Run the special-product formulas backwards: a difference of squares and a perfect square.
Point 4 of 5 in this lesson: 8.3.4 Check it's truly a perfect square
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.
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.
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.
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.
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.
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).
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.
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.
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.
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. ✓
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.
Not every trinomial with square ends is a perfect square. The whole pattern lives or dies on one test: the middle term must equal 2·(√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 |
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.
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.
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.
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.
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).
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.
Work each one out first, then open the answer to check your thinking.
Six questions to lock it in. Tap the answer you think is right.
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.