We Distribute, Yet Things Multiply

Why This Matters

Here is a small puzzle. What is 23 × 27?

Most people reach for the long-multiplication method. But there is a faster way. Notice that 23 × 27 is close to 25 × 25. If you know a little algebra, you can write 23 × 27 as (25 − 2)(25 + 2) = 25² − 2² = 625 − 4 = 621. Done in your head, in seconds.

That trick is not magic. It comes from one simple, powerful idea called distributivity. It is the rule that lets you break a big multiplication into smaller, easy pieces.

You use this idea every day without naming it. When a shop sells 7 notebooks at ₹12 each, you might think “7 tens are 70, 7 twos are 14, so ₹84.” You just split 12 into 10 + 2 and multiplied each part. That is the distributive law.

In this chapter, we turn that everyday move into a tool. We will write it in the language of algebra, prove why it works with pictures, and use it to build the famous identities — like (a + b)² = a² + 2ab + b² — that make hard sums easy. By the end, you will expand brackets with confidence, and you will be able to run the whole thing backwards too. That backwards move is called factorising, and it is one of the most useful skills in all of algebra.

The Big Idea

Multiplication and addition are linked by one rule: to multiply a sum, you can multiply each part separately and then add. In symbols, a(b + c) = ab + ac. This is the distributive law. Everything in this chapter grows from it. When you multiply two brackets, you are just using this rule twice. The special identities — the square of a sum, the square of a difference, the difference of two squares — are simply this rule applied to neat, common cases. And because the rule works both ways, you can also run it in reverse: pulling a common factor back out is called factorising. One idea, many uses.

Let’s Break It Down

We will go step by step. First, the distributive law itself, with a picture that shows why it is true. Then we multiply two brackets (binomials). Then we meet the three star identities, each proved with its own diagram. Finally, we learn factorising — the reverse of expanding.

Before we start, let us quickly refresh the basic algebra words this chapter leans on, so nothing trips you up.

The distributive law: a(b + c) = ab + ac

Let us start with the heart of the chapter. The distributive law says:

a(b + c) = ab + ac

In words: multiplying a by the sum (b + c) gives the same answer as multiplying a by b, multiplying a by c, and adding the two results.

Why is this true? The cleanest way to see it is with area. Remember that the area of a rectangle is its height times its width.

The picture below shows one rectangle, counted two ways.

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Figure 6.1 makes the rule obvious: the big rectangle is the two small ones joined, so its area equals their areas added. The height a multiplies both widths, b and c.

This is not only for nice positive numbers. The law holds for any numbers — including negatives and fractions. We will rely on that freely.

Now, why does this matter so much? Because it lets you split a hard multiply into easy ones. Watch it turn 23 × 27 into something simpler.

That last point is worth a picture. When you multiply 23 by something, adding 1 to that something adds one more group of 23.

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The law also works when there are more than two terms in the bracket, and when the terms themselves have letters. For example, 3a(a − b + 5) = 3a·a − 3a·b + 3a·5 = 3a² − 3ab + 15a. You just multiply the outside term by every term inside, keeping track of the signs.

Multiplying two brackets: (a + m)(b + n)

What if both things being multiplied are sums, like (a + m)(b + n)? We simply use the distributive law twice.

Treat the first bracket (a + m) as a single block. Then distribute it over (b + n):

(a + m)(b + n) = (a + m)b + (a + m)n

Now distribute again inside each piece:

(a + m)b = ab + mb

(a + m)n = an + mn

Putting it together gives our second key result:

(a + m)(b + n) = ab + mb + an + mn

In plain words: multiply every term in the first bracket by every term in the second bracket, then add them all. Two terms times two terms gives four products.

Once again, area shows exactly why. The picture below splits a big rectangle into four parts.

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Figure 6.3 is the picture behind the words “each term times each term.” This handles subtraction too — you just write a minus as adding a negative. For example (a + 1)(b − 1) uses m = 1 and n = −1, giving ab + b − a − 1. Let us see that carefully.

A quick word on the term identity. An identity is an equation between two expressions that is true for all values of the letters, not just one special value. a(b + c) = ab + ac is an identity. So is (a + 1)(b − 1) = ab + b − a − 1. We can check an identity by expanding, or by substituting many numbers and seeing it always balances.

Let us do one more, with both terms having coefficients.

Special case 1: the square of a sum, (a + b)²

Now we meet the first star identity. What is (a + b)²? Remember (a + b)² just means (a + b)(a + b). So we use the rule we just learned, with both brackets the same.

(a + b)(a + b) = a·a + a·b + b·a + b·b = a² + ab + ab + b².

The two middle terms, ab and ba, are like terms (ba is the same as ab), so they add to 2ab. This gives:

(a + b)² = a² + 2ab + b²

Read it slowly: square the first, plus twice the product of the two, plus square the second. That middle 2ab is the part students most often forget.

Why is there a 2ab and not just ab? The area picture answers this perfectly.

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Figure 6.4 shows the secret of the missing term: there are two equal ab rectangles, one along the top-right and one along the bottom-left, so together they give 2ab. This identity makes squaring numbers easy.

Special case 2: the square of a difference, (a − b)²

Next, what is (a − b)²? We can find it the same way, or by a clever shortcut. The cleanest is to write (a − b)² = (a + (−b))² and reuse the previous identity with b replaced by −b:

(a + (−b))² = a² + 2(a)(−b) + (−b)² = a² − 2ab + b².

So:

(a − b)² = a² − 2ab + b²

It looks just like the (a + b)² identity, but the middle term is now minus 2ab. Square the first, minus twice the product, plus square the second.

There is a lovely area argument too. Start with a big square of side a, then cut away what you do not want. The picture shows it.

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Figure 6.5 explains the surprising + b² at the end: the corner gets subtracted twice, so we must add it back once. This identity makes squaring numbers a little below a round figure very easy.

Special case 3: the difference of two squares, (a + b)(a − b)

The third star identity is the most useful trick of all. What happens when you multiply a sum by the matching difference, (a + b)(a − b)?

Expand it term by term: (a + b)(a − b) = a·a − a·b + b·a − b·b = a² − ab + ab − b².

The middle terms −ab and +ab are opposites, so they cancel to 0. We are left with:

(a + b)(a − b) = a² − b²

This is called the difference of two squares. The cross-terms vanish, leaving just the two squares subtracted.

We can also see it. Take a square of side a, cut a small square of side b out of one corner, and rearrange the leftover L-shape into a neat rectangle.

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Figure 6.6 shows the L-shape becoming a rectangle of sides (a + b) and (a − b) — same area, just reshaped. This identity turns awkward products into a quick subtraction of squares.

Factorising: running the law in reverse

Everything so far has been expanding — opening brackets to spread terms out. The truly powerful move is to go the other way. Factorising means taking an expanded expression and writing it back as a product (with brackets). It is expanding in reverse.

The picture below shows the two directions side by side.

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How do you factorise? The first thing to look for is a common factor — something that appears in every term. You pull it out front.

Take ab + ac. Both terms contain a. Pull out a: ab + ac = a(b + c). That is exactly the distributive law read backwards.

Take 6x + 9. Both terms share a factor of 3 (since 6 = 3 × 2 and 9 = 3 × 3). Pull out 3: 6x + 9 = 3(2x + 3).

Let us factorise something with letters in the common part.

We can also factorise using the star identities, run in reverse. If you see a difference of two squares, you can split it instantly.

Here is a tidy summary of the three identities, read both ways, so you can keep them in one place.

Identity (expanded)	In words	Used to factorise
(a + b)² = a² + 2ab + b²	square of a sum	a² + 2ab + b² becomes (a + b)²
(a − b)² = a² − 2ab + b²	square of a difference	a² − 2ab + b² becomes (a − b)²
(a + b)(a − b) = a² − b²	difference of two squares	a² − b² becomes (a + b)(a − b)

Common Mistakes

These three slips are the ones examiners see again and again. Catch them now and you will save easy marks.

Quick Check

Try these quick questions. The explanation appears after you answer, so read it either way.

Practice Problems

Try each one yourself first. Then tap to check your full solution.

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Challenge

Summary

• The distributive law is the heart of the chapter: a(b + c) = ab + ac. Multiplying a sum equals multiplying each part and adding. The area model shows why.
• To multiply two brackets, use the law twice: (a + m)(b + n) = ab + mb + an + mn. Multiply every term by every term, then add.
• An identity is an equation that is true for all values of the letters.
• The three star identities, each provable with an area picture:
  • (a + b)² = a² + 2ab + b² (square of a sum — mind the 2ab).
  • (a − b)² = a² − 2ab + b² (square of a difference).
  • (a + b)(a − b) = a² − b² (difference of two squares — great for fast products like 98 × 102).
• These identities make hard sums easy: 65² = (60 + 5)², 91² = (100 − 9)², 53 × 47 = (50 + 3)(50 − 3).
• Factorising is expanding in reverse: it puts the brackets back. Pull out a common factor (ab + ac = a(b + c)), or spot a difference of squares (x² − 25 = (x + 5)(x − 5)).

What’s Next

You can now open brackets, build the key identities from pictures, and run the whole thing backwards by factorising. These are the basic tools of all algebra — you will use them in every later chapter.

Next, in Chapter 7 — Proportional Reasoning, you will see how quantities grow together in fixed ratios — recipes, maps, speeds and prices. The careful “multiply each part” thinking you built here will make that reasoning feel natural. Onward!