NCERT Solutions Maths Ch-14 Factorisation for Class 8th
Chapter 14: Factorisation from NCERT Mathematics for Class 8
Factorisation : Factorisation is the process of breaking down an expression into a product of its factors.
- Factors: Numbers or expressions that can be multiplied together to get another number or expression.
- Algebraic Expressions: Expressions that involve variables (like xx and yy) and constants (like numbers).
NCERT Solutions of Class 8th Chapter 14 Factorisation Exercise 14.1, 14.2 and 14.3
We try to teach you all Questions in easy way. We solve all chapter wise sums of maths textbook. In every chapter include NCERT solutions. For solutions of Exercise 14.1, 14.2 and 14.3 click on Tabs :
Question 1.
Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28p2q2
(iv) 2x, 3x2, 4
(v) 6abc, 24ab2, 12a2b
(vi) 16x3, -4x2, 32x
(vii) 10pq, 20qr, 30rp
(viii) 3x2y3, 10x3y2, 6x2y2z
Solution:
(i) 12x, 36
(2 × 2 × 3 × x) and (2 × 2 × 3 × 3)
Common factors are 2 × 2 × 3 = 12
Hence, the common factor = 12
(ii) 2y, 22xy
= (2 × y) and (2 × 11 × x × y)
Common factors are 2 × y = 2y
Hence, the common factor = 2y
(iii) 14pq, 28p2q2
= (2 × 7 × p × q) and (2 × 2 × 7 × p × p × q × q)
Common factors are 2 × 7 × p × q = 14pq
Hence, the common factor = 14pq
(iv) 2x, 3x2, 4
= (2 × x), (3 × x × x) and (2 × 2)
Common factor is 1
Hence, the common factor = 1 [∵ 1 is a factor of every number]
(v) 6abc, 24ab2, 12a2b
= (2 × 3 × a × b × c), (2 × 2 × 2 × 3 × a × b × b) and (2 × 2 × 3 × a × a × b)
Common factors are 2 × 3 × a × b = 6ab
Hence, the common factor = 6ab
(vi) 16x3, -4x2, 32x
= (2 × 2 × 2 × 2 × x × x × x), -(2 × 2 × x × x), (2 × 2 × 2 × 2 × 2 × x)
Common factors are 2 × 2 × x = 4x
Hence, the common factor = 4x
(vii) 10pq, 20qr, 30rp
= (2 × 5 × p × q), (2 × 2 × 5 × q × r), (2 × 3 × 5 × r × p)
Common factors are 2 × 5 = 10
Hence, the common factor = 10
(viii) 3x2y2, 10x3y2, 6x2y2z
= (3 × x × x × y × y), (2 × 5 × x × x × x × y × y), (2 × 3 × x × x × y × y × z)
Common factors are x × x × y × y = x2y2
Hence, the common factor = x2y2.
Question 2.
Factorise the following expressions.
(i) 7x – 42
(ii) 6p – 12q
(iii) 7a2 + 14a
(iv) -16z + 20z3
(v) 20l2m + 30alm
(vi) 5x2y – 15xy2
(vii) 10a2 – 15b2 + 20c2
(viii) -4a2 + 4ab – 4ca
(ix) x2yz + xy2z + xyz2
(x) ax2y + bxy2 + cxyz
Solution:
(i) 7x – 42 = 7(x – 6)
(ii) 6p – 12q = 6(p – 2q)
(iii) 7a2 + 14a = 7a(a + 2)
(iv) -16z + 20z3 = 4z(-4 + 5z2)
(v) 20l2m + 30alm = 10lm(2l + 3a)
(vi) 5x2y – 15xy2 = 5xy(x – 3y)
(vii) 10a2 – 15b2 + 20c2 = 5(2a2 – 3b2 + 4c2)
(viii) -4a2 + 4ab – 4ca = 4a(-a + b – c)
(ix) x2yz + xy2z + xyz2 = xyz(x + y + z)
(x) ax2y + bxy2 + cxyz = xy(ax + by + cz)
Question 3.
Factorise:
(i) x2 + xy + 8x + 8y
(ii) 15xy – 6x + 5y – 2
(iii) ax + bx – ay – by
(iv) 15pq + 15 + 9q + 25p
(v) z – 7 + 7xy – xyz
Solution:
(i) x2 + xy + 8x + 8y
Grouping the terms, we have
x2 + xy + 8x + 8y
= x(x + y) + 8(x + y)
= (x + y)(x + 8)
Hence, the required factors = (x + y)(x + 8)
(ii) 15xy – 6x + 5y – 2
Grouping the terms, we have
(15xy – 6x) + (5y – 2)
= 3x(5y – 2) + (5y – 2)
= (5y – 2)(3x + 1)
(iii) ax + bx – ay – by
Grouping the terms, we have
= (ax – ay) + (bx – by)
= a(x – y) + b(x – y)
= (x – y)(a + b)
Hence, the required factors = (x – y)(a + b)
(iv) 15pq + 15 + 9q + 25p
Grouping the terms, we have
= (15pq + 25p) + (9q + 15)
= 5p(3q + 5) + 3(3q + 5)
= (3q + 5) (5p + 3)
Hence, the required factors = (3q + 5) (5p + 3)
(v) z – 7 + 7xy – xyz
Grouping the terms, we have
= (-xyz + 7xy) + (z – 7)
= -xy(z – 7) + 1 (z – 7)
= (-xy + 1) (z – 1)
Hence the required factor = -(1 – xy) (z – 7)
Question 1.
Factorise the following expressions.
(i) a2 + 8a +16
(ii) p2 – 10p + 25
(iii) 25m2 + 30m + 9
(iv) 49y2 + 84yz + 36z2
(v) 4x2 – 8x + 4
(vi) 121b2 – 88bc + 16c2
(vii) (l + m)2 – 4lm. (Hint: Expand (l + m)2 first)
(viii) a4 + 2a2b2 + b4
Solution:
(i) a2 + 8a + 16
Here, 4 + 4 = 8 and 4 × 4 = 16
a2 + 8a +16
= a2 + 4a + 4a + 4 × 4
= (a2 + 4a) + (4a + 16)
= a(a + 4) + 4(a + 4)
= (a + 4) (a + 4)
= (a + 4)2
(ii) p2 – 10p + 25
Here, 5 + 5 = 10 and 5 × 5 = 25
p2 – 10p + 25
= p2 – 5p – 5p + 5 × 5
= (p2 – 5p) + (-5p + 25)
= p(p – 5) – 5(p – 5)
= (p – 5) (p – 5)
= (p – 5)2
(iii) 25m2 + 30m + 9
Here, 15 + 15 = 30 and 15 × 15 = 25 × 9 = 225
25m2 + 30m + 9
= 25m2 + 15m + 15m + 9
= (25m2 + 15m) + (15m + 9)
= 5m(5m + 3) + 3(5m + 3)
= (5m + 3) (5m + 3)
= (5m + 3)2
(iv) 49y2 + 84yz + 36z2
Here, 42 + 42 = 84 and 42 × 42 = 49 × 36 = 1764
49y2 + 84yz + 36z2
= 49y2 + 42yz + 42yz + 36z2
= 7y(7y + 6z) +6z(7y + 6z)
= (7y + 6z) (7y + 6z)
= (7y + 6z)2
(v) 4x2 – 8x + 4
= 4(x2 – 2x + 1) [Taking 4 common]
= 4(x2 – x – x + 1)
= 4[x(x – 1) -1(x – 1)]
= 4(x – 1)(x – 1)
= 4(x – 1)2
(vi) 121b2 – 88bc + 16c2
Here, 44 + 44 = 88 and 44 × 44 = 121 × 16 = 1936
121b2 – 88bc + 16c2
= 121b2 – 44bc – 44bc + 16c2
= 11b(11b – 4c) – 4c(11b – 4c)
= (11b – 4c) (11b – 4c)
= (11b – 4c)2
(vii) (l + m)2 – 4lm
Expanding (l + m)2, we get
l2 + 2lm + m2 – 4lm
= l2 – 2lm + m2
= l2 – Im – lm + m2
= l(l – m) – m(l – m)
= (l – m) (l – m)
= (l – m)2
(viii) a4 + 2a2b2 + b4
= a4 + a2b2 + a2b2 + b4
= a2(a2 + b2) + b2(a2 + b2)
= (a2 + b2)(a2 + b2)
= (a2 + b2)2
Question 2.
Factorise.
(i) 4p2 – 9q2
(ii) 63a2 – 112b2
(iii) 49x2 – 36
(iv) 16x5 – 144x3
(v) (l + m)2 – (l – m)2
(vi) 9x2y2 – 16
(vii) (x2 – 2xy + y2) – z2
(viii) 25a2 – 4b2 + 28bc – 49c2
Solution:
(i) 4p2 – 9q2
= (2p)2 – (3q)2
= (2p – 3q) (2p + 3q)
[∵ a2 – b2 = (a + b)(a – b)]
(ii) 63a2 – 112b2
= 7(9a2 – 16b2)
= 7 [(3a)2 – (4b)2]
= 7(3a – 4b)(3a + 4b)
[∵ a2 – b2 = (a + b)(a – b)]
(iii) 49x2 – 36 = (7x)2 – (6)2
= (7x – 6) (7x + 6)
[∵ a2 – b2 = (a + b)(a – b)]
(iv) 16x5 – 144x3 = 16x3 (x2 – 9)
= 16x3 [(x)2 – (3)2]
= 16x3(x – 3)(x + 3)
[∵ a2 – b2 = (a + b)(a – b)]
(v) (l + m)2 – (l – m)2
= (l + m) – (l – m)] [(l + m) + (l – m)]
[∵ a2 – b2 = (a + b)(a – b)]
= (l + m – l + m)(l + m + l – m)
= (2m) (2l)
= 4ml
(vi) 9x2y2 – 16 = (3xy)2 – (4)2
= (3xy – 4)(3xy + 4)
[∵ a2 – b2 = (a + b)(a – b)]
(vii) (x2 – 2xy + y2) – z2
= (x – y)2 – z2
= (x – y – z) (x – y + z)
[∵ a2 – b2 = (a + b)(a – b)]
(viii) 25a2 – 4b2 + 28bc – 49c2
= 25a2 – (4b2 – 28bc + 49c2)
= (5a)2 – (2b – 7c)2
= [5a – (2b – 7c)] [5a + (2b – 7c)]
= (5a – 2b + 7c)(5a + 2b – 7c)
Question 3.
Factorise the expressions.
(i) ax2 + bx
(ii) 7p2 + 21q2
(iii) 2x3 + 2xy2 + 2xz2
(iv) am2 + bm2 + bn2 + an2
(v) (lm + l) + m + 1
(vi) y(y + z) + 9(y + z)
(vii) 5y2 – 20y – 8z + 2yz
(viii) 10ab + 4a + 5b + 2
(ix) 6xy – 4y + 6 – 9x
Solution:
(i) ax2 + bx = x(ax + 5)
(ii) 7p2 + 21q2 = 7(p2 + 3q2)
(iii) 2x3 + 2xy2 + 2xz2 = 2x(x2 + y2 + z2)
(iv) am2 + bm2 + bn2 + an2
= m2 (a + b) + n2(a + b)
= (a + b)(m2 + n2)
(v) (lm + l) + m + 1
= l(m + 1) + (m + 1)
= (m + 1) (l + 1)
(vi) y(y + z) + 9(y + z) = (y + z)(y + 9)
(vii) 5y2 – 20y – 8z + 2yz
= 5y2 – 20y + 2yz – 8z
= 5y(y – 4) + 2z(y – 4)
= (y – 4) (5y + 2z)
(viii) 10ab + 4a + 5b + 2
= 2a(5b + 2) + 1(5b + 2)
= (5b + 2)(2a + 1)
(ix) 6xy – 4y + 6 – 9x
= 6xy – 4y – 9x + 6
= 2y(3x – 2) – 3(3x – 2)
= (3x – 2) (2y – 3)
Question 4.
Factorise.
(i) a4 – b4
(ii) p4 – 81
(iii) x4 – (y + z)4
(iv) x4 – (x – z)4
(v) a4 – 2a2b2 + b4
Solution:
(i) a4 – b4 – (a2)2 – (b2)2
[∵ a2 – b2 = (a – b)(a + b)]
= (a2 – b2) (a2 + b2)
= (a – b) (a + b) (a2 + b2)
(ii) p4 – 81 = (p2)2 – (9)2
= (p2 – 9) (p2 + 9)
[∵ a2 – b2 = (a – b)(a + b)]
= (p – 3)(p + 3) (p2 + 9)
(iii) x4 – (y + z)4 = (x2)2 – [(y + z)2]2
[∵ a2 – b2 = (a – b)(a + b)]
= [x2 – (y + z)2] [x2 + (y + z)2]
= [x – (y + z)] [x + (y + z)] [x2 + (y + z)2]
= (x – y – z) (x + y + z) [x2 + (y + z)2]
(iv) x4 – (x – z)4 = (x2)2 – [(y – z)2]2
= [x2 – (y – z)2] [x2 + (y – z)2]
= (x – y + z) (x + y – z) (x2 + (y – z)2]
(v) a4 – 2a2b2 + b4
= a4 – a2b2 – a2b2 + b4
= a2(a2 – b2) – b2(a2 – b2)
= (a2 – b2)(a2 – b2)
= (a2 – b2)2
= [(a – b) (a + b)]2
= (a – b)2 (a + b)2
Question 5.
Factorise the following expressions.
(i) p2 + 6p + 8
(ii) q2 – 10q + 21
(iii) p2 + 6p – 16
Solution:
(i) p2 + 6p + 8
Here, 2 + 4 = 6 and 2 × 4 = 8
p2 + 6p + 8
= p2 + 2p + 4p + 8
= p (p + 2) + 4(p + 2)
= (p + 2) (p + 4)
(ii) q2 – 10q + 21
Here, 3 + 7 = 10 and 3 × 7 = 21
q2 – 10q + 21
= q2 – 3q – 7q + 21
= q(q – 3) – 7(q – 3)
= (q – 3) (q – 7)
(iii) p2 + 6p – 16
Here, 8 – 2 = 6 and 8 × 2 = 16
p2 + 6p – 16
= p2 + 8p – 2p – 16
= p(p + 8) – 2(p + 8)
= (p + 8) (p – 2)
Question 1.
Carry out the following divisions.
(i) 28x4 ÷ 56x
(ii) -36y3 ÷ 9y2
(iii) 66pq2r3 ÷ 11qr2
(iv) 34x3y3z3 ÷ 51xy2z3
(v) 12a8b8 ÷ (-6a6b4)
Solution:
Question 2.
Divide the following polynomial by the given monomial.
(i) (5x2 – 6x) ÷ 3x
(ii) (3y8 – 4y6 + 5y4) ÷ y4
(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2
(iv) (x3 + 2x2 + 3x) ÷ 2x
(v) (p3q6 – p6q3) ÷ p3q3
Solution:
Question 3.
Work out the following divisions.
(i) (10x – 25) ÷ 5
(ii) (10x – 25) ÷ (2x – 5)
(iii) 10y(6y + 21) ÷ 5(2y + 7)
(iv) 9x2y2(3z – 24) ÷ 27xy(z – 8)
(v) 96abc(3a – 12) (5b – 30) ÷ 144(a – 4)(b – 6)
Solution:
Question 4.
Divide as directed.
(i) 5(2x + 1) (3x + 5) ÷ (2x + 1)
(ii) 26xy (x + 5)(y – 4) ÷ 13x(y – 4)
(iii) 52pqr(p + q) (q + r) (r + p) ÷ 104pq(q + r)(r + p)
(iv) 20(y + 4)(y2 + 5y + 3) ÷ 5(y + 4)
(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)
Solution:
Question 5.
Factorise the expressions and divide them as directed.
(i) (y2 + 7y + 10) ÷ (y + 5)
(ii) (m2 – 14m – 32) ÷ (m + 2)
(iii) (5p2 – 25p + 20) ÷ (p – 1)
(iv) 4yz(z2 + 6z – 16) ÷ 2y(z + 8)
(v) 5pq(p2 – q2) ÷ 2p(p + q)
(vi) 12xy(9x2 – 16y2) ÷ 4xy(3x + 4y)
(vii) 39y3(50y2 – 98) ÷ 26y2(5y + 7)
Solution:
