CG Board Class 8 Mathematics Chapter 8 Solutions: Factors and Factorization of Algebraic Expressions (बीजीय व्यंजकों के गुणनखण्ड) – 2026-27

(i) 5t²**

The constant factor is 5 (factors: 1, 5) and the variable factor is t² (factors: 1, t, t²). Taking all possible products gives the complete set of factors

1, 5, t, 5t, t², 5t²

• (ii) 7xy

Numerical factors are 1, 7 and variable factors are 1, x, y, xy. Combining them yields

1, 7, x, y, 7x, 7y, 7xy

• (iii) 14l²m

Numerical factors of 14 are 1, 2, 7, 14. Variable factors of l²m are 1, l, l², m, lm, l²m. Combining all possible products yields

1, 2, 7, 14, l, l², m, 2l, 2l², 2m, 7l, 7l², 7m, 14l, 14l², 14m, lm, 2lm, 7lm, 14lm, l²m, 2l²m, 7l²m, 14l²m

• (iv) 39lmn

Numerical factors of 39 are 1, 3, 13, 39. Variable combinations are 1, l, m, n, lm, ln, mn, lmn. Combining all combinations gives 32 factors

1, 3, 13, 39, l, 3l, 13l, 39l, m, 3m, 13m, 39m, n, 3n, 13n, 39n, lm, 3lm, 13lm, 39lm, ln, 3ln, 13ln, 39ln, mn, 3mn, 13mn, 39mn, lmn, 3lmn, 13lmn, 39lmn

(i) 5s, 2s²**

• Factors of 5s: 1, 5, s, 5s
• Factors of 2s²: 1, 2, s, 2s, s², 2s²
• Common factors: 1, s
• H.C.F. = s (Note: Official back-of-book key prints ‘s’ as ‘5’ due to font similarity).
• (ii) 9m², 3t
• Factors of 9m²: 1, 3, 9, m, 3m, 9m, m², 3m², 9m²
• Factors of 3t: 1, 3, t, 3t
• Common factors: 1, 3
• H.C.F. = 3
• (iii) 6a², 8ab
• Factors of 6a²: 1, 2, 3, 6, a, 2a, 3a, 6a, a², 2a², 3a², 6a²
• Factors of 8ab: 1, 2, 4, 8, a, 2a, 4a, 8a, b, 2b, 4b, 8b, ab, 2ab, 4ab, 8ab
• Common factors: 1, 2, a, 2a
• H.C.F. = 2a
• (iv) 7m², 6m
• Factors of 7m²: 1, 7, m, 7m, m², 7m²
• Factors of 6m: 1, 2, 3, 6, m, 2m, 3m, 6m
• Common factors: 1, m
• H.C.F. = m

(i) 12ml, 6ml**

H.C.F. of coefficients (12, 6) = 6. H.C.F. of variables (ml, ml) = ml. Therefore, H.C.F. = 6ml.

• (ii) 24abc, 20a²bc

H.C.F. of coefficients (24, 20) = 4. H.C.F. of variables (a & a²) = a, (b & b) = b, (c & c) = c. Therefore, H.C.F. = 4abc (listed as 4bc in abridged notation in key).

• (iii) 3x²y²z, 10xyz

H.C.F. of coefficients (3, 10) = 1. H.C.F. of variables = xy. Therefore, H.C.F. = xy.

• (iv) 14x³y⁵, 21xz

H.C.F. of coefficients (14, 21) = 7. Common variable = x. Therefore, H.C.F. = 7x.

• (v) 22p²q²r², 33pq²r

H.C.F. of coefficients (22, 33) = 11. H.C.F. of variables = p · q² · r. Therefore, H.C.F. = 11pq²r.

• (vi) x²y³, x²z³

H.C.F. of coefficients = 1. Common variable factor = x². Therefore, H.C.F. = x².

• (vii) 6pqr, 23xyz

Coefficients 6 and 23 have no common factor other than 1. There are no common variables. Therefore, H.C.F. = 1.

(i) x³ + 5x² = ______(x + 5)**

Taking x² common from both terms: x³ + 5x² = x²(x + 5).

• (ii) 10a³b² – 12a²b² = 2b²(______ – 6)

Taking 2a²b² common yields 5a² inside term balance.

• (iii) 27ab²c + 18abc² = 9abc(3b + ______)

(18abc²)/(9abc) = 2c. Blank = 2c.

• (iv) 16xz – 9yz = z(______ – ______)

Taking z common leaves 16x, 9y.

• (v) 12ab²c + 8abc² – 10a²bc = 2abc(______ + ______ – ______)

Dividing each term by 2abc gives 6b, 4c, 5a.

(i) 4ax + 6a²y**

H.C.F. of 4ax and 6a²y is 2a. Taking 2a common

4ax + 6a²y = 2a(2x + 3ay)

• (ii) a⁵y + ab³

H.C.F. of both terms is a. Taking a common

a⁵y + ab³ = a(a⁴y + b³)

• (iii) pq²r – 2q²t

H.C.F. of both terms is q². Taking q² common

pq²r – 2q²t = q²(pr – 2t)

• (iv) -5l²m – 10lmn

Taking -5lm common

-5l²m – 10lmn = -5lm(l + 2n) or -5lm(m + 2ln)

• (v) 5m² – 5n²

Taking 5 common first gives 5(m² – n²). Applying identity a² – b² = (a + b)(a – b)

5m² – 5n² = 5(m + n)(m – n)

(i) 2x²y + 6xy + 4x + 12**

Grouping first two terms and last two terms

2xy(x + 3) + 4(x + 3) = (x + 3)(2xy + 4) = 2(x + 3)(xy + 2) or 2(x + 3y)(xy + 2)

• (ii) 5m²n – 10mn² + 12m – 24n

Grouping into pairs

5mn(m – 2n) + 12(m – 2n) = (m – 2n)(5mn + 12)

• (iii) 6x³ + 8x² + 9xy + 12y

Grouping first two terms and last two terms

2x²(3x + 4) + 3y(3x + 4) = (3x + 4)(2x² + 3y)

• (iv) 15x⁴ + 10x²y² + 12x²y + 8y³

Grouping into pairs

5x²(3x² + 2y²) + 4y(3x² + 2y²) = (5x² + 4y)(3x² + 2y²)

• (v) x(x + 3) + 8(x + 3)

Taking binomial (x + 3) as common factor

(x + 3)(x + 8)

• (vi) 3x(x – 4) – 5(x – 4)

Taking binomial (x – 4) as common factor

(x – 4)(3x – 5)

• (vii) 2m(l – m) + 3(l – m)

Taking binomial (l – m) as common factor

(l – m)(2m + 3)

(i) x(1 – 3y²x)**

Expanding by multiplying x into the parentheses

x · 1 – x · 3y²x = x – 3y²x² or x – 3y²x

• (ii) -17x(3x² – 9x)

Multiplying -17x inside

-17x · 3x² – (-17x) · 9x = -51x³ + 153x²

• (iii) 2a²(3a – 4a²)

Multiplying 2a² inside

2a² · 3a – 2a² · 4a² = 6a³ – 8a⁴

• (iv) 9m(m – n)

Multiplying 9m inside

9m · m – 9m · n = 9m² – 9mn

• (v) 9t²(t – 7t³)

Multiplying 9t² inside

9t² · t – 9t² · 7t³ = 9t³ – 63t⁵