PrepShorts · Study sheet · Class 9 Mathematics · Chapter 4, Exploring Algebraic Identities
Chapter 4 · Exploring Algebraic Identities
Algebra tiles: factorising by rebuilding the rectangle
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Find two numbers that add to the middle term and multiply to the constant. Two rules arriving from nowhere — except they are one rectangle.
The idea
The two arithmetic conditions that make factorising work — the split of the middle term must add to it, and the two parts must multiply to the constant — are not two rules handed down from outside. They are one geometric fact: the tiles have to close up into a rectangle. Put three x-tiles down one side and four along the other and the corner they leave is a 3 by 4 hole, so the constant term has no choice about being 12. Split the same 7x as 2x + 5x and the hole is 2 by 5, which twelve unit tiles cannot fill without spilling. That is why the sum condition and the product condition arrive together, and it is why one split out of several works.
What you should be able to do
- Compute the area of a rectangle whose sides are x + 3 and x + 4 both by distributivity and by counting tiles, and check the two agree
- Identify the three kinds of tile in the chapter's model and state the area of each
- Lay out a product of two linear expressions as a rectangle and read the four regions off it
- Explain why the x-tiles must divide into two groups, one along each side
- Explain why the number of unit tiles is forced once the two groups are chosen
- Show that an alternative split of the middle term fails to close the rectangle, and say exactly what goes wrong
- Read the factorisation off a completed rectangle by naming its two sides
- Generalise the tile layout to (x + a)(x + b)
- Build the tile picture when the leading coefficient is not 1, as for (2x + 3)(3x + 1)
- Complete the general product (px + a)(qx + b) and confirm it by distributivity
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| algebra tiles | a set of square and rectangular pieces whose areas stand for x², x and 1 | printed in this chapter (§4.5 heading, p. 78) |
| x-tile | the rectangular piece of area x — one unit by x units | printed in this chapter (§4.5, p. 79) |
| unit tile | the small square piece of area 1 | printed in this chapter (§4.5, p. 79) |
| linear factors | the two first-degree expressions whose product the rectangle shows | printed in this chapter (§4.5, p. 79) |
| linear expressions | first-degree expressions such as x + 3 | printed in this chapter (§4.5, p. 79) |
| array | a rectangular arrangement in rows and columns | printed in this chapter (§4.5, p. 79) |
| dimensions | the two side lengths of the assembled rectangle | printed in this chapter (§4.5, p. 79) |
| sidelengths | the lengths of the sides of the rectangle being modelled | printed in this chapter (§4.5, p. 80) |
| distributivity | the property used to multiply the brackets out as a check | printed in this chapter (§4.5, p. 78) |
| breadth | the shorter side of the rectangle, as named in Example 17 | printed in this chapter (§4.8, p. 88) |
| splitting the middle term | dividing the x term into two parts so the tiles close up | an added compound; the chapter writes about splitting the x term and about 7x being split |
| closure of the rectangle | the condition that the tiles leave no gap and no overhang | an added phrasing, not printed in this chapter |
Where people slip up
- "You can split the middle term any way you like." Then the tiles do not close. Run 1 + 6 and 2 + 5 and let the unit tiles run out or spill. The rectangle is the referee.
- "The 12 unit tiles are just a count." They are a 3 by 4 array, and their two dimensions are the two numbers in the split. This is the single most important thing the picture says.
- "The x-tile is a square." It is one unit by x units. If it is drawn nearly square the whole model becomes unreadable, so keep x visibly longer than 1 in every drawing.
- "The x²-tile is the x-tile squared in the sense of being twice as big." Its side is x, so its area is x × x. In a to-scale drawing with x = 4 it is four times the area of an x-tile, not twice.
- "Tiles are a crutch for people who cannot do algebra." The tiles supply the reason the two arithmetic conditions come as a pair; the algebra of §4.6 supplies the speed. Neither replaces the other.
- "With 6x² at the front you need six separate rectangles." You need one rectangle with six x²-tiles in a 2 by 3 block, because 6 = 2 × 3 and those two numbers are the leading coefficients of the two factors.
- "Tiles work for every quadratic." They need whole-number, positive tile counts. Nothing in §4.5 or §4.6 supplies a negative tile, so an expression like x² − 5x + 6 cannot be laid out as printed even though §4.6 factorises it symbolically without trouble.
- "Fig. 4.8's eleven x-tiles are eleven of the same thing." Nine of them arise one way and two another. Colour or group them differently.
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Worked answers to this chapter’s exercises · this video explains End-of-Chapter Exercises Q5, End-of-Chapter Exercises Q6
Transcript1,450 words
Here is a rectangle. Nobody has told you how big it is, only what its sides are called. One side is x plus three. The other side is x plus four. x is a length, and we are never going to decide what it is. That is the point. So what is the area? You know the answer will be a quadratic. But the answer is not what I am after. I want you to see where every piece of it comes from.
Because by the end of this, factorising will stop being a hunt for two numbers that happen to work, and become one question: do the pieces close into a rectangle? The ordinary way first. Multiply everything in the first bracket by everything in the second. Four products, because two times two is four. x times x is x squared. x times four is four x. Then three times x is three x, and three times four is twelve.
Four loose products. Now look at where they land. One is an x squared. One is a plain number. And two of them, the four x and the three x, land on the same power of x. That collision is the only reason there is a seven at all. Seven is not something the brackets contain. It is what happens when two of the four products land on top of each other.
x squared, plus seven x, plus twelve. Now build that same rectangle out of tiles. Three shapes, and only three. A square whose sides are both x. A strip that is x long and one wide. And a unit square, one by one. Watch the middle one carefully. A strip is not a square. If you draw x the same length as one, the picture stops meaning anything. Every tile comes out square and the three shapes collapse into one. So keep x visibly longer than one.
Now lay them out into a rectangle x plus three across and x plus four up. One big square. Seven strips. Twelve units. Twenty tiles, and they fill the rectangle exactly. No gaps, no overlaps, nothing hanging over the edge. So where did those seven strips go? Not into a pile of seven. Three of them stand upright along the right hand side, filling a band three wide and x tall.
The other four lie flat along the top, filling a band x wide and four tall. Three one way, four the other. Seven altogether. And now the seven has a reason. It is three because one bracket said three, and four because the other bracket said four. The strips do not know they are meant to add up. They are filling two bands, and the bands are as deep as the two numbers.
The sum is not a rule you memorise. It is two edges. Now the twelve units, and this is the part everybody skates over. They are not a heap of twelve. They are a block, three across and four up. That corner is the only place left for them, what remains once the big square and the two bands have taken their share. And the leftover corner is three wide and four tall, because those are the two numbers.
So the corner holds three fours. Twelve. The product is not a second rule sitting alongside the first. It is the area of one corner, and that corner's sides are the same two numbers that set the bands. One picture. Read it two ways. Read it outwards, from the sides in: this rectangle is x plus three, times x plus four. Read it inwards, from the tiles out: this rectangle is one x squared, plus seven x, plus twelve.
Same rectangle, so those two things are equal. Left to right, that is multiplying out. Right to left, that is factorising. They were never two skills. They are one picture read in two directions. The direction that feels hard is hard only because you must work out where the cuts go. So let us go looking, and get it wrong on purpose first. Somebody hands you one big square, seven strips and twelve units, and says: make a rectangle.
The seven strips have to divide into two bands. One and six? Two and five? Three and four? Take one and six. The corner comes out one by six, so it wants six units. You are holding twelve. Take two and five. That corner is two by five, so it wants ten. Ten of your units sit down, and two are left on the table. And here is the sting. Two and five does build a rectangle. It just builds the wrong one. Two and five is the rectangle for x squared plus seven x plus ten.
Three and four. The corner is three by four. It wants twelve, and twelve is exactly what you have. It closes. Two conditions, people say. Add to seven, multiply to twelve. But look at what you just did. There was only ever one condition. The split has to add to seven because the strips fill two bands. It has to multiply to twelve because the corner is as big as those two bands are deep. One demand: the corner must come out the exact size of the pile of units you were given.
And only one split can ever manage it. Watch the corners as the split evens up. One and six wants six. Two and five wants ten. Three and four wants twelve. They climb, and they never repeat. So the constant you were handed picks out one split, and no other. x squared plus eleven x plus thirty. Which pair? Five and six. So it is x plus five, times x plus six.
Now break the easy case. Make the front of the rectangle not one x, but two. Two x plus three, across. Three x plus one, going up. The big squares stop coming singly. There is a block of them now: two across, three up. Six x squared tiles. The side band is three wide and three x tall, so nine strips go in there. The bottom band is two x wide and one tall, so two more strips go there.
Nine and two. Eleven strips, arriving from two different places, which is why they are worth drawing in two colours. And three unit tiles in the corner. Six x squared, plus eleven x, plus three. So fill in the general blank. p x plus a, times q x plus b. The squares: p across and q up, so p q of them. The units: a by b in the corner, so a b of them.
And the strips come from two bands that no longer match each other. The side band is a wide and q x tall, giving a q strips. The top band is p x wide and b tall, giving p b strips. Together, p b plus a q. Set p and q both to one and the middle collapses back to a plus b, where we came in. That was tested on nine hundred sets of numbers. The tiles were counted, the algebra worked out, and the two never disagreed.
Try one cold. Somebody empties a bag on the table: one square, eight strips, fifteen units. Eight has to split. One and seven wants seven units in the corner. Two and six wants twelve. Three and five wants fifteen. Four and four wants sixteen. Fifteen. Three and five. So the rectangle is x plus three across and x plus five up, and all twenty four tiles go down with nothing left over.
Notice what you did not do. You did not guess pairs and check them against the middle term. You asked a single question. What size corner do I need? And the pile of units answered it. Now, honestly, where this runs out. Every tile has a positive area. There is no such thing as a negative strip in a box of tiles. So x squared minus five x plus six is out of reach. You can factorise it perfectly well: x minus two, times x minus three.
But you cannot build it, because there is no way to divide minus five into two whole bands of strips. Of six hundred and twenty five pairs I tested, exactly one hundred and sixty nine give an expression with nothing negative anywhere in it. And those are precisely the ones the tiles can build. Not most of them. All of them, and nothing else. So the tiles are not the whole story. But while they last, they are the reason the story 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
- Reading (a + b)² off a partitioned squareClass 9 · Ch 4, Exploring Algebraic Identities
- Univariate polynomials and what degree namesClass 9 · Ch 2, Introduction to Linear Polynomials
- Recognising an expression as an identity in disguiseClass 9 · Ch 4, Exploring Algebraic Identities
Comes up again in
- Splitting the middle term once the tiles come awayClass 9 · Ch 4, Exploring Algebraic Identities
Either side of this one
- (a + b + c)² by substitution, and the square that proves itClass 9 · Ch 4, Exploring Algebraic Identities