Looking for authentic SEBA Class 10 Maths Chapter 6 Revision 4 solutions for the 2026–27 academic year? Access complete step-by-step solved exercise problems, textual question answers, and detailed mathematical explanations to help you score top marks in your HSLC board exams.

Exercise R-4

Find common factors of the following :

(i) 14 pq , 28 p²q²

(ii) 16x³ – 4x² , 32x

(iii) 20pq , 30qr , 40rp

(iv) 3x²y³ , 10x³y² , 6x²y²z

(i)

Solution:

14pq = 2 × 7 × p × q

28p²q² = 2 × 2 × 7 × p × p × q × q

Common factors = 2 × 7 × p × q = 14pq

Ans: 14pq

(ii)

Solution: 16x³ = 2 × 2 × 2 × 2 × x × x × x

-4x² = -1 × 2 × 2 × x × x

32x = 2 × 2 × 2 × 2 × 2 × x

Common factors = 2 × 2 × x = 4x

Ans: 4x

(iii)

Solution: 20pq = 2 × 2 × 5 × p × q

30qr = 2 × 3 × 5 × q × r

40rp = 2 × 2 × 2 × 5 × r × p

Common factors = 2 × 5 = 10

Ans: 10

(iv)

Solution: 3x²y³ = 3 × x × x × y × y × y

10x³y² = 2 × 5 × x × x × x × y × y

6x²y²z = 2 × 3 × x × x × y × y × z

Common factors = x × x × y × y = x²y²

Ans: x²y²

(i)

Solution:

4a² + 8a³

= 4a²(1) + 4a²(2a)

= 4a²(1 + 2a)

Ans: 4a²(1 + 2a)

(ii)

Solution:

7x²y – 21xy²

= 7xy(x) – 7xy(3y)

= 7xy(x – 3y)

Ans: 7xy(x – 3y)

(iii)

Solution:

a²bc + ab²c + abc²

= abc(a) + abc(b) + abc(c)

= abc(a + b + c)

Ans: abc(a + b + c)

(iv

Solution:

a³ – a²b²

= a²(a) – a²(b²)

= a²(a – b²)

Ans: a²(a – b²)

SEBA Class 10 Maths Revision 4 Exercise Solutions

HSLC Class 10 Maths Chapter 6 Revision 4 Textual Solutions

Factorise

(i) x² + xy + 6x + 6y

(ii) xy + x + y + 1

(iii) 24x²y + 12x² – 12xy – 6x

(iv) z – 7 + 7xy – xyz

(i)

Solution:

x² + xy + 6x + 6y

= x(x + y) + 6(x + y)

= (x + y)(x + 6)

Ans: (x + y)(x + 6)

(ii)

Solution:

xy + x + y + 1

= x(y + 1) + 1(y + 1)

= (y + 1)(x + 1)

Ans: (y + 1)(x + 1)

(iii)

Solution:

24x²y + 12x² – 12xy – 6x

= 6x(4xy + 2x – 2y – 1)

= 6x [2x(2y + 1) – 1(2y + 1)]

= 6x(2y + 1)(2x – 1)

Ans: 6x(2y + 1)(2x – 1)

(iv)

Solution:

z – 7 + 7xy – xyz

= (z – 7) + 7xy – xyz

= 1(z – 7) – xy(z – 7)

= (z – 7)(1 – xy)

Ans: (z – 7)(1 – xy)

SEBA Class 10 Maths Chapter 6 Revision 4 Solutions

Class 10 Maths Chapter 6 Revision 4 Question Answer SEBA 2026-27

Express in Factors

(i) 4x² + 12x + 9

(ii) 25m² + 30m + 9

(iii) x² – 10x + 25

(iv) 121b² – 88bc + 16c²

(v) 9p² – 16q²

(vi) (l + m)² – (l – m)²

(vii) x² – 13x – 30

(viii) y² – 5y – 36

(ix) 4y² + 25y – 21

(x) 3x⁶ – 6x²y – 45x²y²

(i)

Solution:

4x² + 12x + 9

= 4x² + (6 + 6)x + 9

= 4x² + 6x + 6x + 9

= 2x(2x + 3) + 3(2x + 3)

= (2x + 3)(2x + 3)

Ans: (2x + 3)(2x + 3)

(ii)

Solution:

25m² + 30m + 9

 = 25m² + (15 + 15)m + 9

= 25m² + 15m + 15m + 9

= 5m(5m + 3) + 3(5m + 3)

= (5m + 3)(5m + 3)

Ans: (5m + 3)(5m + 3)

(iii) Given

Solution:

x² – 10x + 25

= x² – (5 + 5)x + 25

= x² – 5x – 5x + 25

= x(x – 5) – 5(x – 5)

= (x – 5)(x – 5)

Ans: (x – 5)(x – 5)

(iv)

Solution:

121b² – 88bc + 16c²

= (11b)² – 2(11b)(4c) + (4c)²

= (11b – 4c)²

= (11b – 4c)(11b – 4c)

Ans: (11b – 4c)(11b – 4c)

(v)

Solution:

9p² – 16q²

= (3p)² – (4q)²

= (3p – 4q)(3p + 4q)

Ans: (3p – 4q)(3p + 4q)

(vi)

Solution:

(l + m)² – (l – m)²

= [(l + m) – (l – m)][(l + m) + (l – m)]

= (l + m – l + m)(l + m + l – m)

= (2m)(2l) = 4lm

Ans: 4lm

(vii)

Solution:

x² – 13x – 30

= x² – 15x + 2x – 30

= x(x – 15) + 2(x – 15)

= (x – 15)(x + 2)

Ans: (x – 15)(x + 2)

(viii)

Solution:

y² – 5y – 36

= y² – 9y + 4y – 36

= y(y – 9) + 4(y – 9)

= (y – 9)(y + 4)

Ans: (y – 9)(y + 4)

(ix)

Solution: 4y² + 25y – 21

= 4y² + 28y – 3y – 21

= 4y(y + 7) – 3(y + 7)

= (y + 7)(4y – 3)

Ans: (y + 7)(4y – 3)

(x)

Solution:

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