Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 5 · Chapter companion part
Multiply and divide algebraic fractions using complete factors and correct nonzero conditions.
Class 8 English Version: part C of the chapter 5 companion. Eight sections, 30 written tasks and a notebook project. Selected companion coverage, not complete textbook coverage; use all four parts. Numerical/coefficient answers are checked; equivalent symbolic forms, domains and proofs need guardian or teacher review.
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For B,D nonzero, (A/B)(C/D)=AC/(BD). Unlike addition, multiplication does not require a common denominator. For (3/4)(8/9), cancel complete common factors 3 with 9 and 8 with 4, obtaining 2/3. The original grid shows 2/3 of 3/4 of one whole: six of twelve cells, or one half. It is an area model of a product, not the sum of the two fractions.
In [6a²b/(5c)] × [10c/(9ab)], cancel whole factors across the product to get 4a/3, for a,b,c nonzero. A factor in a numerator may cancel a nonzero factor in either denominator because the entire expression is one product. It cannot be cancelled across an addition sign unless the whole sum factors first.
For [(x²−9)/(x²−5x+6)] × [(x−2)/(x+3)], factor to [(x−3)(x+3)/((x−2)(x−3))] × [(x−2)/(x+3)]. It reduces to 1, but the original restrictions x≠2,3,−3 remain. A constant final result is not permission to use a formerly forbidden input.
The product (yz/x²)(zx/y²)(xy/z²) is x²y²z²/(x²y²z²)=1, for xyz≠0. A fourth or later factor follows the same rule. Keep exponents until the full factor count is clear. A zero numerator makes a product zero when every denominator is defined; multiplication does not require every numerator to be nonzero.
The reciprocal of C/D is D/C, requiring both D≠0 for the original fraction and C≠0 to reverse it. Division A/B ÷ C/D means A/B × D/C. In the grouping bar, 3/4 contains three groups of size 1/4, so (3/4)÷(1/4)=3. Dividing by a fraction can increase a positive result; it is not the same operation as multiplying by that fraction.
For [(x²−1)/(x−2)] ÷ [(x−1)/(x+3)], exclude x=2 and −3 for the original denominators, and x=1 because it makes the divisor zero. Factor and reverse the second fraction to get (x+1)(x+3)/(x−2), on that original domain. x=−1 is allowed: the dividend is zero while the divisor is defined and nonzero.
For (1+1/x) ÷ (1−1/x²), first write (x+1)/x and (x²−1)/x². Reverse the whole second fraction, then factor x²−1. The result is x/(x−1), with x≠0,1,−1. Do not reverse only the last term 1/x², and do not restore x=−1 simply because it cancels.
For [(a²−b²)/(a²+ab+b²)] ÷ [(a+b)/(a³−b³)], reverse the divisor and use a³−b³=(a−b)(a²+ab+b²). The result is (a−b)² on the original domain: a²+ab+b²≠0, a³−b³≠0, a+b≠0. Multiplying the same two fractions gives a different expression. Write the operation before cancelling and check a permitted pair such as a=5,b=2.
For [(x²−1)/(x−2)] ÷ [(x−1)/(x+3)], first exclude x=2,−3,1. Multiply (x−1)(x+1)/(x−2) by (x+3)/(x−1). The result is (x+1)(x+3)/(x−2) on that original domain. At x=4, the original 15/2 ÷ 3/7 gives 17.5, agreeing with 5×7/2.
A defined zero factor makes a product zero. A divisor must be nonzero; these are different rules.
The reverse fraction may look defined at an input where the original divisor was not. Preserve the original conditions.
In these rules B and D are originally nonzero.
| Operation | Rule | Additional condition |
|---|---|---|
| Multiply | (A/B)(C/D)=AC/(BD) | C may be zero |
| Divide | (A/B)÷(C/D)=AD/(BC) | C must be nonzero |
| Reciprocal | 1/(C/D)=D/C | Both C and D nonzero |
A zero numerator is allowed for a multiplier but not for a divisor. At C=0 the original division is undefined even if later factors look cancellable.
Example: [(x²−1)/(x−2)] ÷ [(x−1)/(x+3)].
| Original part | Forbidden value | Why |
|---|---|---|
| Dividend denominator x−2 | 2 | Original fraction undefined |
| Divisor denominator x+3 | −3 | Original divisor undefined |
| Divisor numerator x−1 | 1 | Division by a zero fraction |
The value x=−1 is allowed: the dividend is zero, while the divisor is defined and nonzero. Do not forbid every numerator root.
Multiplication can cancel complete factors across the product.
[(x²−9)/(x²−5x+6)] × [(x−2)/(x+3)] = [(x−3)(x+3)/((x−2)(x−3))] × [(x−2)/(x+3)] = 1, x≠2,3,−3
A final constant result does not restore excluded inputs.
Start by converting both bracketed expressions into fractions.
(1+1/x) ÷ (1−1/x²) = [(x+1)/x] × [x²/(x²−1)] = x/(x−1), x≠0,1,−1
Keep x=−1 excluded even though x+1 cancels. Reversing just the last term is invalid.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact-area unit rectangle. Positive numerical fractions illustrate multiplication; signed algebra is justified by its rules, not negative physical areas.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact equal-part bar. It models positive numerical measurement division; the algebraic quotient rule carries its stated nonzero conditions.
Show your working on paper. Check your answer, then choose Next.
2
24/36 = 2/3.
3
The x factors cancel and 6/2 = 3.
8
The product reduces to 4a/3, giving 8.
4
Numerator powers give x⁵ divided by x, leaving x⁴.
3
The combined y denominator power is 5, reduced by numerator power 2.
1
All factors cancel at this permitted input.
1
Each variable appears twice in both numerator and denominator.
0
A numerator is zero while both denominators are nonzero.
5
The reciprocal reverses numerator and denominator.
1.5
Multiply x/3 by 9/(2x): the result is 3/2 = 1.5.
1
The quotient is 45x³/(8ay²) = 360/360 = 1.
4
Its numerator vanishes at 4, while the denominator there is nonzero.
3
Exclude 2 and −3 for denominators, and 1 for the zero divisor.
1.5
The reduced expression is x/(x−1), giving 3/2.
9
The expression reduces to (a−b)² = 9.
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Divide one whole into 3 rows and 4 columns. Select two rows and three columns; their overlap is 6 of 12 equal cells, hence 1/2.
Show equal cell areas and distinguish the overlap product from an addition.
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4a/3, requiring a,b,c≠0. Coefficients reduce 60/45 to 4/3; b,c and one a factor cancel.
Retain variables excluded by either original denominator.
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1 for x≠2,3,−3. Factor the quadratic denominator as (x−2)(x−3), then cancel complete factors.
A final 1 must still carry all original exclusions.
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1/(x²+xy+y²), on x³+y³≠0 and x³−y³≠0. Use both cube factorizations and x²−y²=(x−y)(x+y).
Keep the original domain rather than only the final quadratic restriction.
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The numerator and denominator are both x²y²z²; cancellation is allowed for x,y,z≠0.
Count every exponent and explain why a zero variable is forbidden.
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(x−2)²(x+4)/[(x−3)²(x+3)], with x≠1,3,4,−3. Factor all six quadratics and cancel complete factors. x=2 is allowed and gives zero.
Track restrictions from original denominators before cancelling; numerator roots are not automatically exclusions.
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There are three quarter-sized groups in three quarters. Algebra gives (3/4)×4=3. The divisor 1/4 is nonzero.
The grouping picture and reciprocal calculation must agree.
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45x³/(8ay²), requiring a,x,y≠0. Reverse the whole second fraction to 15x/(4y²).
Include the zero-divisor condition y≠0 as well as original denominators.
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(x+1)(x+3)/(x−2), with x≠2,−3,1. The extra exclusion 1 makes the divisor zero; −1 is allowed.
Distinguish zero dividend from zero divisor.
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(x²+3x+9)(x+6)/[(x−1)(x+3)], with x≠1,6,−6,3,−3. Reverse the second fraction, factor all pieces, and retain the divisor-zero roots ±3.
Original and zero-divisor restrictions both belong in the final answer.
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(x²+xy+y²)/(x²−xy+y²), requiring x³+y³≠0, x+y≠0 and x²−y²≠0. The last condition makes the divisor nonzero.
Use the cube identities and retain the cancelled x−y restriction.
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x/(x−1), for x≠0,1,−1. Write the two bracketed fractions separately before reversing the divisor.
Do not invert only one term inside the divisor.
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(a−b)², requiring a²+ab+b²≠0, a³−b³≠0 and a+b≠0. The reciprocal turns the cube difference into numerator factors.
Show every factor and the nonzero numerator of the original divisor.
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The product is 8/15. The quotient is (2/3)(5/4)=5/6. Multiplication uses 4/5 directly; division uses its reciprocal.
A numerical countercheck should show these are different operations.
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The simplified form is (x+3)/(x+2), but x≠−2,−3,1. At 1 the original divisor is zero, so the original division is undefined.
Explain why the divisor condition survives cancellation.
Create a multiplication/division notebook with one numeric grid, one polynomial product and two quotients.
Keep your work on paper or in a local file. This page does not upload it or automatically grade the project.
This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 5, printed pages 77–100. It does not replace the complete chapter or its practical teaching.
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