Visual walkthrough — Multiplication of algebraic expressions — monomial × polynomial, polynomial × polynomial

This page rebuilds the parent result Multiplication of algebraic expressions from absolute zero, one picture at a time. We lean on the Distributive Property, Area Models for Algebra, Combining Like Terms, and Exponent Rules as we go — each is introduced the moment we need it.

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Step 1 — What does "length times length" even mean?

WHAT. Before any letters, let us agree on the most basic fact: if a rectangle is $3$ units wide and $2$ units tall, its area is $3\times 2=6$ little unit squares. You can literally count the squares.

WHY. We start here because area is the machine that turns a multiplication into a picture. Every symbol we meet later — $x$, $x^2$, $ab$ — will be an area of some tile. If we anchor multiplication to area now, nothing later will be abstract.

PICTURE. Look at the grid below: the width is broken into $3$ green ticks, the height into $2$ yellow ticks, and the interior fills with $6$ blue unit squares.

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Step 2 — Letting a side become unknown: the monomial $\times$ term

WHAT. Now replace a known width by an unknown length. Call it $x$ — a stick whose length we do not yet know. A rectangle that is $x$ wide and $3$ tall has area $3x$.

WHY. We use a letter because algebra's job is to reason about lengths before we measure them. The symbol $x$ is not "a mystery number"; it is a stick of unknown length. Multiplying it by $3$ stacks that stick $3$ times — that is what $3x$ is.

PICTURE. Below, one blue stick of length $x$ sits along the top; the height $3$ makes three copies of it. The shaded slab is the area $3x$.

$\underset{\text{height: 3 copies}}{\underbrace{3}}\times \underset{\text{unknown width}}{\underbrace{x}}=\underset{\text{area of the slab}}{\underbrace{3x}}$

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Step 3 — A sum on one side: the distributive property as tiling

WHAT. Take a width that is a sum of two pieces, $(x+2)$, and a fixed height $3$. What is the area?

WHY. This is the first real theorem: the Distributive Property. We want to show why $3(x+2)=3x+6$ instead of just asserting it. The picture proves it with no algebra at all.

PICTURE. The top edge is cut into a blue part of length $x$ and a red part of length $2$. Drop a vertical line at the cut, and the big rectangle splits into two tiles: a $3\times x$ slab and a $3\times 2$ block.

$3\underset{\text{whole width}}{\underbrace{(x+2)}}=\underset{\text{blue tile}}{\underbrace{3x}}+\underset{\text{red tile}}{\underbrace{6}}$

• $3x$ — the area over the blue length $x$.
• $6$ — the area over the red length $2$, since $3\times 2=6$.

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Step 4 — Both sides are sums: the four-tile rectangle

WHAT. Now the flagship case. Let the width be $(a+b)$ and the height be $(c+d)$. Cut the width at the $a/b$ boundary and the height at the $c/d$ boundary. Two cuts across a rectangle always make four tiles.

WHY. This is where "every term times every term" comes from — and the picture is the reason. Each tile is the meeting of one width-piece with one height-piece. There are $2\times 2=4$ meetings, so there are $4$ tiles, so there are $4$ products. Nothing is chosen; the geometry forces it.

PICTURE. A grid split into four coloured tiles. Read each tile's area off its two sides.

$(a+b)(c+d)=\underset{\text{top-left}}{\underbrace{ac}}+\underset{\text{bottom-left}}{\underbrace{ad}}+\underset{\text{top-right}}{\underbrace{bc}}+\underset{\text{bottom-right}}{\underbrace{bd}}$

• $ac$ — width-piece $a$ meets height-piece $c$.
• $ad$ — width-piece $a$ meets height-piece $d$.
• $bc$ — width-piece $b$ meets height-piece $c$.
• $bd$ — width-piece $b$ meets height-piece $d$.

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Step 5 — Worked case with numbers: $(2x+3)(x-4)$

WHAT. Apply the four-tile picture to $(2x+3)(x-4)$. Width pieces: $2x$ and $3$. Height pieces: $x$ and $-4$.

WHY. We use a subtraction on one side to see how a negative length-piece behaves — its tile has negative area (it is subtracted, drawn hatched below).

PICTURE. The $-4$ height-piece is drawn hatched to mean "area to remove."

$(2x+3)(x-4)=\underset{2x\cdot x}{\underbrace{2x^2}}\underset{2x\cdot (-4)}{\underbrace{-8x}}+\underset{3\cdot x}{\underbrace{3x}}\underset{3\cdot (-4)}{\underbrace{-12}}$

Now Combining Like Terms: $-8x$ and $+3x$ live on tiles of the same shape (both are a length $\times$ a length giving an $x$-slab), so they merge:

$-8x+3x=-5x⟹(2x+3)(x-4)=2x^2-5x-12$

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Step 6 — When a side has three pieces: $(x+2)(x^2-3x+5)$

WHAT. Extend the rectangle: width $(x+2)$ has $2$ pieces, height $(x^2-3x+5)$ has $3$ pieces. Two cuts one way, three the other → a $2\times 3=6$-tile grid.

WHY. This shows the rule scales with no new idea: $m$ pieces times $n$ pieces always makes an $m\times n$ grid, hence $m\times n$ products. "Every term with every term" is just "every column meets every row."

PICTURE. A $2\times 3$ grid; each cell labelled with its product.

$$\begin{array}{cccc}& x^2& -3x& +5\\ x& x^3& -3x^2& 5x\\ 2& 2x^2& -6x& 10\end{array}$$

Collect tiles of matching shape (same power of $x$):

$x^3+\underset{-x^2}{\underbrace{(-3x^2+2x^2)}}+\underset{-x}{\underbrace{(5x-6x)}}+10=x^3-x^2-x+10$

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Step 7 — The degenerate & special cases (never leave a gap)

WHAT. Three edge cases that the tile picture handles automatically.

WHY. A derivation is only trustworthy if it survives the extreme inputs. Here they are, each as its own tile-story.

PICTURE. Three mini-rectangles: a perfect square, a difference of squares (with the cancelling strips), and the zero-piece collapse.

(a) Perfect square $(a+b{)}^2$. Same length on both sides → a square split into four tiles, but the two off-diagonal tiles are identical ($ab$ and $ba$): $(a+b{)}^2=a^2+\underset{2ab}{\underbrace{ab+ba}}+b^2=a^2+2ab+b^2$ This is why $(x+3{)}^2=x^2+6x+9$, not $x^2+9$ — the two green strips of area $3x$ are real tiles.

(b) Difference of squares $(a+b)(a-b)$. Here one height-piece is $+a$ and one is $-b$. The $+ab$ tile and the $-ab$ tile have the same size and opposite sign, so they annihilate: $(a+b)(a-b)=a^2+ab-ab-b^2=a^2-b^2$

(c) A zero piece. If one width-piece has length $0$ — say $(x+0)(c+d)$ — its whole column of tiles has zero area and simply disappears, leaving $x(c+d)=xc+xd$. The rule degrades gracefully back to Step 3.

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The one-picture summary

Every multiplication on this page is one instruction: cut the rectangle where each side is added, then add up all the tiles. Monomial $\times$ polynomial is a $1\times n$ strip; polynomial $\times$ polynomial is an $m\times n$ grid; special products are grids whose tiles happen to double up or cancel.

Recall Feynman retelling — say it to a friend with no algebra

Picture a garden plot. One side of the plot is measured in two stretches — an unknown stretch $a$ and a known stretch $b$ — and the other side in two stretches $c$ and $d$. If you stand at the corner and draw one line across and one line down at the measurement joins, the garden falls into four smaller rectangles. Each little rectangle's area is its width times its height, and the whole garden is just those four areas added together. That's it — that's $(a+b)(c+d)=ac+ad+bc+bd$. When a side has three stretches, you get six little rectangles instead of four; the recipe never changes. Two rectangles the same shape? Push them together (that's combining like terms). A stretch of length zero? Its rectangles are flat and vanish. A plus-and-minus pair the same size? One adds and one subtracts, so they cancel — that's the difference of squares. You never memorise a rule; you just count tiles.

Recall Quick self-check

Why does $(a+b)(c+d)$ give exactly four products? ::: Two cuts across a rectangle make $2\times 2=4$ tiles, and each tile is one width-piece times one height-piece. In $(x+3{)}^2$, where does the $6x$ come from geometrically? ::: From the two identical $3\times x$ strips ($3x+3x=6x$) that sit off the diagonal of the square. Why do the middle tiles in $(a+b)(a-b)$ disappear? ::: One is $+ab$ and the other is $-ab$; equal size, opposite sign, so they cancel, leaving $a^2-b^2$.

Related deep tools: Factoring Polynomials runs this whole picture backwards (given the tiles, rebuild the sides), Quadratic Expressions is what a $2\times 2$ grid produces, and Pascal's Triangle counts the tiles for higher powers like $(a+b{)}^n$.