The square of a sum


The square of a sum with two terms, like $\rc{a} + \bc{b}$ squared, can be rewritten by the following formula

\[(\rc{a} + \bc{b})^2 = \rc{a}^2 + 2\rc{a}\bc{b} + \bc{b}^2 \p\]

For example

\[16 = (\rc{1} + \bc{3})^2 = \rc{1}^2 \,+\, 2\cdot\rc{1} \cdot\bc{3} \,+\, \bc{3}^2 = 1 + 6 + 9\p\]

This is an important rule that comes up a lot, so it’s good to get familiar with it.


We can derive this rule simply from the distributive property of multiplication.

Imagine that you want to buy apples for a children’s birthday party of 4 boys and 5 girls. Each child should get two apples. How many apples should you buy?

Do you add up the boys and girls first and then multiply by two, or do you multiply the number of girls by two, the number of boys by two and then add these together?

The answer is that it doesn’t matter, of course, because

\[(4 + 5) \cdot \gc{2} = 4 \cdot \gc{2}\,+\, 5 \cdot \gc{2} \p\]

We say that the multiplication by $\gc{2}$ distributes over the summation of 4 and 5. For three numbers $a$, $b$ and $\gc{x}$ he know that

\[(a + b)\gc{x} = a\gc{x}\, +\, b\gc{x} \p\]

With this knowledge, we can work out what happens if $\gc{x}$ itself is also a sum: $\gc{x} = \gc{c} + \gc{d}$

\[\begin{align*} (a + b)\gc{x} &= a\gc{x} + b\gc{x} \\ &= a\gc{(c + d)} + b\gc{(c + d)} \p\\ \end{align*}\]

Now, we can just apply distribution again on both terms of this final sum to get

\[a\gc{(c + d)} + b\gc{(c + d)} = a\gc{c} + b\gc{d} + b\gc{c} + b\gc{d} \p\\\]

How does this relate to our original problem of $(\rc{a} + \bc{b})^2$? That one is just a special case of $(\rc{a} + \bc{b})(c + d)$ because $(\rc{a} + \bc{b})^2 = (\rc{a}+ \bc{b})(\rc{a} + \bc{b})$. Following the same derivation as before, we get

\[\begin{align} (\rc{a} + \bc{b})(\rc{a} + \bc{b}) &= (\rc{a} + \bc{b})\rc{a} + (\rc{a} + \bc{b})\bc{b} \\ &= \rc{a}\rc{a} + \bc{b}\rc{a} + \rc{a}\bc{b} + \bc{b}\bc{b} \\ &= \rc{a}^2 + 2\rc{a}\bc{b} + \bc{b}^2 \p \end{align}\]


This particular formula can be illustrated very neatly with a simple diagram. Note that the reason we call the operation of raising something to the second power a square, is because $x^2$ gives us the area of a square with sides of length $x$.

That suggests that we can draw a square for $(\rc{a} + \bc{b})^2$ with sides of length $\rc{a} + \bc{b}$ and its area should sum up to $\rc{a}^2 + 2\rc{a}\bc{b} + \bc{b}^2$.

An illustration of the sum/square rule.

question We showed above that the formula for the square of a sum was a specific case of the rule

\((\rc{a} + \bc{b})(c + d) = \rc{a}c + \rc{a}d + \bc{b}c + \bc{b}d\).

That suggests that this rule could be similarly illustrated with a diagram. What would this diagram look like?


Binomial theorem

What if we have $(\rc{a} + \bc{b})^3$ or $(\rc{a} + \bc{b})^5$? We can work these out step by step with the distributive property, but it gets a bit hairy, and it becomes very easy to make a mistake. Happily, there’s a general rule called the binomial theorem which can tell us exactly how to get rid of the brackets.

If the sum itself contains more than two terms, things get even more complicated. In that case we need the Multinomial theorem.