You have the distributive rule locked in: a(b + c) = ab + ac. So at some point, staring at a(bc), a very tempting thought shows up: if I can spread a multiplier across a sum, surely I can spread it across a product too? And you reach for a(bc) = (ab)(ac). This article is about why that move is always wrong — and what the right rule for a(bc) actually looks like.

The temptation, in one line

The familiar law is

a(b + c) = ab + ac.

The wrong generalisation is

a(b \times c) \;\stackrel{?}{=}\; (ab)(ac).

Let's test it on numbers. Set a = 2, b = 3, c = 4.

The right side is 48, exactly twice the left. So the two are not equal. Why: the right-hand side applies the multiplier a twice — once to b and once to c — and then multiplies the results. But applying a "twice" by multiplication means multiplying by a \times a = a^2, not by a. That extra factor of a is the gap between the two sides.

What actually happens when you "spread" over multiplication

If you stare at the failed rule a moment longer, the actual pattern emerges.

(ab)(ac) \;=\; a \cdot b \cdot a \cdot c \;=\; a^2 \cdot bc \;=\; a \cdot [a (bc)].

So (ab)(ac) equals a^2 \cdot bc, not a \cdot bc. The right-hand side has one extra copy of a that is not supposed to be there. That is the exact thing distributivity over multiplication would have to do — but it isn't a thing, and this is why.

The right rule for a(bc)

What does work for a \times (b \times c) is the associative law, not distributivity.

a \times (b \times c) \;=\; (a \times b) \times c \;=\; abc

You do not spread a across both b and c. You slide the brackets around so you can combine the numbers pairwise. a and b combine into the single product ab; then that is multiplied by c. The original a appears once, not twice.

This is a big enough distinction that it is worth separating the two rules in your head:

Expression Rule that applies Result
a(b + c) Distributivity ab + ac — two products, added
a(b \cdot c) Associativity abc — one product of three factors

They look similar on the page, but they are two different rules answering two different questions.

Why distributivity only goes one way

Distributivity is a link between two operations: multiplication and addition. It says the multiplication can be pushed across the addition. Nothing in arithmetic says one operation can be pushed across itself.

Think about the geometric meaning. The distributive rule is the picture of a rectangle of height a and width b + c being cut into two smaller rectangles of widths b and c. The total area is either a(b + c) or ab + ac — same area, two descriptions.

Distributive rule and associative rule as two different picturesOn the left, a rectangle of height a and width b plus c is split by a vertical line into two pieces: a by b, and a by c. The areas are labelled a b and a c, and their sum equals a times the quantity b plus c. On the right, a three-dimensional rectangular box of side lengths a, b, and c is shown, with the volume a b c. An arrow between the two pictures is crossed out, indicating that these are different structures and you cannot swap distributive with associative. a b a c b c a a(b + c) = ab + ac distributive — a SUM splits abc a(bc) = abc associative — a PRODUCT chains
Left: distributivity. A rectangle whose width is a sum $b + c$ splits into two smaller rectangles. Right: associativity. A box whose side lengths are $a$, $b$, $c$ has one volume $abc$ — no splitting, just a rearrangement of which pair you multiply first. The red $\times$ between them says: these are two different rules for two different structures, and you cannot swap one for the other.

But what would the analogous picture be for a(bc)? It would have to be a rectangle of height a and width b \times c — and the area of that rectangle is just abc, a single product, not two. There is no way to cut it into two pieces whose areas are ab and ac, because ab + ac is not equal to abc in general. The geometry has no splitting to do.

The \log connection (for later)

When you meet logarithms later, you will see a rule that does look like distributivity over multiplication:

\log(bc) \;=\; \log b + \log c.

That is not distributivity of \log over \cdot. It is the logarithm converting multiplication into addition — a completely different mechanism, valid because logs are the inverse of exponentials. The superficial resemblance to distributivity can be misleading. You cannot run the same trick with an arbitrary multiplier: a(bc) is abc, not ab + ac.

What each operation actually respects

Once you separate the ideas, the table becomes clean.

Every law has a direction and a pair of operations it applies to. Distributivity specifically connects one multiplication to one addition. Nothing in it says anything about two multiplications.

The exam-room reflex

When you face an expression like 2(3x \cdot 5y), do not distribute. Associate:

2 \cdot (3x \cdot 5y) \;=\; 2 \cdot 3 \cdot 5 \cdot x \cdot y \;=\; 30xy.

You rearranged the factors using commutativity and associativity, and multiplied the constants. There is no "spreading" because there is no addition to spread over.

If the expression had been 2(3x + 5y) — with a plus sign — then distributivity fires:

2(3x + 5y) \;=\; 6x + 10y.

The difference between these two lines is a single symbol. The + sign is the only thing that licences distribution. Without it, you are in associative-and-commutative-only territory.

Related: Operations and Properties · Commutative, Associative, Distributive — Three Laws as Three Rearrangements · Algebraic Identities · Exponents and Powers