Multiplying two two-term expressions, and where the four terms come from

0:00Multiply two brackets, each holding two terms, and four things come out. Not three. Not two. Most people learn that as a rule, with a name and crossing arrows. It is not a rule. It is one rectangle, cut twice. Here is a rectangle a plus u tall and b plus v wide. Cut it across at the height a, and cut it down at the width b. Two cuts. Four pieces. Nowhere for a fifth to hide, and no way for one of the four to go missing.

0:33Everything here is that sentence, said carefully. Here is the argument, and it takes two steps. The first step treats the whole first bracket as if it were one thing, and breaks the second bracket apart. So a plus u, times b plus v, becomes a plus u times b, plus a plus u times v. That is just multiplying a sum: split it into parts and multiply each. Now do it again to each half. The first gives a b plus u b. The second gives a v plus u v.

1:11Four products, and every one of them is a term from the first bracket against a term from the second. The bracket was never opened by a special rule. It was opened twice by the same one. Now read the four off the picture and check them against the four off the algebra. The big block is a by b. The tall one is a by v. The wide one is u by b. The corner is u by v.

1:41Four blocks, four products. Every term of the first bracket has met every term of the second, exactly once. That is the reason there is no fifth piece: you cannot invent a pair that is not already on the list. And none of the four is optional. Drop the corner, substitute numbers, and the two sides come apart. So counting pieces is not a memory trick. It is a count of pairings.

2:12What if one of the brackets has a minus in it? Nothing changes. A minus is not a new operation; it is a term carrying a negative number. So a plus u, times b minus v, is a b minus a v plus b u minus u v. The same four pairs. The signs came out of ordinary sign rules, unaided. Put the minus on the other bracket and you get a b plus a v minus b u minus u v. On both, and you get a b minus a v minus b u plus u v.

2:49Three answers, one procedure, and no extra rule anywhere. Every one of those was checked by substituting whole numbers, positive, negative and zero, into both sides. Now look back at that argument and notice what it never did. It never counted how many terms a bracket holds. So it does not stop working when one holds three. Two terms against three gives six products. Three against three gives nine. In general you get one product per pair, and the number of pairs is the two counts multiplied.

3:28A seventeen-term bracket against a two-term one gives thirty-four products, and nobody would want to draw that rectangle. You do not have to. The argument is the same one, and it was never about the drawing. Here is a worked one with a fraction and a letter out in front. Multiply three a over two by the bracket a minus b plus one fifth. One term against three, so three products. Form all three first, tidy afterwards.

4:02Three a over two times a is three halves of a times a, which is three halves a squared. Three a over two times b is three halves a b, subtracted. Three a over two times one fifth is three tenths a. So the answer is three halves a squared, minus three halves a b, plus three tenths a. Three terms. Can any of them be put together? Look at the first and the last. Three halves a squared, and three tenths a.

4:35They both involve a. They are not the same thing. One measures a times a, the other measures a. You cannot add a length to an area. Two terms add only when they are built from exactly the same letters raised to exactly the same powers. Those are like terms. The number in front has nothing to do with it. Two a b and five a b are like terms; two a b and two a squared b are not.

5:04In this expansion, no two of the three are alike, so three products stay three terms and the work is finished. Now square a two-term expression. Multiply a plus b by a plus b. Two against two, so four products, as before. Nothing about squaring is special. The four are a times a, a times b, b times a, and b times b. But a times b and b times a are the same term.

5:36So collecting is not a formality this time. It merges two of the four, and four products become three terms. The answer is a squared, plus two a b, plus b squared, and that two is a count of how many products landed on the same block. That is why the middle term is doubled, and it is worth having rather than remembering. Take that answer and multiply it by a plus b again.

6:06Two terms against three, so six products, and this time four of them collide in pairs. Forming them first: a cubed, two a squared b, a b squared, a squared b, two a b squared, b cubed. Now collect. One a squared b and two more make three. One a b squared and two more make three. Six products, four terms: a cubed, plus three a squared b, plus three a b squared, plus b cubed.

6:37Those threes were not chosen. They are counts of collisions the multiplication had already decided. Substituting numbers confirms it is a plus b, cubed. Here is a place you have been doing this since you were small. Write a two-digit number as ten a plus b, and another as ten c plus d. Multiply the brackets and the four products are one hundred a c, ten a d, ten b c, and b d.

7:08Those are the partial products of long multiplication, and the hundreds and tens in front are place value doing what the letters said. Take twenty-three times twenty-seven. The four pieces are four hundred, one hundred and forty, sixty, and twenty-one. Add them and you get six hundred and twenty-one. This was checked against ordinary multiplication for all eight thousand one hundred two-digit pairs, and agreed every time. This idea is much older than the notation.

7:46It was in use, without ever being stated, in Egyptian arithmetic, in Mesopotamian tablets, in Greek geometry, in Chinese calculation and in Indian algebra. Euclid uses it as areas, Aryabhata algebraically. Both use it constantly; neither writes it down as a rule. The first person known to have stated it outright was Brahmagupta, in the seventh century. He called it khanda-gunanam, which means multiplication by parts. Break the multiplier into parts that add back to it, multiply the other number by each part, and add the results.

8:30For two parts, that sentence is exactly the rule this whole video has been using. One last thing, and it is the best argument for forming products first. Multiply a minus b by a plus b. Four products, two cancel, and what is left is a squared minus b squared. Now by a squared plus a b plus b squared. Six products, four cancel, and you get a cubed minus b cubed.

9:01Again with a four-term bracket: eight products, six cancel, a to the fourth minus b to the fourth. You can see the next one before doing it. A five-term bracket should give a to the fifth minus b to the fifth. Expand it and that is what happens: ten products, eight of them cancel, two survive. The pattern was never in the answers. It was in the pairings, and the pairings were visible from the start.

9:32Which is the whole point. Collecting is the last step, not the first, and almost everything interesting happens before it.