One routine that picks the right method, plus what factoring unlocks next.
Point 3 of 5 in this lesson: 8.6.3 Proving divisibility and simplifying numbers
Across 8.1 through 8.5 you collected a toolbox: pull out a common factor, run the special-product formulas backward, cross-multiply a trinomial, group four terms in pairs. The trouble with a full toolbox is knowing which tool to reach for. This lesson is the decision routine that answers that question every time — and then shows you the three big payoffs that make all this factoring worth the effort. We keep the steady color habit one more time: a factor (a side length, a piece pulled out, the finished factored form) is teal; an expanded term (a product, a piece of area) is amber; a sign trap or "not done yet" warning is red; and the blue frame is the structure holding it together.
By the end you will run one ordered checklist that names the right first move on any factoring problem, factor a number to evaluate it faster than a calculator, prove that quantities are divisible without dividing, and take the very first step of canceling a fraction — the doorway to rational expressions in Stage 9.
Almost every wrong answer in factoring comes not from a hard step but from doing the steps in the wrong order — or skipping the first one. So memorize the order, not a pile of special cases. There are only four moves, and you always make them in this sequence.
Watch the routine run on 3x2−12. (1) GCF: both terms carry a 3, so factor it out → 3(x2−4). (2) Count: the bracket has 2 terms. (3) Method: x2−4 is a difference of squares, (x)2−(2)2, so it becomes (x−2)(x+2). (4) Check: neither (x−2) nor (x+2) factors further. The finished answer is 3(x−2)(x+2). Expand it back if you doubt it: 3·(x2−4) = 3x2−12. ✓
Always in this order: GCF first, then count the terms to pick the method (2 → difference of squares, 3 → perfect square / cross, 4 → grouping), then check every factor is fully factored.
Tell the machine how many terms the polynomial has and whether the terms share a common factor. It names the recommended first move — exactly what the routine above would say.
The first payoff is pure arithmetic speed. A sum is awkward to compute; a product is easy. So when a number is hiding a factorable shape, rewriting it as a product can beat a calculator. The shape to watch for is the difference of squares, a2−b2=(a−b)(a+b).
Take 992−1. Squaring 99 is a chore. But 1 is 12, so this is a difference of squares with a = 99 and b = 1: 992−1 = (99−1)(99+1) = 98·100 = 9800. Two of those factors are round numbers — the multiplication is instant. Or take 782−222 = (78−22)(78+22) = 56·100 = 5600: the "plus" factor lands on a clean 100, so again no real work.
Evaluate 512−492 without squaring anything.
Here a = 51, b = 49. Difference of squares: 512−492 = (51−49)(51+49) = 2·100 = 200. The "minus" factor is tiny (2) and the "plus" factor is round (100) — that pairing is exactly why this trick is so fast.
Pick any two numbers a and b (with a larger than b). See how factoring turns the long subtraction a2−b2 into the quick product (a−b)(a+b) — same answer, far less work.
The second payoff is that factoring turns a question about arithmetic into a question you can answer just by looking. Once an expression is written as a product, every factor is automatically a divisor — you can read off what divides it without dividing anything.
Classic example: is n2−n always even? Factor it: n(n−1). Those are two consecutive integers, and of any two integers in a row one must be even — so their product carries that factor of 2. The expression is always even, no cases to check. (The same reasoning shows n2+n = n(n+1) is always even.)
Factoring also reveals divisibility of big numbers. We already wrote 992−1 = (99+1)(99−1) = 100·98; that product is visibly divisible by 100 and by 98, with no long division. And powers behave the same way: 39−38 = 38(3−1) = 2·38 — pull out the shared 38 and the answer is plainly even, and plainly a multiple of 38.
Show that 39−38 is even.
Both terms carry 38, so pull it out: 39−38 = 38(3 − 1) = 2·38. The factor 2 is sitting right there — the number is even (in fact it equals 2 · 6561 = 13122). No exponent ever had to be expanded to see it.
The third payoff is the big one — it is the reason factoring earns a whole stage. When a fraction has a polynomial on top and bottom, you cannot simplify it as written. But factor the top and factor the bottom, and any factor they share cancels, just like reducing 6/8 to 3/4 by canceling a shared 2.
Look at x2−1x+1. The top is a difference of squares, (x+1)(x−1); the bottom is already (x+1). They share the factor (x+1), which cancels, leaving simply x−1. One factored shared piece, gone.
A slightly bigger one shows the same move: x2−9x2+6x+9. Factor each part — the top is a difference of squares, (x−3)(x+3); the bottom is a perfect square, (x+3)2 = (x+3)(x+3). One (x+3) cancels top against bottom, leaving x−3x+3.
You may cancel only a piece shared by the whole top and the whole bottom. In x2−9x+3 you may not "cancel the 9 and the 3," or strike the x2 against the x. Factor first: (x−3)(x+3) ÷ (x+3) = x − 3 — not "x2−3." A small caveat for later: canceling (x+3) quietly assumes x ≠ −3, since that would make the original bottom zero. Keep it in the back of your mind; Stage 9 makes it precise.
Step through four ready-made fractions. Each one factors top and bottom; watch the shared factor cancel and the fraction collapse to something simple. (This is the whole idea behind Stage 9.)
You now have a single decision routine and three reasons to trust it. Before you write a final answer, run this quick self-check — the same checklist whether the problem is a number or a polynomial.
The deeper habit underneath all of it: read the structure before you compute. One glance — "two terms," "a shared 3," "a perfect square on the bottom" — usually tells you the move before you write a single line. That same structural eye is exactly what makes the next stage feel natural.
One routine handles every factoring problem: GCF first, then count the terms — 2 terms means a difference of squares, 3 terms a perfect square or cross-multiplication, 4 terms grouping — then check every factor is fully factored, and confirm by expanding. Factoring then pays off three ways. It speeds up arithmetic: 992−1 = 98·100 = 9800. It proves divisibility by reading: n2−n = n(n−1) is always even, and 39−38 = 2·38. And it opens canceling: factor top and bottom, then cancel the shared factor, so x2−1x+1 = x−1 — only ever canceling a factor of the whole top and whole bottom, never a stray term.
You just took the first step of canceling. Next you meet rational expressions head-on: what they are, when they are defined, and how to reduce, multiply, and divide them — all built on the factoring you just mastered.
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 capstone lesson pulls Stage 8 together under A-SSE.B.3a (choose and produce an equivalent form of an expression to reveal its structure) and reaches forward into A-APR.D.6 (rewrite simple rational expressions in different forms), previewing the canceling work of Stage 9. It rests on the proportional-reasoning and number-structure foundations of 7.EE and A-SSE.A.2. The single most damaging misconception here is cancelling across sums instead of factors — students see (x2−9)/(x+3) and "cancel the 9 and the 3" to get a wrong "x2−3," or strike x2 against x. The antidote is a one-line rule, drilled in Section 8.6.4 and the quiz: you may cancel only a factor shared by the WHOLE top and the WHOLE bottom — so factor first, then cancel. A close second misconception is forgetting Step 1 of the routine (pulling out the GCF) and so producing an only-partly-factored answer; the cure is the four-step checklist and the unbreakable habit of expanding the answer back to confirm it equals the original.