The square identity, drawn: cut a small b² from the corner of an a×a square, slide the leftover strip, and the L-shape becomes a clean (a+b) by (a−b) rectangle. So a² − b² = (a+b)(a−b) — no algebra required, just scissors.
In Lesson 8.2 you learned the first move of every factoring problem: pull out the piece every term carries. That move alone rarely finishes the job. This lesson hands you three patterns — shapes a polynomial can fall into — that let you factor on sight. They are the multiplication formulas you already met, simply read backwards: where multiplying turned (a+b)(a−b) into a²−b², factoring takes the a²−b² and hands you back the two sides. We keep the steady color habit of Stage 8: a factor (a side length, the answer you are building) is teal, an expanded term (a product, a piece of area) is amber, a sign trap or "not yet finished" warning is red, and the blue frame is the structural skeleton.
By the end you will recognize a difference of squares and factor it instantly, see through disguises where the squares are hidden inside coefficients and higher powers, spot and factor a perfect-square trinomial, run the quick 2ab test to be sure a square is genuine, and — most important — remember to pull out the common factor before you go pattern-hunting.
8.3.1 Difference of squares
The first formula is the most useful single line in all of factoring: a² − b² = (a + b)(a − b). In words, a square minus a square always splits into the sum of the roots times the difference of the roots. The hero figure above is the whole proof: a big a×a square with a small b×b square cut from the corner has area a² − b²; slide the leftover L into a rectangle and you measure that same area as (a+b) across by (a−b) tall.
You can also just multiply the right side out to check: (a+b)(a−b) = a² − ab + ab − b² = a² − b². The two middle terms −ab and +ab are opposites, so they erase each other — that cancellation is exactly why a difference of squares has no middle term, and why a sum or difference with a middle term is not this pattern.
Multiply (a+b)(a−b) and the middle terms −ab and +ab destroy each other, leaving the bare a² − b². Read that line right-to-left and you have factored.
Key idea
A difference of squares factors as the sum of the roots times the difference of the roots: a² − b² = (a + b)(a − b). Name a and b out loud, and you are done.
🎮 Try itCut the square, watch it become a rectangle
Set a and b (with a larger than b). The teal region is the leftover area a²−b²; the readout shows it rearranged into the (a+b)(a−b) rectangle and checks the numbers.
a (big side)
b (small square)
8.3.2 Seeing a disguised difference of squares
Real problems rarely arrive wearing the letters a and b. The skill is to rewrite each term as a square first, then read off what a and b are. A coefficient like 4 is the square of 2; a power like x⁴ is the square of x²; the number 9 is 3². Once both terms are dressed as squares, the formula does the rest.
Take 4x² − 9. Rewrite it as (2x)² − 3². Now a = 2x and b = 3, so it factors as (2x − 3)(2x + 3). Or 25 − x² = 5² − x² = (5 − x)(5 + x). The pattern does not care which term comes first, only that one square is subtracted from another.
Worked example — factor twice
Factor x⁴ − 16. First, x⁴ − 16 = (x²)² − 4² = (x² − 4)(x² + 4). But you are not finished: the first factor x² − 4 is itself a difference of squares, x² − 2² = (x − 2)(x + 2). The second factor x² + 4 is a sum of squares and will not factor over the reals. So the full answer is (x − 2)(x + 2)(x² + 4).
Keep cutting until nothing splits further: x⁴ − 16 → (x²−4)(x²+4) → (x−2)(x+2)(x²+4). The sum of squares x²+4 is a dead end over the real numbers.
A sum of squares does not factor
a² + b² has no real factorization. There is no pair of real numbers whose squares add and give a middle term of zero, so x² + 4, x² + 9, and x² + 1 are already as factored as they get. Only a minus sign between the squares opens the door.
8.3.3 Perfect-square trinomials
The second and third formulas run together. Squaring a binomial gives three terms:
a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)².
The outer two terms are perfect squares; the middle term is twice the product of the two roots, and its sign tells you whether the binomial is a sum or a difference.
The picture is a square of side a + b, chopped into four tiles: a big a² corner, a small b² corner, and two identical a·b strips. Those two strips together are the 2ab middle term — there are two of them, which is why the 2 is there.
A square of side a+b holds an a², a b², and two equal ab strips. The two strips are the 2ab middle term — that is where the 2 comes from.
Reading the formula backwards: if a trinomial has two perfect-square ends and a middle term equal to twice their roots' product, it collapses into a single squared binomial. x² + 6x + 9 = x² + 2·x·3 + 3² = (x + 3)². And with a minus middle: 4x² − 12x + 9 = (2x)² − 2·(2x)·3 + 3² = (2x − 3)².
Worked example — name a and b
Factor 4x² − 12x + 9. The ends are squares: √(4x²) = 2x and √9 = 3, so try a = 2x, b = 3. Check the middle: 2ab = 2·(2x)·3 = 12x, and the given middle is −12x — magnitudes match, sign is minus, so it is a difference square: (2x − 3)². Expand to confirm: (2x−3)² = 4x² − 12x + 9. ✓
🎮 Try itBuild the (a+b)² area square
Set a and b. Watch the square split into the four tiles — the two ab strips are what make the 2ab middle term. The readout writes the trinomial and its factored square.
a
b
8.3.4 Check it's truly a perfect square
Not every trinomial with square ends is a perfect square. The whole pattern lives or dies on one test: the middle term must equal2·(√first)·(√last). If it does not, the trinomial is something else and this formula does not apply.
Look at x² + 5x + 9. The ends are squares — √(x²) = x and √9 = 3 — so it is tempting. But the test gives 2·1·3 = 6, and the actual middle is 5. Since 5 ≠ 6, this is not a perfect square; it is not (x + 3)² (which would be x² + 6x + 9). Always run the 2ab check before you write a squared binomial.
Trinomial
√first · √last
2·(√first)·(√last)
actual middle
verdict
x² + 6x + 9
x · 3
6x
6x
= → (x+3)²
4x² − 12x + 9
2x · 3
12x
−12x
= → (2x−3)²
x² + 5x + 9
x · 3
6x
5x
≠ → not a square
The square-ends trap
Two perfect-square ends are not enough. You must also confirm the middle term is exactly twice the product of the roots. Skip that check and you will "factor" x² + 5x + 9 as (x + 3)² — which is simply wrong.
🎮 Try itThe perfect-square tester
Set the three coefficients of Ax² + Bx + C (with A and C perfect squares). The tester computes 2·√A·√C and tells you whether B matches — and if so, the squared binomial.
A (= a²)
B (middle)
C (= c²)
8.3.5 Common factor first, then the formula
Here is the habit that ties Stage 8 together: pull out the common factor before you reach for any pattern. A polynomial can hide a clean difference of squares behind a shared number, and if you skip the GCF you will either miss the pattern or stop short of a full factorization.
Watch 2a² − 2b². The naked eye sees a difference, but the roots are not whole until you pull the 2: 2a² − 2b² = 2(a² − b²) = 2(a − b)(a + b). Likewise 3x² − 12 = 3(x² − 4) = 3(x − 2)(x + 2), and x³ − x = x(x² − 1) = x(x − 1)(x + 1). In each case the GCF comes off first, and then the leftover bracket is a textbook difference of squares.
Key idea — order of operations for factoring
GCF first, formula second. Pull out every shared factor, look at what remains, and only then match it to a difference of squares or a perfect square. The pulled-out factor stays in the final answer — do not lose it.
Two clean steps: strip the 3, then split the x² − 4. Skip the first step and the difference of squares never shows itself cleanly.
★ The big ideas, in one breath
Three patterns let you factor on sight. A difference of squares is the sum of the roots times their difference: a² − b² = (a + b)(a − b) — and you may have to dress each term as a square first (4x² − 9 = (2x−3)(2x+3)) or split twice (x⁴ − 16 = (x−2)(x+2)(x²+4)). A sum of squares like x² + 4 does not factor over the reals. A perfect-square trinomial collapses to one squared binomial — a² ± 2ab + b² = (a ± b)², so x² + 6x + 9 = (x+3)² — but only if the middle term really equals 2·(√first)·(√last), which is why x² + 5x + 9 is not a square. And always pull the common factor first: 3x² − 12 = 3(x−2)(x+2).
Coming up next — 8.4
Not every trinomial is a perfect square. Next you will learn cross-multiplication for the general quadratic x² + px + q: find two numbers that add to the middle and multiply to the end, and the cross diagram that checks both at once.
✎ Exercises 8.3
Work each one out first, then open the answer to check your thinking.
Factor x² − 16.
Show answer
A difference of squares with a = x, b = 4: (x − 4)(x + 4). Check: x² + 4x − 4x − 16 = x² − 16. ✓
Factor 9x² − 25.
Show answer
Dress as squares: (3x)² − 5², so a = 3x, b = 5 → (3x − 5)(3x + 5). Check: 9x² + 15x − 15x − 25 = 9x² − 25. ✓
Factor x² + 10x + 25.
Show answer
Ends are x² and 5²; middle test 2·x·5 = 10x ✓ — a perfect square: (x + 5)². Check: (x+5)² = x² + 10x + 25. ✓
Factor x² − 8x + 16.
Show answer
Ends x² and 4², middle 2·x·4 = 8x with a minus → (x − 4)². Check: (x−4)² = x² − 8x + 16. ✓
Factor 49 − y².
Show answer
7² − y² with a = 7, b = y → (7 − y)(7 + y). Check: 49 + 7y − 7y − y² = 49 − y². ✓
Is x² + 4 factorable over the real numbers?
Show answer
No. It is a sum of squares, and a sum of squares has no real factorization. Only a difference (a minus sign) splits.
Fully factor 2x² − 18.
Show answer
GCF first: 2(x² − 9), then a difference of squares → 2(x − 3)(x + 3). Check: 2(x²−9) = 2x² − 18. ✓
(x²)² − 1² = (x² − 1)(x² + 1); the first factor splits again into (x − 1)(x + 1), the second is a sum of squares. Answer: (x − 1)(x + 1)(x² + 1). Check: (x−1)(x+1) = x²−1, times (x²+1) = x⁴ − 1. ✓
Six questions to lock it in. Tap the answer you think is right.
§ For teachers and parents
This lesson develops A-SSE.A.2 (use the structure of an expression to identify ways to rewrite it) and A-SSE.B.3a (factor a quadratic to reveal its structure), built on the special-product identities students met when multiplying polynomials in Stage 7. The difference of squares and the perfect-square trinomials are simply those identities read right-to-left. Three misconceptions dominate. First, students try to "factor" a sum of squares such as x² + 9 — emphasize that only a difference splits over the reals. Second, they label x² + 5x + 9 a perfect square just because its ends are squares — the antidote is the explicit 2ab test: the middle term must equal twice the product of the roots, here 2·1·3 = 6 ≠ 5. Third, they stop at x⁴ − 16 = (x²−4)(x²+4) without re-factoring x²−4 — insist on "factor all the way down." The single best habit, reinforced in Section 8.3.5, is to name a and b out loud and then expand the answer back to confirm it equals the original.
eastmath.com · Stage 8 · 8.3 Factoring with the Formulas · Intuition before notation
eastmath.com · 8.3 Factoring with the Multiplication Formulas · 8.3.2 Seeing a disguised difference of squares