Simplifying a rational expression, and the factor you must not cancel

What to assume they know

• Factorising a quadratic by splitting the middle term (Splitting the middle term once the tiles come away)
• The square identities and the difference of squares (Reading (a + b)² off a partitioned square, (a + b + c)² by substitution, and the square that proves it)
• The cube identities and the three-term cubic identity (Cubes: (a ± b)³ from a cube cut into eight pieces, Sum and difference of cubes, and the three-term cubic identity)
• Taking a numerical common factor out of every term of a sum
• That division by zero is undefined (What "rational" means, and why the denominator cannot be zero)
• Equivalent fractions, and that cancelling a common factor is dividing top and bottom by the same non-zero quantity

What they 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

Where it usually goes wrong

• "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.

Questions to check understanding

• Simplify a quotient of two quadratics, stating the condition that makes the cancellation valid
• Simplify a quotient needing a cube identity on one side and a square identity on the other
• Factorise an expression in two letters by treating it as a quadratic in one of them
• State the values at which a given rational expression is undefined
• Complete a partially factorised simplification, as in the p. 87 Think and Reflect
• Identify and correct an invalid cancellation

Examples worth working on the board

Inputs, not answers. Values marked Verified are worked out here; this book prints no answer key.

• Example 16 (§4.8, p. 86). Simplify (x² − 7x + 12)/(5x² + 5x − 100), and the page states before doing anything that 5x² + 5x − 100 is not zero. Inputs: the two expressions and that condition. Verified, numerator: a + b = −7 and ab = 12 give a = −3 and b = −4, so it is (x − 3)(x − 4). Verified, denominator: all three terms are multiples of 5, so it is 5(x² + x − 20); then a + b = 1 and ab = −20 give 5 and −4, so it is 5(x − 4)(x + 5). Verified, result: the shared bracket x − 4 goes, leaving (x − 3) over 5(x + 5). Two things worth slowing down. First, taking the 5 out first is the cheap route rather than a precondition: it is not that the middle term cannot otherwise be split, since splitting the product 5 × (−100) = −500 into 25 and −20 gives 5x(x + 5) − 20(x + 5) = (5x − 20)(x + 5) = 5(x − 4)(x + 5), the same answer by more work. Say cheaper, not necessary. Second, the page shows three expressions joined by two equals signs — the original, the version with 5 extracted, and the fully factorised version — before any cancelling happens. Cancel too early and the licence has not been established yet.
• The licence, stated by the page (§4.8, p. 86). The page says the common factor x − 4 may be cancelled because it is not zero, and gives the reason: we know that 5x² + 5x − 100 is not zero. Inputs: that chain of reasoning, which runs in one direction only. From "a product of three things is not zero" it follows that no one of the three is zero, so x − 4 ≠ 0 and also x + 5 ≠ 0. The converse would be worthless: knowing that one factor is non-zero says nothing about the product.
• The two expressions are not identical objects. Not printed; this is added here. The original is undefined at x = 4 and at x = −5, since either makes its denominator zero. The simplified form (x − 3)/(5(x + 5)) is still undefined at x = −5 but is perfectly happy at x = 4, where it evaluates to 1/45. Verified: (4 − 3)/(5 × 9) = 1/45. So the two expressions agree at every value the first one accepts, and the second one accepts one extra value. That is what the assumption in the problem statement is protecting.
• Think and Reflect (§4.8, p. 87). The reader must simplify (36s² − 12st + t²)/(t² + 2ts − 48s²). The box prints the numerator already factorised as (6s − t)² and leaves the denominator as four blanks in the form (__ + __)(__ + __), with a hint to factorise t² + 2ts − 48s² and to assume it is not zero. Verified: treat it as a quadratic in t — the two numbers with product −48 and sum 2 are 8 and −6 — giving (t + 8s)(t − 6s). To cancel you must then notice that (6s − t)² is the same as (t − 6s)², since squaring kills the overall sign; the simplified expression is (t − 6s)/(t + 8s). That sign step is the trap in the item and the chapter does not warn about it.
• Exercise Set 4.5 (p. 87), six items, with the instruction that the denominators are not zero. (i) (3p² − 3pq − 18q²)/(p² + 3pq − 10q²). Verified: the numerator is 3(p − 3q)(p + 2q) and the denominator is (p + 5q)(p − 2q); they share no factor, so as printed the item does not simplify. See the note below. (ii) (n³ − 3n²m + 3nm² − m³)/(5m² − 10mn + 5n²). Verified: the numerator is (n − m)³ and the denominator is 5(m − n)², which is the same as 5(n − m)², so the answer is (n − m)/5. The sign trap of the Think and Reflect box appears again here. (iii) (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx). Verified: reading the numerator with a = w, b = −v, c = x makes it a³ + b³ + c³ − 3abc, so it is (w − v + x)(w² + v² + x² + wv + vx − wx); the denominator is (w − v + x)²; the answer is (w² + v² + x² + wv + vx − wx)/(w − v + x). This is the only item in the chapter that needs the three-term cubic identity and the three-term square in the same breath, and it is the best argument for §4.8 being placed last. (iv) (4y² − 20yz + 25z²)/(25z² − 4y²). Verified: (2y − 5z)² over (5z − 2y)(5z + 2y), which is (5z − 2y)/(5z + 2y). (v) ((x² + x − 6)(x² − 7x + 12))/((x² − 6x + 8)(x² − 9)). Verified: the four quadratics factorise as (x + 3)(x − 2), (x − 3)(x − 4), (x − 2)(x − 4) and (x − 3)(x + 3), so everything cancels and the answer is 1. But it is 1 only where the original is defined — not at x = 2, 3, 4 or −3. Look at what the factorisation above actually shows: numerator and denominator are the same product of the same four brackets, so every one of those four values kills both at once and all four are of the shape zero over zero, not just two. At x = 2 the numerator reads 0 × 2 over a denominator 0 × (−5); at x = 4 it is 14 × 0 over 0 × 7. This is the best item in the chapter for making the domain point stick, and it is stronger read this way — there is no second kind of failure to distinguish. (vi) (p⁴ − 16)/(p² − 4p + 4). Verified: p⁴ − 16 is (p² − 4)(p² + 4) and then (p − 2)(p + 2)(p² + 4), while the denominator is (p − 2)², so the answer is (p + 2)(p² + 4)/(p − 2). Note that one factor of (p − 2) survives in the denominator, so the expression is still undefined at p = 2 after simplifying — the opposite of what happened in Example 16.
• End-of-Chapter Q4 (p. 89), three items, with a printed note that the denominators are not zero. Verified: (i) (4x² + 4x + 1)/(4x² − 1) = (2x + 1)²/((2x − 1)(2x + 1)) = (2x + 1)/(2x − 1); (ii) 9(3a³ − 24b³)/(9a² − 36b²) = 27(a³ − 8b³)/(9(a² − 4b²)) = 3(a − 2b)(a² + 2ab + 4b²)/((a − 2b)(a + 2b)) = 3(a² + 2ab + 4b²)/(a + 2b); (iii) (s³ + 125t³)/(s² − 2st − 35t²) = (s + 5t)(s² − 5st + 25t²)/((s − 7t)(s + 5t)) = (s² − 5st + 25t²)/(s − 7t). Item (ii) is the one to work, because it needs a numerical factor pulled out of both top and bottom before either identity is visible.
• The chapter's closing statement (p. 90, last bullet of the summary). It says that these expressions may be simplified by factorising and removing common factors, provided such a factor exists and is not zero. Both conditions are doing work: Exercise Set 4.5 (i) is a case where no shared factor exists, and every other item depends on the non-zero clause.

Figures to have open

• §4.8 prints no figures. Checked on pp. 86, 87 and 88 — the only drawing anywhere near this section is the one Example 17 asks the reader to make, which belongs to Algebra tiles: factorising by rebuilding the rectangle. So every visual here is added here.
• A number line marked with the forbidden values of a rational expression, used twice: once for Example 16 and once for Exercise Set 4.5 (v). Standard schematic, and the topic's most important image.
• A dependency map for section 12 — the chapter's identities on one side, the §4.8 items that need them on the other. Standard schematic.
• A side-by-side card for (6s − t)² and (t − 6s)². Standard schematic.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 4, "Exploring Algebraic Identities": §4.8 "Simplifying Rational Expressions", pp. 86–87, Example 16.
• Think and Reflect, p. 87, with its printed hint.
• Exercise Set 4.5, p. 87, all six items, printed with the assumption that the denominators are not zero.
• End-of-Chapter Exercises Q4 (i)–(iii), p. 89, with the printed note about the denominators.
• Chapter summary, p. 90, final bullet, which carries both conditions — that a common factor exists and that it is not zero.
• Companion topics: Splitting the middle term once the tiles come away for the middle-term split this section leans on, Sum and difference of cubes, and the three-term cubic identity for the cubic identities two of these items require.