PrepShorts · Study sheet · Class 8 Mathematics · Chapter 6, We Distribute, Yet Things Multiply
Chapter 6 · We Distribute, Yet Things Multiply
Why the product of a sum and the matching difference is a² − b², and the patterns that follow
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A sum times the matching difference gives four products, like any two brackets. It only looks like two because two of them cancel.
The idea
A sum times its matching difference has four products like any other pair of brackets, but two of them are the same rectangle counted once as added and once as removed, so they annihilate and only two squares survive. That is why the answer is short, and the area picture says the same thing without algebra: cut a strip off a square and slide it, and a square becomes a rectangle. The real payoff is that the identity runs backwards. Reading a² − b² as a product turns a hard square into two easy multiplications and turns two numerical patterns — which look like coincidences on the page — into theorems, and it is the expansion, not the list of instances, that does the turning.
What you should be able to do
- Observe a numerical pattern in doubled sums of squares and identify the two numbers hidden in each result
- Prove that twice a sum of two squares equals the square of the sum plus the square of the difference, by adding Identities 1A and 1B
- Read a list of numerical instances as a conjecture, state the conjecture in letters, and test it by expanding
- Expand
(a + b)(a − b)and explain which two terms cancel and why - Give an area argument for the same identity by cutting a strip from a square and moving it
- Use Identity 1C forwards, to compute a product of two numbers straddling a round number
- Use Sridharacharya's rearrangement of the identity to square a number by multiplying two easier numbers
- Choose the shift that makes one of the two factors round
- Decide whether the two patterns depend on the numbers being counting numbers, and justify the decision
- Derive consequences: the square of the middle of three consecutive numbers against the product of its neighbours, and writing a given number as a difference of two squares
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| Identity 1C | the chapter's label for a sum times the matching difference giving a difference of squares | printed in this chapter (Part I, §6.2, p.148) |
| identity | an equality of two expressions that holds for every substitution | printed in this chapter (Part I, §6.1, p.139) |
| like terms | terms built from exactly the same letters to the same powers | printed in this chapter (Part I, §6.1, p.141) |
| natural number | one of the numbers used for counting, starting at one | printed in this chapter (Part I, §6.2, p.148) |
| counting number | the chapter's alternative wording for the same thing | printed in this chapter (Part I, §6.2, p.149) |
| difference of two squares | one square subtracted from another | printed in this chapter (Part I, §6.2, p.149) |
| distributive property | the rule that multiplying a sum is the same as multiplying each part and adding | printed in this chapter (Part I, §6.1, p.137) |
| sidelength | the length of a side of a square or rectangle, written by the chapter as one word | printed in this chapter (Part I, §6.2, p.145) |
| cross term | one of the two products formed from one term of each bracket | an added phrasing; the chapter writes the two products out and names them only as terms |
| straddling pair | two numbers the same distance either side of a chosen middle number | an added phrasing; the chapter sets such pairs without naming the arrangement |
Where people slip up
- "
a² − b²is(a − b)²." The commonest confusion in the chapter, and the reason the previous topic comes first. One is a product of two different brackets, the other a square. Substitute 5 and 3 and get 16 against 4. - "A sum times a difference should have four terms." It does have four, before collecting. Two of them are one rectangle added and the same rectangle removed. Show them cancelling rather than asserting that they do.
- "Four instances make a proof." Pattern 2 is printed as four lines that all work, and the chapter's very next move is to ask whether it is a true identity — because four lines cannot tell you. This is the most transferable idea in the section.
- "
(a − b)²and(b − a)²are different." Squaring destroys the sign, so neither is greater. Students expect a size question and the answer is that the question dissolves. Part I p.149 no. 1 is built on this. - "45 times 55 has to be multiplied out." Both sit five away from 50. The skill is spotting the middle, not remembering the identity.
- "43 times 45 is not a sum-and-difference pair." It is, around 44 — and unlike 98 and 102 nothing on the page tells you so.
- "The patterns are facts about whole numbers." The proof is algebra and never mentions what kind of numbers the letters are, which is exactly what Part I p.149 no. 4 is asking the student to notice.
- "Sridharacharya's method needs a shift of one." Choose the shift so that one factor lands on a round number — the chapter's own second example takes a shift of 3 to reach 200.
- "Cutting and moving a piece could change the area." It cannot, and saying so out loud is worth a sentence: the area argument and the algebra are two accounts of one fact, not two facts that happen to agree.
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Worked answers to this chapter’s exercises · this video explains Figure it Out · 6.2 Q1, Figure it Out · 6.2 Q2, Figure it Out · 6.2 Q3, Figure it Out · 6.2 Q4, Figure it Out · 6.4 Q1, Figure it Out · 6.4 Q7, Figure it Out · 6.4 Q9
Transcript1,335 words
Two squared plus one squared, doubled, comes to ten. And ten is three squared plus one squared. Three squared plus one squared, doubled, comes to twenty. Twenty is four squared plus two squared. Six squared plus five squared, doubled, comes to a hundred and twenty-two. Eleven squared plus one squared. Five squared plus three squared, doubled, comes to sixty-eight. Eight squared plus two squared. Four lines, and every one of them works.
Double a total of two squares and you seem to get another total of two squares. So where do the new numbers come from? Put the pair going in beside the pair coming out. Two and one go in. Three and one come out. Three is two plus one. One is two minus one. Three and one go in, four and two come out. Four is three plus one, two is three minus one.
Six and five give eleven and one. Five and three give eight and two. Every time, the two new numbers are the total of the pair and the difference of the pair. That is a conjecture, and four lines cannot tell you whether it is true. So write it in letters and settle it. Twice a squared plus b squared, we think, should be a plus b squared, plus a minus b squared.
You already have both of those. The square of the total is a squared, plus two a b, plus b squared. The square of the difference is a squared, minus two a b, plus b squared. Add them line by line. Two a squared. Two b squared. And two a b together with minus two a b, which is nothing at all. The middle terms annihilate and what is left is exactly twice the total of the squares. The pattern was never a coincidence. It is those two identities, added.
Here is a second pattern, and this time each line starts as a difference. Nine times nine, less one times one, is eighty. And eighty is ten times eight. Eight times eight, less six times six, is twenty-eight. And twenty-eight is fourteen times two. Seven times seven, less two times two, is forty-five, which is nine times five. Ten times ten, less four times four, is eighty-four, which is fourteen times six.
Every time, the two factors are the total of the pair and the difference of the pair. The same two numbers as before. Four lines again. And four lines are still not a proof. So expand it. A plus b, times a minus b. Four products, exactly as any two brackets of two terms give. Not two. A times a is a squared. A times minus b is minus a b.
B times a is plus a b. B times minus b is minus b squared. Now look at the middle two. Minus a b, and plus a b. That is one rectangle, taken away once and added once. It annihilates, and the two squares are all that survive. So the answer is short, and not because two terms went missing. Two of them cancelled, and you can watch them do it.
There is a way to see the same thing without any algebra. Start with a square of side a, and cut a square of side b out of one corner. What is left is an L, and its area is a squared minus b squared. Cut the L into two pieces. A band across the top, a wide and a minus b deep. A strip beside the hole, a minus b wide and b deep.
Now turn the strip a quarter turn, and stand it against the end of the band. The two pieces make a rectangle, a plus b across and a minus b down. Nothing was added and nothing was thrown away, so the L and the rectangle have the same area. The picture and the algebra are two accounts of one fact, not two facts that happen to agree. Now use it. Ninety-eight times a hundred and two.
Both of them sit two away from a hundred. So the product is a hundred squared less two squared, which is nine thousand nine hundred and ninety-six. Nothing was multiplied out. Forty-five times fifty-five. Both are five away from fifty. Fifty squared less five squared. Two thousand four hundred and seventy-five. Forty-three times forty-five looks like nothing special, until you notice that both are one away from forty-four. Forty-four squared, less one. One thousand nine hundred and thirty-five. The skill was never the identity. It is spotting the middle.
Now run the identity the other way. A squared minus b squared is a plus b, times a minus b. Move the b squared across. A squared is a plus b, times a minus b, plus b squared. Read that as a recipe for squaring. Shift the number up by some amount and down by the same amount, multiply, and add the square of the shift. Thirty-one squared. Shift by one: thirty-two times thirty, plus one, is nine hundred and sixty-one.
A hundred and ninety-seven squared. Shift by three: two hundred times a hundred and ninety-four, plus nine, is thirty-eight thousand eight hundred and nine. That rearrangement is Sridharacharya's, and it is nothing but the identity with one term moved across. The shift is a choice, and choosing it well is the whole method. A hundred and forty-five. Shift by five and one factor lands on a hundred and fifty. A hundred and fifty times a hundred and forty, plus twenty-five, is twenty-one thousand and twenty-five.
A thousand and ninety-seven. Shift by three and one factor lands on eleven hundred. Eleven hundred times a thousand and ninety-four, plus nine, is one million two hundred and three thousand four hundred and nine. Four hundred and six. Shift by six and a factor lands on four hundred exactly: four hundred and twelve times four hundred, plus thirty-six. Any shift at all gives the right answer. Only one of them saves you the work.
Here is a question that looks like it is about size. Which is greater: a minus b, squared, or b minus a, squared? Expand the first one. A squared, minus two a b, plus b squared. Expand the second one. B squared, minus two a b, plus a squared. Those are the same three terms. So neither is greater. They are equal, always, for every a and every b. The two brackets are each other's negative, and squaring destroys the sign. The question has no bigger side. It dissolves.
Do these patterns need whole numbers? Look back at what the proof actually did. It multiplied out two brackets and collected like terms. It never once said what kind of numbers the letters stood for. So take minus four and seven. The difference of their squares is minus thirty-three, and their total times their difference is three times minus eleven, which is minus thirty-three. Take three halves and one half. The difference of their squares is two, and two times one is two.
Both work, and they work because nothing in the argument ever depended on counting. That is worth noticing. A proof tells you its own range, and this one never narrowed it. One last consequence, and it is a warning. Take three numbers in a row. Square the middle one, and take away what the outer two multiply to. Four squared is sixteen. Three times five is fifteen. The gap is one.
Nine squared is eighty-one. Eight times ten is eighty. The gap is one again. Twenty triples in a row, and every single one of them gives one. But step the neighbours out to two apart and it stops. Now the gap is four. Expand it and you can see why. The gap is the square of the step, and one was only ever one because the step was one. Twenty lines that work will tell you what to conjecture. Only the expanding tells you whether it is true.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- Multiplying two two-term expressions, and where the four terms come fromClass 8 · Ch 6, We Distribute, Yet Things Multiply
- (a + b)² and (a − b)²: why there is a middle term at allClass 8 · Ch 6, We Distribute, Yet Things Multiply
Comes up again in
- Many different-looking expressions for one growing patternClass 8 · Ch 6, We Distribute, Yet Things Multiply