SEBA Class 10 Maths Revision Exercise R-4 Solutions : Factorisation | Revision Chapter
Get free and reliable SEBA Class 10 Maths Revision Exercise R-4 Solutions: Factorisation from the latest SCERT Assam textbook 2026.
See More:
- SEBA Class 10 Maths Chapter R-1 Solutions: Square and Square Root
- SEBA Class 10 Maths Chapter R-2 Solutions: Cube and Cube Root
- SEBA Class 10 Maths Chapter R-3 Solutions: Indices and Power
- SEBA Class 10 Maths Chapter R-4 Solutions: Factorisation
- SEBA Class 10 Maths Chapter R-5 Solutions: Congruence of Triangles
1. Find common factors of the following:
(i) $14pq, \; 28p^2q^2$
$\mathbf{Sol^n.}$
$14pq = 2 \times 7 \times p \times q$
$28p^2q^2 = 2 \times 2 \times 7 \times p \times p \times q \times q$
∴ Common factors $= 2 \times 7 \times p \times q = 14pq$
(ii) $16x^3, \; -4x^2, \; 32x$
$\mathbf{Sol^n.}$
$16x^3 = 2 \times 2 \times 2 \times 2 \times x \times x \times x$
$-4x^2 = -1 \times 2 \times 2 \times x \times x$
$32x = 2 \times 2 \times 2 \times 2 \times 2 \times x$
∴ Common factors $= 2 \times 2 \times x = 4x$
(iii) $20pq, \; 30qr, \; 40rp$
$\mathbf{Sol^n.}$
$20pq = 2 \times 2 \times 5 \times p \times q$
$30qr = 2 \times 3 \times 5 \times q \times r$
$40rp = 2 \times 2 \times 2 \times 5 \times r \times p$
∴ Common factor $= 2 \times 5 = 10$
(iv) $3x^2y^3, \; 10x^3y^2, \; 6x^2y^2z$
$\mathbf{Sol^n.}$
$3x^2y^3 = 3 \times x^2 \times y^3$
$10x^3y^2 = 2 \times 5 \times x^3 \times y^2$
$6x^2y^2z = 2 \times 3 \times x^2 \times y^2 \times z$
$\therefore$ Common factor is $x^2y^2$.
2. Factorise:
(i) $4a^2 + 8a^3$
$\mathbf{Sol^n.}$
$4a^2 + 8a^3$
$= 4a^2(1 + 2a)$
(ii) $7x^2y – 21xy^2$
$\mathbf{Sol^n.}$
$7x^2y – 21xy^2$
$= 7xy(x – 3y)$
(iii) $a^2bc + ab^2c + abc^2$
$\mathbf{Sol^n.}$
$a^2bc + ab^2c + abc^2$
$= abc(a + b + c)$
(iv) $a^3 – a^2b^2$
$\mathbf{Sol^n.}$
$a^3 – a^2b^2$
$= a^2(a – b^2)$
3. Factorise:
(i) $x^2 + xy + 6x + 6y$
$\mathbf{Sol^n.}$
$x^2 + xy + 6x + 6y$
$= x(x + y) + 6(x + y)$
$= (x + y)(x + 6)$
(ii) $xy + x + y + 1$
$\mathbf{Sol^n.}$
$xy + x + y + 1$
$= x(y + 1) + 1(y + 1)$
$= (y + 1)(x + 1)$
(iii) $24x^2y + 12x^2 – 12xy – 6x$
$\mathbf{Sol^n.}$
$24x^2y + 12x^2 – 12xy – 6x$
$= 6x(4xy + 2x – 2y – 1)$
$= 6x[2x(2y + 1) – 1(2y + 1)]$
$= 6x(2y + 1)(2x – 1)$
(iv) $z – 7 + 7xy – xyz$
$\mathbf{Sol^n.}$
$z – 7 + 7xy – xyz$
$= z – xyz – 7 + 7xy$
$= z(1 – xy) – 7(1 – xy)$
$= (1 – xy)(z – 7)$
(or $(xy – 1)(7 – z)$)
4. Express in factors:
(i) $4x^2 + 12x + 9$
$\mathbf{Sol^n.}$
$4x^2 + 12x + 9$
$= (2x)^2 + 2(2x)(3) + (3)^2$
$= (2x + 3)^2$
$= (2x + 3)(2x + 3)$
(ii) $25m^2 + 30m + 9$
$\mathbf{Sol^n.}$
$25m^2 + 30m + 9$
$= (5m)^2 + 2(5m)(3) + (3)^2$
$= (5m + 3)^2$
$= (5m + 3)(5m + 3)$
(iii) $x^2 – 10x + 25$
$\mathbf{Sol^n.}$
$x^2 – 10x + 25$
$= (x)^2 – 2(x)(5) + (5)^2$
$= (x – 5)^2$
$= (x – 5)(x – 5)$
(iv) $121b^2 – 88bc + 16c^2$
$\mathbf{Sol^n.}$
$121b^2 – 88bc + 16c^2$
$= (11b)^2 – 2(11b)(4c) + (4c)^2$
$= (11b – 4c)^2$
$= (11b – 4c)(11b – 4c)$
(v) $9p^2 – 16q^2$
$\mathbf{Sol^n.}$
$9p^2 – 16q^2$
$= (3p)^2 – (4q)^2$
$= (3p + 4q)(3p – 4q)$
(vi) $(l + m)^2 – (l – m)^2$
$\mathbf{Sol^n.}$
Using identity: $a^2 – b^2 = (a + b)(a – b)$
Here, $a = (l + m)$ and $b = (l – m)$
$= [(l + m) + (l – m)][(l + m) – (l – m)]$
$= (l + m + l – m)(l + m – l + m)$
$= (2l)(2m)$
$= 4lm$
(vii) $x^2 – 13x – 30$
$\mathbf{Sol^n.}$
$x^2 – 13x – 30$
$= x^2 – (15 – 2)x – 30$
$= x^2 – 15x + 2x – 30$
$= x(x – 15) + 2(x – 15)$
$= (x – 15)(x + 2)$
(viii) $y^2 – 5y – 36$
$\mathbf{Sol^n.}$
$y^2 – 5y – 36$
$= y^2 – (9 – 4)y – 36$
$= y^2 – 9y + 4y – 36$
$= y(y – 9) + 4(y – 9)$
$= (y – 9)(y + 4)$
(ix) $4y^2 + 25y – 21$
$\mathbf{Sol^n.}$
$4y^2 + 25y – 21$
$= 4y^2 + (28 – 3)y – 21$
$= 4y^2 + 28y – 3y – 21$
$= 4y(y + 7) – 3(y + 7)$
$= (y + 7)(4y – 3)$
(x) $3x^6 – 6x^2y – 45x^2y^2$
$\mathbf{Sol^n.}$
$3x^6 – 6x^2y – 45x^2y^2$
$= 3x^2(x^4 – 2y – 15y^2)$