Visual walkthrough — Factoring — common factor extraction, grouping, using identities

The parent note lists the difference of squares identity as a "memorized recipe." This page refuses to memorize. We are going to watch it happen — cut real squares out of paper, slide the pieces around, and let the picture tell us why the factorization is forced to be true. Everything is built from the meaning of one word: area.

We link this back to the parent factoring note and to the Distributive Property it secretly relies on.

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Step 1 — What does $a^2$ even mean? Draw the big square

WHAT. We start with two lengths. Call the bigger one $a$ and the smaller one $b$, where $a$ is longer than $b$ (so $a>b>0$ for now — other cases come later). We draw the square built on side $a$.

WHY. The whole identity has the symbol $a^2$ in it. Before we are allowed to touch it, we must know what it is. It is the area of a square whose side is $a$. No area, no meaning.

PICTURE. Look at the big teal square below.

$a^2=\underset{\text{side going right}}{\underbrace{a}}\times \underset{\text{side going up}}{\underbrace{a}}$

Each $a$ is one edge of the square. Multiplying the two edges counts every unit box inside — that count is the area, written $a^2$.

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Step 2 — Bite a small square out of the corner

WHAT. From the corner of the big $a$-square we remove a small square of side $b$. Its area is $b^2$. The leftover shape — a big square with a square bite taken out — has area exactly:

$a^2-b^2$

WHY. This leftover is the left-hand side of our identity. The minus sign is not abstract: it is literally the missing corner. If we can measure the area of this L-shaped leftover a second way — by chopping and rearranging — the two measurements must agree, and that agreement is the factorization.

PICTURE. The plum square is the bite; the orange L-shape is what remains.

$\underset{\text{whole big square}}{\underbrace{a^2}}-\underset{\text{corner we removed}}{\underbrace{b^2}}=\text{area of the orange L}$

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Step 3 — Slice the L-shape into two rectangles

WHAT. We make one straight cut across the L, splitting it into a tall rectangle and a short rectangle.

WHY. An L-shape is awkward — we cannot write its area as one clean product. But a rectangle's area is just $\text{width}\times \text{height}$. So we break the awkward shape into pieces we can measure with a single multiplication each.

PICTURE. The dotted line is the cut. It separates a tall piece (top) from a wide piece (bottom).

Reading the tall top rectangle: $\text{tall piece}=\underset{\text{width: full }a\text{ minus the bite }b}{\underbrace{(a-b)}}\times \underset{\text{full height}}{\underbrace{a}}$

Reading the wide bottom rectangle: $\text{wide piece}=\underset{\text{full width}}{\underbrace{a}}\times \underset{\text{height: only up to the bite}}{\underbrace{b}}$

Wait — those two do not yet share a common edge we can factor. That is why the next step re-cuts more cleverly.

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Step 4 — The clever cut: two rectangles that share the edge $(a-b)$

WHAT. Instead of the cut in Step 3, slide the small square's boundary and cut the L into two rectangles that both have width $(a-b)$. One is $(a-b)$ by $a$; the other is $(a-b)$ by $b$.

WHY. For factoring we need a shared dimension — a common factor we can pull out front, exactly like the Distributive Property pulls a common term out of a sum. If both rectangles have the same width $(a-b)$, we can glue them along that shared side and add only their heights.

PICTURE. Both orange rectangles are the same width $(a-b)$. The heights are $a$ (top) and $b$ (bottom).

$\underset{\text{top rectangle}}{\underbrace{(a-b)\cdot a}}+\underset{\text{bottom rectangle}}{\underbrace{(a-b)\cdot b}}=a^2-b^2$

Every term is a real rectangle you can point to.

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Step 5 — Glue the pieces: factor out the shared width

WHAT. Both rectangles carry the factor $(a-b)$. Pull it out front:

$a^2-b^2=(a-b)\cdot a+(a-b)\cdot b=(a-b)(a+b)$

WHY. This is common-factor extraction — the exact reverse of the distributive property. Stacking a rectangle of height $a$ on top of one of height $b$, both of width $(a-b)$, gives one rectangle of width $(a-b)$ and height $(a+b)$. Its single area is the product $(a-b)(a+b)$.

PICTURE. The two orange rectangles are stacked into one clean rectangle.

$\underset{\text{width: shared edge pulled out}}{\underbrace{(a-b)}}\times \underset{\text{height: }a\text{ stacked on }b}{\underbrace{(a+b)}}$

Same L-shape, measured two ways. The areas must be equal, so:

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Step 6 — Edge case: what if $a=b$?

WHAT. Let the small square grow until it fills the big one, so $b=a$.

WHY. A formula that only works for "nice" numbers is untrustworthy. We must check the boundary where the shapes degenerate.

PICTURE. The bite swallows the whole square — nothing is left.

$a^2-a^2=0,(a-a)(a+a)=0\cdot 2a=0✓$

The width $(a-b)$ shrinks to zero, so the stacked rectangle collapses to a line of zero area. Both sides give $0$. The identity survives the degenerate case.

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Step 7 — Sign case: what if $b>a$?

WHAT. Now let the "bite" be bigger than the square, e.g. $a=3,b=5$.

WHY. Areas were our whole picture, and areas are never negative — so does the algebra still hold when the picture can't be drawn as a positive L? We check with numbers.

PICTURE. We reinterpret: the leftover "area" becomes negative, but the algebra doesn't care about which square is bigger.

$a^2-b^2=9-25=-16$ $(a-b)(a+b)=(3-5)(3+5)=(-2)(8)=-16✓$

The factor $(a-b)$ simply turns negative, and the product stays correct. The identity is true for all real $a,b$ — the picture is a proof for $a>b>0$, and the algebra extends it everywhere.

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The one-picture summary

One glance: big square minus corner square = the L = one stacked rectangle of size $(a-b)$ by $(a+b)$. Left count and right count of the same region can never disagree — that equality is the factorization.

Recall

Feynman retelling — say it to a 12-year-old. Take a big square tile, side $a$. Its area is $a^2$. Now snap off a smaller square, side $b$, from one corner — you removed area $b^2$, so the leftover chunk is $a^2-b^2$. That leftover is an L-shape. Slice the L into two strips that are both exactly $(a-b)$ wide. Stack them: one is $a$ tall, the other $b$ tall, so together they make a single neat rectangle that is $(a-b)$ wide and $(a+b)$ tall. Its area is $(a-b)(a+b)$. But it's the same paper as the L-shape — you only cut and slid, never added or lost any. Same area, two names: $a^2-b^2=(a-b)(a+b)$. If the corner square is as big as the whole tile ($a=b$) nothing is left, and both sides are $0$. If the corner is somehow "bigger" ($b>a$), the algebra just flips the sign and stays honest.

Related tools once you have roots in factored form: Zero Product Property, and for quadratics that don't factor by inspection, Quadratic Formula and Completing the Square. Over the complex numbers even $a^2+b^2$ factors — see Complex Numbers.