Visual walkthrough — Polynomials — degree, types (monomial, binomial, trinomial)

Before we start, three plain-word promises about the symbols:

• $x$ is a number we do not yet know — a length we can slide longer or shorter. Picture it as one side of a square tile.
• $p$ and $q$ are two fixed, known numbers (like $2$ and $3$). Picture each as a short fixed strip of length $p$ or $q$.
• A term is one "word" of an algebraic sentence, e.g. $x^2$, or $5x$, or $6$. (See Algebraic Expressions.)

Everything below is one long picture of an area being cut into pieces. Follow the coloured regions.

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Step 1 — Draw the two things we are multiplying

WHAT. We want the product of two binomials: $(\underset{\text{unknown side}}{\underbrace{x}}+\underset{\text{fixed strip}}{\underbrace{p}})\times (\underset{\text{unknown side}}{\underbrace{x}}+\underset{\text{fixed strip}}{\underbrace{q}})$ The first factor is a length: $x$ plus a little extra $p$. The second factor is another length: $x$ plus a little extra $q$.

WHY start with lengths? Because multiplying two lengths is finding the area of a rectangle. If I can draw a rectangle whose width is $(x+p)$ and whose height is $(x+q)$, then its area is exactly the answer we want. We have turned an algebra question into a geometry question we can literally see.

PICTURE. The horizontal side is split into a blue part of length $x$ and an orange part of length $p$. The vertical side is split into a blue part of length $x$ and a green part of length $q$.

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Step 2 — The whole rectangle is the product

WHAT. The area of the entire rectangle equals width $\times$ height: $\text{Area}=(x+p)(x+q)$ Here $(x+p)$ is the full bottom edge and $(x+q)$ is the full left edge.

WHY is this legal? Area of any rectangle is always (one side) $\times$ (the other side) — that never changes, whatever the sides are made of. So this single big area is a true, exact copy of our product. Now all we have to do is measure the same area a second, smarter way and set the two measurements equal.

PICTURE. The whole outlined rectangle is highlighted; a bracket labels the bottom edge $x+p$ and the left edge $x+q$.

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Step 3 — Cut the rectangle into 4 smaller pieces

WHAT. The horizontal cut at height $x$ and the vertical cut at width $x$ slice the big rectangle into four smaller rectangles. Their sizes are:

Piece	Width	Height	Area
top-left (blue)	$x$	$x$	$x\cdot x=x^2$
top-right (orange)	$p$	$x$	$p\cdot x=px$
bottom-left (green)	$x$	$q$	$x\cdot q=qx$
bottom-right (red)	$p$	$q$	$p\cdot q=pq$

WHY cut here and nowhere else? Because we already split each edge at exactly the spot where the unknown $x$ ends and the known strip begins. Cutting straight across from those two split points is the only way to get pieces whose sides are each purely one thing ($x$, $p$, or $q$) — so every piece's area is a clean single term.

PICTURE. Four coloured tiles, each labelled with its dimensions and its area right inside it.

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Step 4 — Add the four pieces back up

WHAT. The four pieces together must add up to the whole (nothing was lost or added by cutting): $\underset{\text{big blue}}{\underbrace{x^2}}+\underset{\text{orange}}{\underbrace{px}}+\underset{\text{green}}{\underbrace{qx}}+\underset{\text{red}}{\underbrace{pq}}$

WHY. This is the "second measurement" promised in Step 2. Same rectangle, counted as four parts instead of one. This is exactly the FOIL Method (First $x\cdot x$, Outer $x\cdot q$, Inner $p\cdot x$, Last $p\cdot q$) — but now you can see that FOIL is just "four tiles," not a rule to memorise.

PICTURE. The same four tiles, each with an arrow pulling its area out to a running sum below.

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Step 5 — Combine the two "middle" tiles (like terms)

WHAT. The orange tile $px$ and the green tile $qx$ are like terms — both are "some number of $x$'s." We can add them: $px+qx=(p+q)x$ where $(p+q)$ is just how many $x$-strips we have in total.

WHY are we allowed? Because $px$ means "$p$ copies of length $x$" and $qx$ means "$q$ copies of length $x$"; together that is $p+q$ copies of $x$. (This is Combining Like Terms.) The blue $x^2$ tile has area units, and the red $pq$ tile is a pure number — neither is a bunch of $x$-strips, so neither can join this merge. That is why only the middle two combine.

PICTURE. The orange and green strips are slid together into one wider strip of total length $p+q$, labelled $(p+q)x$; blue and red stay put.

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Step 6 — Read off the finished trinomial

WHAT. Setting the whole area (Step 2) equal to the summed pieces (Steps 4–5): $(x+p)(x+q)=\underset{\text{degree }2}{\underbrace{x^2}}+\underset{\text{sum}}{\underbrace{(p+q)}}x+\underset{\text{product}}{\underbrace{pq}}$

WHY this is the whole point. Look at what controls the answer:

• The coefficient of the middle term $x$ is the sum $p+q$.
• The constant term (no $x$) is the product $pq$.

So the numbers $5$ and $6$ from the parent note are no mystery: with $p=2,q=3$ we get sum $2+3=5$ and product $2\cdot 3=6$, giving $x^2+5x+6$. Three tiles survived (blue $x^2$, merged middle $5x$, red $6$) — that is why the result is a trinomial.

PICTURE. The three final tiles with the equation written underneath, sum-arrow to the middle coefficient and product-arrow to the constant.

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Step 7 — Edge case: when the strips are negative ($p$ or $q<0$)

WHAT. What if $q$ is negative, e.g. $(x+2)(x-3)$? Here $p=2,q=-3$. The formula still holds: $(x+2)(x-3)=x^2+(2+(-3))x+(2)(-3)=x^2-x-6$

WHY does the picture survive going negative? A length can't be negative, so we can't draw a strip of length $-3$. Instead, a negative strip means we remove area instead of adding it. The green tile becomes a piece we subtract, and the red product $pq=-6$ is subtracted too. The algebra doesn't care about the picture's sign — it just adds a negative number.

PICTURE. Same rectangle but the $q$-strip is drawn hatched and marked "$-$" (subtracted); the running sum shows the signs.

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Step 8 — Degenerate case: when the middle tiles cancel or a strip is zero

WHAT. Two special collapses:

1. Strip is zero ($q=0$): $(x+p)(x+0)=(x+p)\cdot x=x^2+px$. The green and red tiles have zero area (a side of length $0$), so only two tiles survive → a binomial.
2. Opposite strips ($q=-p$): $(x+p)(x-p)=x^2+(p-p)x+(p)(-p)=x^2-p^2$. Here $p+q=0$, so the whole middle strip vanishes, again leaving a binomial — the famous difference of squares.

WHY show this? It proves the trinomial result is not magic: kill the sum $p+q$ and the middle term dies; kill a strip's length and a whole column dies. Every count of terms is accounted for.

PICTURE. Left panel: $q=0$ column collapsed to a line. Right panel: the $p+q=0$ middle strip greyed out, leaving $x^2-p^2$.

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The one-picture summary

One rectangle, cut into four tiles, three of which survive as a trinomial. The bottom edge is the first binomial, the left edge is the second, the four tiles are FOIL, and combining the two $x$-tiles gives the $(p+q)x$ middle term. Read the sum on the middle, the product on the corner.

Recall Feynman retelling (say it in plain words)

I wanted to multiply two two-part lengths, $(x+p)$ and $(x+q)$. Multiplying lengths is just finding the area of a rectangle, so I drew one. I split each side where the unknown $x$ stops and the fixed strip starts. That chopped the rectangle into four tiles. The big corner tile is $x\times x=x^2$. Two skinny tiles are strips of $x$'s — one has $p$ of them, one has $q$ of them — so together they're $(p+q)$ copies of $x$, i.e. $(p+q)x$. The last little corner tile is $p\times q=pq$, a plain number. Add the survivors: $x^2+(p+q)x+pq$ — a trinomial. That's why, to un-multiply (factor) $x^2+bx+c$, I hunt for two numbers that add to $b$ and multiply to $c$. If a strip is negative I subtract instead of add; if a strip is zero or the strips are opposite, the middle or a whole column dies and I'm left with a binomial.

See also: Polynomials — degree, types (monomial, binomial, trinomial) (index 2.1.13), Exponent Rules, Linear Equations, Polynomial Functions.