign, before you can take the next step. If one reads (x + y) and the other reads (x − y), they are not a match.
8.5.3 Pull out the shared bracket
Once the bracket matches, treat (x + y) as a single bundle — one object — that both terms carry. It is a common factor just like a number or a letter, so pull it out front exactly the way you pulled out a GCF back in Lesson 8.2. What is left behind is a from the first piece and b from the second:
a(x + y) + b(x + y) = (x + y)(a + b)
And there it is: four scattered terms have become one clean product. Reading the rectangle confirms it — the block is a + b tall and x + y wide, so its area is (a + b)(x + y), which is precisely the four terms we started with. Sides times sides equals piece plus piece plus piece plus piece.
🎮 Try itGroup the 2×2 block
Set the four side-lengths a, b, x, y. Watch the rectangle's four pieces (the amber terms) collapse, row by row, into the two teal sides — and read off the factored product.
a
b
x
y
The same routine works with letters and powers, not just clean two-letter products. Watch a cubic fall apart. In x3 + x2 + x + 1, group the first two and the last two: x3 + x2 gives x2(x + 1), and x + 1 gives 1(x + 1) — yes, you deliberately factor out a 1 so the bracket shows. The shared bracket (x + 1) pulls out, leaving (x + 1)(x2 + 1).
Worked example — a cubic with a sign
Factor 2x3 − 6x2 + x − 3. Group: (2x3 − 6x2) + (x − 3). Factor each pair:2x2(x − 3) + 1(x − 3). Pulling out 2x2 from the first pair leaves x − 3; the second pair already isx − 3, so factor a +1 to make the bracket visible. Pull out the bracket: (x − 3)(2x2 + 1). Check by expanding: (x − 3)(2x² + 1) = 2x³ + x − 6x² − 3 = 2x³ − 6x² + x − 3 ✓.
🎮 Try itWatch a four-term polynomial group
Step through five worked four-term polynomials. The widget shows the grouping, the factor pulled from each pair, the matching bracket, and the finished product — then re-expands to prove it.
Example
8.5.4 Be ready to regroup
Grouping is not magic; it depends on pairing the right terms together. Sometimes the order you are handed hides the structure, and the first pairing leaves two brackets that do not match. When that happens, do not panic and do not declare the polynomial unfactorable — just shuffle the terms and try a different pairing. Because addition can be reordered, you are always allowed to.
Take ac + bd + ad + bc. Pairing it as handed gives (ac + bd) + (ad + bc): the first pair shares nothing, and the brackets refuse to match. So regroup — gather the terms that share a letter:
ac + ad + bc + bd = a(c + d) + b(c + d) = (c + d)(a + b)
Same four terms, smarter pairing, clean factorization. A good rule of thumb: scan for which terms share a common letter or factor, and put those together.
The #1 trap — the invisible +1 and −1
The most common grouping mistake is forgetting to factor a +1 or −1 from the second pair so the bracket lines up. In 2x3 − 6x2 + x − 3 the second pair x − 3 is already x − 3 — you must write it as +1(x − 3), not pretend the bracket appears "for free." And watch signs: in x3 + 2x2 − 5x − 10 the second pair is −5x − 10, so you must factor out −5 to get −5(x + 2) — pulling out a positive 5 would leave (−x − 2) and the brackets would clash.
🎮 Try itDoes this grouping work?
Step through several four-term polynomials. The widget reports whether the given pairing produces a matching bracket — and if not, shows the regrouping that does.
Case
8.5.5 Grouping into a difference of squares
Grouping does not always split four terms into 2 + 2. Sometimes three of the terms quietly form a perfect-square trinomial, and grouping those three together turns the whole expression into a difference of squares — the formula you met in Lesson 8.3, now hiding inside a longer polynomial.
Look at m2 − 6m + 9 − n2. The first three terms m2 − 6m + 9 are a perfect square: they collapse to (m − 3)2. Now the whole thing reads (m − 3)2 − n2 — a square minus a square. Apply A² − B² = (A + B)(A − B) with A = m − 3 and B = n:
Three terms make the perfect square (m − 3)2; subtracting n2 leaves a difference of squares, which factors into (m − 3 + n)(m − 3 − n). Grouping and the special-product formulas, working together.
The same trick handles x2 + 2xy + y2 − z2: the first three terms are (x + y)2, so the expression is (x + y)2 − z2 = (x + y + z)(x + y − z). The lesson: before you force a 2 + 2 split, glance at whether three terms are secretly a square. When they are, group 3 + 1 and reach for the difference-of-squares formula.
Worked example — 3 + 1 grouping
Factor x2 + 2xy + y2 − z2. Spot the square: the first three terms are (x + y)2. Rewrite:(x + y)2 − z2, a difference of squares with A = x + y, B = z. Apply the formula: (x + y + z)(x + y − z) = (x + y + z)(x + y − z). Check a piece: the product's first two terms multiply to x² + 2xy + y² − z² ✓ (expand to confirm).
★ The big ideas, in one breath
When a polynomial has four terms with no common factor in all four, group them in pairs and factor each pair: ax + ay + bx + by = a(x + y) + b(x + y). When the same bracket appears in both pieces, pull it out front: = (x + y)(a + b). Always factor a deliberate +1 or −1 from the second pair so the brackets line up sign for sign. If the first pairing gives no shared bracket, regroup — addition lets you reorder freely. And keep an eye out for the slick 3 + 1 case, where a perfect-square trinomial pairs with a fourth term as a difference of squares: m2 − 6m + 9 − n2 = (m − 3)2 − n2 = (m − 3 + n)(m − 3 − n). Every time, confirm the answer by multiplying it back out.
Coming up next — 8.6
You now own five separate factoring tools: pull the GCF, difference of squares, perfect squares, cross-multiplication, and grouping. Next, in 8.6 Putting Factoring to Work, you will learn the one master routine that picks the right tool every time — and see what factoring unlocks: fast mental arithmetic, divisibility proofs, and the canceling that opens the door to rational expressions.
✎ Exercises 8.5
Work each one out first, then open the answer to check your thinking.
Factor 3x3 + x2 − 6x − 2 (mind the sign on the second pair).
Show answer
x2(3x + 1) − 2(3x + 1) = (3x + 1)(x2 − 2). Factor out −2, not +2, so the bracket matches. Check: (3x+1)(x²−2) = 3x³−6x+x²−2 ✓.
🎯 Quick check
Six questions to lock it in. Tap the answer you think is right.
§ For teachers and parents
Factoring by grouping develops A-SSE.A.2 (use the structure of an expression to rewrite it) and rests on A-APR.A.1 (polynomial arithmetic), giving students a method for four-term polynomials and a bridge to factoring general quadratics ax² + bx + c by the "split the middle term" technique. The single most common error is not factoring the second group so the bracket matches in sign: in 2x³ − 6x² + x − 3 a student must deliberately write +1(x − 3) rather than expect the bracket "for free," and in a polynomial like 3x³ + x² − 6x − 2 must factor out −2 (not +2) from the second pair so that both brackets read (3x + 1). A close cousin is a sign slip when the third term is negative. The antidote is mechanical and reliable: always factor a deliberate +1 or −1 from each pair, insist that the two brackets be identical sign-for-sign before pulling one out, and finish by multiplying the factored form back out to confirm it reproduces the original four terms.
eastmath.com · Stage 8 · 8.5 Factoring by Grouping · Intuition before notation
eastmath.com · 8.5 Factoring by Grouping · 8.5.5 Grouping into a difference of squares