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Chapter 4 · Exploring Algebraic Identities
Simplifying a rational expression, and the factor you must not cancel
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Simplifying a fraction with letters in it is not tidying, it is dividing. And there is exactly one number you may never divide by.
The idea
Cancelling a common factor is division, and dividing by zero is not a thing you are allowed to do. So the sentence the chapter prints before it starts — that the denominator is not zero — is not throat-clearing, it is the licence for the whole calculation. Example 16 makes the reasoning explicit: because the whole denominator is known to be non-zero, the factor x − 4 cannot be zero either, and only then may it go. The simplified expression is a different object from the one you started with — it is defined at values the original was not — and the two agree exactly where the original makes sense.
What you should be able to do
- State what has to be true before a common factor may be cancelled
- Factorise both the numerator and the denominator of a rational expression, taking numerical common factors out first
- Identify the common factor and cancel it, quoting the reason it is not zero
- Derive "this factor is not zero" from "the whole denominator is not zero"
- State the values at which the original expression is undefined, and note that the simplified form is defined there
- Factorise a quadratic in the letter that is squared, when the expression contains two letters
- Simplify expressions requiring the cube identities, the three-term cubic identity, or a difference of squares in disguise
- Recognise when two brackets differ only by an overall sign, and rewrite one to match the other
- Explain why this section is placed last in the chapter
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| rational expression | a quotient of two algebraic expressions | printed in this chapter (§4.8, p. 86) |
| rational algebraic expression | the same thing, as the chapter's summary names it | printed in this chapter (§4.8, p. 86; Chapter summary, p. 90) |
| numerator | the expression on top of the quotient | printed in this chapter (§4.8, p. 86) |
| denominator | the expression underneath | printed in this chapter (§4.8, p. 86) |
| cancel | to divide numerator and denominator by a shared non-zero factor | printed in this chapter (§4.8, p. 86) |
| common factor | a factor shared by numerator and denominator | printed in this chapter (§4.3, p. 73; used this way in §4.8, p. 86) |
| factorisation | rewriting an expression as a product | printed in this chapter (§4.3 heading, p. 72; §4.8, p. 86) |
| zero | the value the denominator is assumed not to take | printed in this chapter (§4.8, p. 86) |
| simplify | to rewrite in a form with the common factors removed | printed in this chapter (§4.8, p. 86) |
| undefined value | an input at which the expression has no value at all | an added term; the chapter states the non-zero assumption and never names what happens if it fails |
| domain | the set of values a letter is allowed to take | an added term, not printed in this chapter, though the word is used elsewhere in the book |
Where people slip up
- "You can cancel anything that appears top and bottom." Only factors, and only non-zero ones. Show a student trying to cancel the x² from (x² − 7x + 12)/(5x² + 5x − 100) and watch the value change: at x = 1 the original is 6/(−90), while the illegally cancelled version is nothing like it.
- "The condition that the denominator is not zero is boilerplate." It is the reason the next line is allowed. Show it and keep it there through the cancellation.
- "Once the factor cancels, the value it forbade is fine." Sometimes it is, sometimes it is not. In Example 16 the simplified form does accept x = 4; in Exercise Set 4.5 (vi) a factor of (p − 2) is left behind and p = 2 is still forbidden. The general rule is that the original's forbidden values stay forbidden for the original, whatever the simplified form can do.
- "An expression that simplifies to 1 is equal to 1." Exercise Set 4.5 (v) simplifies to 1 and yet has four values where it has no value at all — every one of them of the shape zero over zero, since the same four brackets sit above and below. Equality of expressions is only ever equality where both sides make sense.
- "(6s − t)² and (t − 6s)² are different." They are the same, because squaring removes the overall sign. Two exercises in this section turn on exactly this and neither warns about it.
- "5x² + 5x − 100 has a leading coefficient of 5, so split the middle term with 5 in mind." Take the 5 out first and the problem becomes the easy kind. The page does this and it is the reason the example is tractable.
- "p⁴ − 16 needs a special identity." It needs the difference of squares twice, once on p⁴ − 16 and once on p² − 4. Nothing new.
- "If it does not simplify, I must have made a mistake." Sometimes the expression genuinely has no shared factor. The chapter's own summary allows for it, and Exercise Set 4.5 (i) as printed is such a case.
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Worked answers to this chapter’s exercises · this video explains Exercise Set 4.5 Q1, End-of-Chapter Exercises Q4
Transcript1,449 words
Here is a fraction with letters in it, and the instruction is to simplify. x squared minus seven x plus twelve, over five x squared plus five x minus a hundred. Which sounds like tidying up. It is not. Simplifying means finding something that is a factor of the top and a factor of the bottom, and dividing both by it. Dividing. Hold on to that word, because it is going to matter more than anything else here.
There is exactly one number you are never allowed to divide by. And an expression with a letter in it can quietly become that number. So before any of this is allowed, something has to promise that the bottom is not nought. Take the top first. x squared minus seven x plus twelve. You want two numbers that add to minus seven and multiply to twelve. Search every whole number pair, and exactly one comes back. Minus four and minus three.
So the top is x minus three, times x minus four. Multiply that out again and it lands on what you started with, term for term. That check costs nothing. Do it every time. Now the bottom. Five x squared plus five x minus a hundred. Do not go straight at the middle term. Five, five and a hundred are all multiples of five. Lift the five out and what is left inside is x squared plus x minus twenty, which is the easy kind.
Two numbers adding to one and multiplying to minus twenty. Minus four and five. So the bottom is five, times x minus four, times x plus five. Taking the five out first was cheaper, and cheaper is not the same as necessary. Go at the middle term directly and you split minus five hundred into minus twenty and twenty five, and you land on exactly the same bottom. More work, same answer.
Put them side by side and the shared bracket is impossible to miss. Compare the two lists of brackets and they have exactly one thing in common. That one. So cross it off, top and bottom, and what is left is x minus three over five, times x plus five. Except stop. You just divided by x minus four. How do you know that was not nought? Go back to the promise made at the very start, before a single bracket was drawn.
The bottom is not nought. That sentence looks like throat clearing. It is the licence for everything that follows. Without it the cancelling is not a shortcut, it is an error. And notice where it sits: before the working, not after. Cancel first and look for permission later, and you have already done the thing you were not allowed to do. But the promise is about the whole bottom. The cancelling needs one bracket.
So walk it across: the bottom is a product of five, x minus four and x plus five. If a product is not nought, then not one of the things multiplied is nought either. Check that everywhere in range and only two cases ever turn up. Either no bracket vanishes and the bottom is fine, or exactly one vanishes and the bottom is nought. There is never a third case.
Now run the arrow backwards, and watch it fail. At x equals minus five, the bracket x minus four is perfectly healthy. It is minus nine. And the bottom is nought anyway, because the other bracket did it. One factor behaving tells you nothing about the product. The product behaving tells you about every factor. Here is the part nobody says out loud. The answer is not the same object as the question.
The original has no value at four, and no value at minus five. Both of those kill its bottom. The simplified version still refuses minus five. But at x equals four it is perfectly happy, and it equals one forty-fifth. Run both across seventeen whole numbers and count. They agree at fifteen. At one value neither has anything to say. At exactly one, the second has an answer and the first does not.
So they are not equal. They agree everywhere the first makes sense, and the second accepts one extra value. That gap is precisely what the promise at the start was protecting. Now the thing that goes wrong most. Cancelling something that is not a factor. Look again: x squared is sitting on top and on the bottom, begging to be struck out. Strike them and you get minus seven x plus twelve, over five x minus a hundred.
Test it at x equals one. The real thing is minus one fifteenth. The struck one is minus one nineteenth. Not close. Just different. Across the range the struck version is right once and wrong fourteen times. The one time it is right is at x equals nought, where the x squared it removed contributed nothing anyway. A coincidence is not a defence. You may cancel factors. A term is not a factor.
Here is a trap that costs marks quietly. Thirty six s squared minus twelve s t plus t squared, over t squared plus two t s minus forty eight s squared. The top is six s minus t, all squared. The bottom, read as a quadratic in t, needs two numbers adding to two and multiplying to minus forty eight. Eight and minus six. So the bottom is t plus eight s, times t minus six s.
And now look. Upstairs it is six s minus t. Downstairs it is t minus six s. Those are not the same bracket. One is the other one's negative. But the top is squared, and squaring throws the overall sign away, so those two squares are the same expression. Rewrite it that way and one copy cancels, leaving t minus six s over t plus eight s. On a grid wide enough to see it, the original refuses seven pairs and the simplified one refuses three. Four values got freed, and nothing else changed.
Try this one. Four quadratics, two on top and two underneath. Factorise all four and something startling happens. The same four brackets appear above and below. x plus three, x minus two, x minus three, x minus four. Upstairs and downstairs, the identical product. Subtract one from the other and there is nothing left at all. So it simplifies to one. And yet it is not the number one. It has no value at minus three, at two, at three, and at four.
At every single one of those four the top is nought as well. Four cases of nought over nought, and the same brackets caused both. There is no second kind of failure to sort out. Thirteen values where it is one, four where it is nothing. An expression that simplifies to one is not equal to one. Do not learn the wrong lesson from the first example. p to the fourth minus sixteen, over p squared minus four p plus four.
The top is a difference of squares twice over. p squared minus four, times p squared plus four. Then p minus two, times p plus two, times p squared plus four. The bottom is p minus two, squared. So one p minus two cancels. And one p minus two is still standing underneath. So p equals two was forbidden before, and it is forbidden afterwards. Nothing was freed. Put the two side by side. The first example gained a value. This one gained none.
The rule is not that cancelling frees the value. It is that the original's forbidden values stay forbidden for the original, whatever the tidier version can do. One more, and it is the reason this comes at the end. w cubed minus v cubed plus x cubed plus three w v x, over the six-term expression underneath. Read the top with minus v in the middle slot and it is a cubed plus b cubed plus c cubed minus three a b c.
So the top is w minus v plus x, times that long second bracket. And the bottom is the very same w minus v plus x, squared. One copy goes. You needed the cubic identity and the three-term square in the same breath, which is why nothing here could have come sooner. Last thing, because it is easy to assume otherwise. Sometimes there is no shared factor and the thing simply does not simplify.
Of six worked quotients here, five share a bracket and one shares nothing at all. The condition is not only that the factor is not nought. It is also that there is one.
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
- Splitting the middle term once the tiles come awayClass 9 · Ch 4, Exploring Algebraic Identities
- Reading (a + b)² off a partitioned squareClass 9 · Ch 4, Exploring Algebraic Identities
- (a + b + c)² by substitution, and the square that proves itClass 9 · Ch 4, Exploring Algebraic Identities
- Cubes: (a ± b)³ from a cube cut into eight piecesClass 9 · Ch 4, Exploring Algebraic Identities
- Sum and difference of cubes, and the three-term cubic identityClass 9 · Ch 4, Exploring Algebraic Identities
- What "rational" means, and why the denominator cannot be zeroClass 9 · Ch 3, The World of Numbers
Either side of this one
- Turning an observation into a definition: the circle as a locusClass 9 · Ch 5, I’m Up and Down, and Round and Round