Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 4 · Chapter companion part
Factor first, choose minimum or maximum powers correctly, verify divisibility and review mixed chapter reasoning.
Class 8 English Version: part D of the chapter 4 companion. Eight sections and 30 written tasks. Selected companion coverage, not complete textbook coverage; use all four parts and compare the mapped source methods. The software checks stated numerical values or coefficients; symbolic work, equivalent forms and proofs need guardian or teacher review.
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The highest common factor (HCF), also called greatest common divisor (GCD), contains every factor shared by all inputs at the smallest exponent present in them. Use the school integer-coefficient convention: extract positive numerical content, the greatest common divisor of each polynomial’s integer coefficients. A leading coefficient alone is not necessarily content: 2x + 1 has content 1, not an extractable factor 2 in integer coefficients.
For 12a³b²c, 18a²bc² and 30a⁵b⁴c³, numerical HCF is 6. The common exponents are a², b and c, giving 6a²bc. A missing factor has exponent zero and cannot appear in the HCF of all inputs. The source’s highest common power wording must be read with divisibility: using a larger exponent would fail to divide an input.
For x² − 3x, x² − 9 and x² − 4x + 3, write x(x − 3), (x − 3)(x + 3) and (x − 3)(x − 1). The HCF is x − 3. For a quadratic like 6x² − x − 1, first factor (3x + 1)(2x − 1); its primitive binomial factors already contain their leading coefficients, so an extra factor 6 must not be inserted into the LCM.
The least common multiple (LCM) is divisible by every input. Choose the numerical LCM of their contents and the largest required exponent of every distinct primitive polynomial factor, including factors absent from some inputs. For 6a²b, 10ab³ and 15a³b², LCM is 30a³b³. Highest powers do not mean only factors common to all three.
For 2x(x + 1), 3(x + 1)² and 4(x² − 1), factor the last as 4(x − 1)(x + 1). The LCM is 12x(x + 1)²(x − 1). Cubic factors may reveal additional quadratics: x³ − 1 = (x − 1)(x² + x + 1). A table recording each exponent prevents missing an input’s unique factor.
For A = 6x²(x − 1) and B = 4x(x − 1)², H = 2x(x − 1) and L = 12x²(x − 1)². Each A/H and B/H is a polynomial; each L/A and L/B is a polynomial. Also HL = AB = 24x³(x − 1)³ under these normalised signs. The two-input product relation does not extend unchanged to three inputs: HCF(2,3,6) × LCM(2,3,6) = 6, while 2 × 3 × 6 = 36.
Factors x − 2 and 2 − x differ by −1. They represent the same nonconstant factor after consistent sign normalisation; do not list them as unrelated LCM factors. Algebraic HCF followed by numerical substitution can differ from numerical HCF after substitution. At x = 2, A = 24 and B = 8 in the preceding example; algebraic HCF evaluates to 4 while numerical HCF is 8. Substitution can create extra numerical divisibility.
If A = (x − 4)(x − 2) and B = (x − 4)(x + 1), the proposed LCM (x − 4)(x − 2) omits x + 1 and fails the second divisibility check. The true LCM is (x − 4)(x − 2)(x + 1). This is the same error type as source sample 3’s deliberately false LCM assertion. In a real-number setting, also reject impossible reciprocal premises rather than force a result. Keep proof questions and equivalently ordered factors under human review.
For contents 6 and 4, x powers 2 and 1, and (x − 1) powers 1 and 2, read minimums for HCF and maximums for LCM. The resulting 2x(x − 1) and 12x²(x − 1)² pass all four polynomial quotient checks.
Highest common factor refers to divisibility, not the greatest value after choosing a number for x. Least common multiple is likewise a polynomial relationship.
These numerical checks are practice feedback. They do not automatically grade a polynomial proof, an alternative valid factor order or an entire written exam.
Original inputs A = 6x²(x − 1) and B = 4x(x − 1)².
| Content or factor | Input A | Input B | HCF choice | LCM choice |
|---|---|---|---|---|
| Numerical content | 6 | 4 | 2 | 12 |
| x exponent | 2 | 1 | 1 | 2 |
| x − 1 exponent | 1 | 2 | 1 | 2 |
Reading minimum powers gives 2x(x − 1). Reading maximum powers and numerical LCM gives 12x²(x − 1)². Numerical content is the common coefficient factor, not automatically the leading coefficient.
Use H = 2x(x − 1), L = 12x²(x − 1)² for the same two inputs.
| Quotient | Polynomial result | Meaning |
|---|---|---|
| A/H | 3x | H divides A |
| B/H | 2(x − 1) | H divides B |
| L/A | 2(x − 1) | A divides L |
| L/B | 3x | B divides L |
Each result is a polynomial after exact factor cancellation. This verifies divisibility as a polynomial relationship, not merely at one numerical input.
Assume nonzero integer-coefficient polynomials, fully factored with positive numerical content and consistently normalised overall signs.
HCF = gcd(contents) × product of common minimum powers LCM = lcm(contents) × product of all required maximum powers For two inputs: HCF × LCM = A × B, up to an overall sign
The source’s highest common power wording does not permit maximum powers in HCF. The two-input product equality is not a three-input rule.
The stated quadratic has nonzero real roots; divide only after noting x ≠ 0.
2x² − 7x + 3 = 0 2x + 3/x = 7 (2x + 3/x)² = 4x² + 12 + 9/x² 4x² + 9/x² = 49 − 12 = 37
Numerical examples are checked values. A full proof of the transformation and the condition still needs written reasoning.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original symbolic cards. Exponents represent repeated factors, not a ranking or a numerical value for x.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original factor-set schematic. Positions are categorical; they do not encode magnitude or geometric overlap.
Show your working on paper. Check your answer, then choose Next.
6
Six is the largest positive integer dividing all three.
2
Use the minimum exponent, 2.
1
Use min(2,1,4) = 1.
30
30 is divisible by each and uses the needed prime powers 2 × 3 × 5.
3
Use the maximum exponent, 3.
3
5 − 2 = 3. This asks for the value of the polynomial HCF.
15
3 × 5 × 1 = 15.
4
2 × 2 × 1 = 4.
48
12 × 4 × 1 = 48.
8
Both values are divisible by 8; this differs from the polynomial HCF value 4.
1
The minimum required exponent is 1.
3
The maximum required exponent is 3.
12
Use 2² × 3 = 12.
1
A constant 1 divides both; neither linear factor divides the other.
30
Their product is 6 × 60 = 360; 360/12 = 30.
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6a²bc. Use gcd(12,18,30) = 6 and the minimum exponents 2,1,1.
Divide each original monomial by the answer and check that every quotient is a polynomial.
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30a³b³. The numerical contents need 30 and the maximum exponents are 3 for both variables.
Check that the result is divisible by all three original monomials.
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Their factorizations are x(x − 3), (x − 3)(x + 3) and (x − 3)(x − 1); HCF is x − 3.
Show all three factorizations; do not compare additive terms as if they were factors.
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The first cube difference and the second quartic both contain x² + xy + y². Numerical HCF is 3; common monomial is xy². HCF is 3xy²(x² + xy + y²).
Factor the quartic as (x² + xy + y²)(x² − xy + y²) and verify minimum exponents.
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12x(x + 1)²(x − 1). The factor x − 1 occurs only in the third expression but is required in the LCM.
Use content LCM 12 and a union of factor powers.
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(x − 1)³(x + 1)²(x² + x + 1).
Show cube and square differences and take the highest exponent of each distinct factor.
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2 − x = −(x − 2). After normalising signs they are the same primitive linear factor. Choose HCF and LCM x − 2 under the stated sign convention; quotients may be −1.
Preserve the sign in each division instead of silently identifying unequal expressions.
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Polynomial HCF is 2x(x − 1), giving 4. The evaluated inputs are 24 and 8, whose numerical HCF is 8. Substitution has introduced extra numerical common factors.
Explain the order of operations and why the two questions differ.
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H = 2x(x − 1), L = 12x²(x − 1)², so HL = 24x³(x − 1)³. Multiplying A and B gives the same product.
Also verify the quotient conditions; a product coincidence alone does not identify both correctly.
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The three-input HCF is 1 and LCM is 6, whose product is 6. The product of the inputs is 36.
Use all three inputs consistently; the failed equality disproves the claimed generalisation.
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The proposed multiple lacks x + 1 and is not divisible by the second polynomial. The correct LCM is (x − 4)(x − 2)(x + 1).
Identify the missing factor and show both quotient checks.
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Factor as (3x + 1)(2x − 1), (3x + 1)(x + 2) and (2x − 1)(x + 2). LCM is (3x + 1)(2x − 1)(x + 2). Each polynomial has numerical content 1.
Do not multiply by another 6 merely because the leading coefficients include 6,3,2.
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Divide by x: 2x + 3/x = 7. Square to get 4x² + 12 + 9/x² = 49, so the requested value is 37. Roots 3 and 1/2 both give 37.
Explain why no choice between the roots is needed and why division by x is allowed.
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Contents 6 and 4 give HCF 2 and LCM 12. Powers of x are 2,1; powers of x − 1 are 1,2. Thus HCF 2x(x − 1) and LCM 12x²(x − 1)².
Treat numerical content and polynomial factors separately and verify divisibility.
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Complete the square: (x − √3/2)² + 1/4 = 0. The left side is positive for every real x, so no real solution exists. Over complex numbers the formal reciprocal consequences are sum √3, square sum 1 and cube sum 0, but that is a different number domain.
Do not use the formal manipulation as evidence of a feasible real x.
Build a common-factor inventory and a short mixed written paper with worked corrections.
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This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 4, printed pages 48–76. It does not replace the complete chapter or its practical teaching.
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