We Distribute, Yet Things Multiply

How much bigger is 23 x 27 than 23 x 26? The direct way is to compute both products and subtract, but there is a faster route. Increasing only the second number by 1 increases the product by exactly the first number, 23. This is the distributive property in action: a(b + c) = ab + ac. Picture a rectangle of a rows and (b + c) columns, split into two smaller rectangles of a rows by b columns, and a rows by c columns. The whole rectangle's area, a(b+c), is obviously just the sum of the two smaller areas, ab + ac, because splitting a shape into pieces never changes its total area. This picture is not a special case for small, friendly numbers, it holds for every a, b and c, which is exactly why the distributive property can be trusted for letter-numbers too.

What if BOTH numbers in a product change? Suppose a becomes a + m and b becomes b + n. Treating (a + m) as one term and expanding step by step: (a+m)(b+n) = (a+m)b + (a+m)n = ab + mb + an + mn. This is Identity 1, and it quietly contains every increase-or-decrease case at once, since m and n can be positive, negative, or zero. Decreasing b by 1 is the same as n = -1, so (a+1)(b-1) becomes ab + b - a - 1 by simply substituting m=1, n=-1 into Identity 1, without redoing the whole expansion from scratch. This is the real payoff of an identity: work it out once in general, and every specific case becomes substitution, not new work.

Multiplying a number by 11 has a well-known shortcut: write down the number, then add each pair of neighbouring digits, carrying where needed. Why does this work? A number like dcba (with digits d, c, b, a) can be written using place value, and multiplying by 11 means multiplying by (10 + 1): dcba x 11 = dcba x 10 + dcba. Writing dcba x 10 as dcba0 and lining it up under dcba for addition is exactly long addition, and it is precisely the distributive property, 10 + 1 being split apart, that justifies why the shortcut is safe to use rather than just a coincidence. Similar shortcuts exist for multiplying by 101 (= 100 + 1), 1001 (= 1000 + 1), and so on, all justified the same way. Indian mathematicians already knew this: Brahmagupta, in his 7th-century work Brahmasphutasiddhanta, described multiplying by breaking a number into parts and adding the results, a technique he called ista-gunana, fast multiplication by parts.

A square plot of side 65 can be built from a square of side 60, a square of side 5, and two rectangles of sides 60 and 5, fitted together. Its area is therefore 60^2 + 5^2 + 2(60 x 5) = 3600 + 25 + 600 = 4225, and this matches the direct way of finding it: multiplying (60+5) by (60+5) using the distributive property term by term gives exactly the same four pieces. Writing this in general, for any a and b, (a+b)(a+b) = a x a + a x b + b x a + b x b = a^2 + 2ab + b^2, since ab and ba are the same and can be combined. This is Identity 1A, and it is not a new rule to memorise so much as the distributive property applied to a square, seen through a picture.

A square of side 55 sits snugly inside a square of side 60. Its area, 55^2, can be found by taking away the two 60-by-5 rectangles from the big square, but that removes the small 5-by-5 corner square twice over, so it must be added back once: 55^2 = 60^2 - (60x5) - (5x60) + 5^2 = 3600 - 300 - 300 + 25 = 3025. In general, (a-b)^2 = a^2 - 2ab + b^2, Identity 1B, which can also be reached by treating (a-b) as (a + (-b)) and reusing Identity 1A directly, since algebra does not care whether a letter-number happens to represent a positive or a negative value.

Look at this pattern: 9x9 - 1x1 = 10x8, and 8x8 - 6x6 = 14x2, and 7x7 - 2x2 = 9x5. The pattern seems to be a^2 - b^2 = (a+b)(a-b). Expanding the right side using the distributive property: (a+b)(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2, since -ab and +ba cancel out. This confirms Identity 1C. Sridharacharya, an Indian mathematician working around 750 CE, used a clever rearrangement of this same identity, a^2 = (a+b)(a-b) + b^2, to square numbers quickly: choosing b to make (a+b) or (a-b) land on a round number. For 31^2, choosing a=31 and b=1 gives 31^2 = (32)(30) + 1 = 960 + 1 = 961, replacing a harder square with an easier multiplication.

A growing pattern of circles can be counted in several genuinely different ways: as (k+1)^2 - 1, or as k^2 + 2k, or as k(k+1) + k, or as k(k+2), depending on how the picture is mentally split up at step k. These four expressions look nothing alike on paper, yet they must all describe the exact same count of circles, so they must all simplify to the same thing. Expanding each one out confirms it: every single one reduces to k^2 + 2k. This is a genuinely useful checkpoint for algebra: whenever two different, correctly-reasoned methods for the same real problem give expressions that look different, expanding and simplifying both must make them match, since there is only one true answer for any given step k.