The NCERT Solutions aim to facilitate students in mastering fundamental concepts and gaining a comprehensive understanding of the diverse question types encountered in CBSE Class 8 Mathematics Examinations.
Students can access the NCERT solutions for class 8 maths chapter 14 Factorisation. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Factorise the following expressions.
(i) 7x–42
(ii) 6p–12q
(iii) 7a2+ 14a
(iv) -16z+20 z3
(v) 20l2m+30alm
(vi) 5x2y-15xy2
(vii) 10a2-15b2+20c2
(viii) -4a2+4ab–4 ca
(ix) x2yz+xy2z +xyz2
(x) ax2y+bxy2+cxyz
(vii) 10a2-15b2+20c2
10a2 = 2×5×a×a
– 15b2 = -1×3×5×b×b
20c2 = 2×2×5×c×c
Common factor of 10 a2 , 15b2 and 20c2 is 5
10a2-15b2+20c2 = 5(2a2-3b2+4c2 )
(viii) – 4a2+4ab-4ca
– 4a2 = -1×2×2×a×a
4ab = 2×2×a×b
– 4ca = -1×2×2×c×a
Common factor of – 4a2 , 4ab , – 4ca are 2, 2, a i.e. 4a
So,
– 4a2+4 ab-4 ca = 4a(-a+b-c)
(ix) x2yz+xy2z+xyz2
x2yz = x×x×y×z
xy2z = x×y×y×z
xyz2 = x×y×z×z
Common factor of x2yz , xy2z and xyz2 are x, y, z i.e. xyz
Now, x2yz+xy2z+xyz2 = xyz(x+y+z)
(x) ax2y+bxy2+cxyz
ax2y = a×x×x×y
bxy2 = b×x×y×y
cxyz = c×x×y×z
Common factors of a x2y ,bxy2 and cxyz are xy
Now, ax2y+bxy2+cxyz = xy(ax+by+cz)
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
Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14 pq, 28p2q2
(iv) 2x, 3x2, 4
(v) 6 abc, 24ab2, 12a2b
(vi) 16 x3, – 4x2 , 32 x
(vii) 10 pq, 20qr, 30 rp
(viii) 3x2y3 , 10x3y2 , 6x2y2z
(i) Factors of 12x and 36
12x = 2×2×3×x
36 = 2×2×3×3
Common factors of 12x and 36 are 2, 2, 3
and , 2×2×3 = 12
(ii) Factors of 2y and 22xy
2y = 2×y
22xy = 2×11×x×y
Common factors of 2y and 22xy are 2, y
and ,2×y = 2y
(iii) Factors of 14pq and 28p2q2
14pq = 2x7xpxq
28p2q2 = 2x2x7xpxpxqxq
Common factors of 14 pq and 28 p2q2 are 2, 7 , p , q
and, 2x7xpxq = 14pq
(iv) Factors of 2x, 3x2and 4
2x = 2×x
3x2= 3×x×x
4 = 2×2
Common factors of 2x, 3x2 and 4 is 1.
(v) Factors of 6abc, 24ab2 and 12a2b
6abc = 2×3×a×b×c
24ab2 = 2×2×2×3×a×b×b
12 a2 b = 2×2×3×a×a×b
Common factors of 6 abc, 24ab2 and 12a2b are 2, 3, a, b
and, 2×3×a×b = 6ab
(vi) Factors of 16x3 , -4x2and 32x
16 x3 = 2×2×2×2×x×x×x
– 4x2 = -1×2×2×x×x
32x = 2×2×2×2×2×x
Common factors of 16 x3 , – 4x2 and 32x are 2,2, x
and, 2×2×x = 4x
(vii) Factors of 10 pq, 20qr and 30rp
10 pq = 2×5×p×q
20qr = 2×2×5×q×r
30rp= 2×3×5×r×p
Common factors of 10 pq, 20qr and 30rp are 2, 5
and, 2×5 = 10
(viii) Factors of 3x2y3 , 10x3y2 and 6x2y2z
3x2y3 = 3×x×x×y×y×y
10x3 y2 = 2×5×x×x×x×y×y
6x2y2z = 3×2×x×x×y×y×z
Common factors of 3x2y3, 10x3y2 and 6x2y2z are x2, y2
and, x2×y2 = x2y2
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
(i) a2+8a+16
= a2+2×4×a+42
= (a+4)2
Using the identity (x+y)2 = x2+2xy+y2
(ii) p2–10p+25
= p2-2×5×p+52
= (p-5)2
Using the identity (x-y)2 = x2-2xy+y2
(iii) 25m2+30m+9
= (5m)2+2×5m×3+32
= (5m+3)2
Using the identity (x+y)2 = x2+2xy+y2
(iv) 49y2+84yz+36z2
=(7y)2+2×7y×6z+(6z)2
= (7y+6z)2
Using the identity (x+y)2 = x2+2xy+y2
(v) 4x2–8x+4
= (2x)2-2×4x+22
= (2x-2)2
Using the identity (x-y)2 = x2-2xy+y2
(vi) 121b2-88bc+16c2
= (11b)2-2×11b×4c+(4c)2
= (11b-4c)2
Using the identity (x-y)2 = x2-2xy+y2
(vii) (l+m)2-4lm (Hint: Expand (l+m)2 first)
Expand (l+m)2 using the identity (x+y)2 = x2+2xy+y2
(l+m)2-4lm = l2+m2+2lm-4lm
= l2+m2-2lm
= (l-m)2
Using the identity (x-y)2 = x2-2xy+y2
(viii) a4+2a2b2+b4
= (a2)2+2×a2×b2+(b2)2
= (a2+b2)2
Using the identity (x+y)2 = x2+2xy+y2
Factorise.
(i) 4p2–9q2
(ii) 63a2–112b2
(iii) 49x2–36
(iv) 16x5–144x3 differ
(v) (l+m)2-(l-m) 2
(vi) 9x2y2–16
(vii) (x2–2xy+y2)–z2
(viii) 25a2–4b2+28bc–49c2
(i) 4p2–9q2
= (2p)2-(3q)2
= (2p-3q)(2p+3q)
Using the identity x2-y2 = (x+y)(x-y)
(ii) 63a2–112b2
= 7(9a2 –16b2)
= 7((3a)2–(4b)2)
= 7(3a+4b)(3a-4b)
Using the identity x2-y2 = (x+y)(x-y)
(iii) 49x2–36
= (7x)2 -62
= (7x+6)(7x–6)
Using the identity x2-y2 = (x+y)(x-y)
(iv) 16x5–144x3
= 16x3(x2–9)
= 16x3(x2–9)
= 16x3(x–3)(x+3)
Using the identity x2-y2 = (x+y)(x-y)
(v) (l+m) 2-(l-m) 2
= {(l+m)-(l–m)}{(l +m)+(l–m)}
Using the identity x2-y2 = (x+y)(x-y)
= (l+m–l+m)(l+m+l–m)
= (2m)(2l)
= 4 ml
(vi) 9x2y2–16
= (3xy)2-42
= (3xy–4)(3xy+4)
Using the identity x2-y2 = (x+y)(x-y)
(vii) (x2–2xy+y2)–z2
= (x–y)2–z2
Using the identity (x-y)2 = x2-2xy+y2
= {(x–y)–z}{(x–y)+z}
= (x–y–z)(x–y+z)
Using the identity x2-y2 = (x+y)(x-y)
(viii) 25a2–4b2+28bc–49c2
= 25a2–(4b2-28bc+49c2 )
= (5a)2-{(2b)2-2(2b)(7c)+(7c)2}
= (5a)2-(2b-7c)2
Using the identity x2-y2 = (x+y)(x-y) , we have
= (5a+2b-7c)(5a-2b+7c)
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
(i) ax2+bx = x(ax+b)
(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 = lm+m+l+1 = m(l+1)+(l+1) = (m+1)(l+1)
(vi) y(y+z)+9(y+z) = (y+9)(y+z)
(vii) 5y2–20y–8z+2yz = 5y(y–4)+2z(y–4) = (y–4)(5y+2z)
(viii) 10ab+4a+5b+2 = 5b(2a+1)+2(2a+1) = (2a+1)(5b+2)
(ix) 6xy–4y+6–9x = 6xy–9x–4y+6 = 3x(2y–3)–2(2y–3) = (2y–3)(3x–2)
Factorise.
(i) a4–b4
(ii) p4–81
(iii) x4–(y+z) 4
(iv) x4–(x–z) 4
(v) a4–2a2b2+b4
(i) a4–b4
= (a2)2-(b2)2
= (a2-b2) (a2+b2)
= (a – b)(a + b)(a2+b2)
(ii) p4–81
= (p2)2-(9)2
= (p2-9)(p2+9)
= (p2-32)(p2+9)
=(p-3)(p+3)(p2+9)
(iii) x4–(y+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}
= (x–y–z)(x+y+z) {x2+(y+z)2}
(iv) x4–(x–z) 4 = (x2)2-{(x-z)2}2
= {x2-(x-z)2}{x2+(x-z)2}
= { x-(x-z)}{x+(x-z)} {x2+(x-z)2}
= z(2x-z)( x2+x2-2xz+z2)
= z(2x-z)( 2x2-2xz+z2)
(v) a4–2a2b2+b4 = (a2)2-2a2b2+(b2)2
= (a2-b2)2
= ((a–b)(a+b))2
= (a – b)2 (a + b)2
Factorise the following expressions.
(i) p2+6p+8
(ii) q2–10q+21
(iii) p2+6p–16
(i) p2+6p+8
We observed that 8 = 4×2 and 4+2 = 6
p2+6p+8 can be written as p2+2p+4p+8
Taking Common terms, we get
p2+6p+8 = p2+2p+4p+8 = p(p+2)+4(p+2)
Again, p+2 is common in both the terms.
= (p+2)(p+4)
This implies that p2+6p+8 = (p+2)(p+4)
(ii) q2–10q+21
We observed that 21 = -7×-3 and -7+(-3) = -10
q2–10q+21 = q2–3q-7q+21
= q(q–3)–7(q–3)
= (q–7)(q–3)
This implies that q2–10q+21 = (q–7)(q–3)
(iii) p2+6p–16
We observed that -16 = -2×8 and 8+(-2) = 6
p2+6p–16 = p2–2p+8p–16
= p(p–2)+8(p–2)
= (p+8)(p–2)
So, p2+6p–16 = (p+8)(p–2)
Carry out the following divisions.
(i) 28x4 ÷ 56x
(ii) –36y3 ÷ 9y2
(iii) 66pq2r3 ÷ 11qr2
(iv) 34x3y3z3 ÷ 51xy2z3
(v) 12a8b8 ÷ (– 6a6b4)
(i)28x4 = 2×2×7×x×x×x×x
56x = 2×2×2×7×x
Divide the given polynomial by the given monomial.
(i)(5x2–6x) ÷ 3x
(ii)(3y8–4y6+5y4) ÷ y4
(iii) 8(x3y2z2+x2y3z2+x2y2z3)÷ 4x2 y2 z2
(iv)(x3+2x2+3x) ÷2x
(v) (p3q6–p6q3) ÷ p3q3
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)
(i) (10x–25) ÷ 5 = 5(2x-5)/5 = 2x-5
(ii) (10x–25) ÷ (2x–5) = 5(2x-5)/( 2x-5) = 5
(iii) 10y(6y+21) ÷ 5(2y+7) = 10y×3(2y+7)/5(2y+7) = 6y
(iv) 9x2y2(3z–24) ÷ 27xy(z–8) = 9x2y2×3(z-8)/27xy(z-8) = xy
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)
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)
(i) (y2+7y+10)÷(y+5)
First, solve the equation (y2+7y+10)
(y2+7y+10) = y2+2y+5y+10 = y(y+2)+5(y+2) = (y+2)(y+5)
Now, (y2+7y+10)÷(y+5) = (y+2)(y+5)/(y+5) = y+2
(ii) (m2–14m–32)÷ (m+2)
Solve for m2–14m–32, we have
m2–14m–32 = m2+2m-16m–32 = m(m+2)–16(m+2) = (m–16)(m+2)
Now, (m2–14m–32)÷(m+2) = (m–16)(m+2)/(m+2) = m-16
(iii) (5p2–25p+20)÷(p–1)
Step 1: Take 5 common from the equation, 5p2–25p+20, we get
5p2–25p+20 = 5(p2–5p+4)
Step 2: Factorise p2–5p+4
p2–5p+4 = p2–p-4p+4 = (p–1)(p–4)
Step 3: Solve original equation
(5p2–25p+20)÷(p–1) = 5(p–1)(p–4)/(p-1) = 5(p–4)
(iv) 4yz(z2 + 6z–16)÷ 2y(z+8)
Factorising z2+6z–16,
z2+6z–16 = z2-2z+8z–16 = (z–2)(z+8)
Now, 4yz(z2+6z–16) ÷ 2y(z+8) = 4yz(z–2)(z+8)/2y(z+8) = 2z(z-2)
(v) 5pq(p2–q2) ÷ 2p(p+q)
p2–q2 can be written as (p–q)(p+q) using the identity.
5pq(p2–q2) ÷ 2p(p+q) = 5pq(p–q)(p+q)/2p(p+q) = 5q(p–q)/2
(vi) 12xy(9x2–16y2) ÷ 4xy(3x+4y)
Factorising 9x2–16y2 , we have
9x2–16y2 = (3x)2–(4y)2 = (3x+4y)(3x-4y) using the identity p2–q2 = (p–q)(p+q)
Now, 12xy(9x2–16y2) ÷ 4xy(3x+4y) = 12xy(3x+4y)(3x-4y) /4xy(3x+4y) = 3(3x-4y)
(vii) 39y3(50y2–98) ÷ 26y2(5y+7)
st solve for 50y2–98, we have
50y2–98 = 2(25y2–49) = 2((5y)2–72) = 2(5y–7)(5y+7)
Now, 39y3(50y2–98) ÷ 26y2(5y+7) =
1. 4(x–5) = 4x–5
2. x(3x+2) = 3x2+2
3. 2x+3y = 5xy
4. x+2x+3x = 5x
5. 5y+2y+y–7y = 0
6. 3x+2x = 5x2
7. (2x) 2+4(2x)+7 = 2x2+8x+7
8. (2x) 2+5x = 4x+5x = 9x
9. (3x + 2) 2 = 3x2+6x+4
10. Substituting x = – 3 in
(a) x2 + 5x + 4 gives (– 3) 2+5(– 3)+4 = 9+2+4 = 15
(b) x2 – 5x + 4 gives (– 3) 2– 5( – 3)+4 = 9–15+4 = – 2
(c) x2 + 5x gives (– 3) 2+5(–3) = – 9–15 = – 24
11.(y–3)2 = y2–9
12. (z+5) 2 = z2+25
13. (2a+3b)(a–b) = 2a2–3b2
14. (a+4)(a+2) = a2+8
15. (a–4)(a–2) = a2–8
16. 3x2/3x2 = 0
17. (3x2+1)/3x2 = 1 + 1 = 2
18. 3x/(3x+2) = ½
19. 3/(4x+3) = 1/4x
20. (4x+5)/4x = 5
1.4(x- 5)= 4x – 20 ≠ 4x – 5 = RHS
The correct statement is 4(x-5) = 4x–20
2.LHS = x(3x+2) = 3x2+2x ≠ 3x2+2 = RHS
The correct solution is x(3x+2) = 3x2+2x
3.LHS= 2x+3y ≠ R. H. S
The correct statement is 2x+3y = 2x+3 y
4.LHS = x+2x+3x = 6x ≠ RHS
The correct statement is x+2x+3x = 6x
5.LHS = 5y+2y+y–7y = y ≠ RHS
The correct statement is 5y+2y+y–7y = y
6.LHS = 3x+2x = 5x ≠ RHS
The correct statement is 3x+2x = 5x
7.LHS = (2x) 2+4(2x)+7 = 4x2+8x+7 ≠ RHS
The correct statement is (2x) 2+4(2x)+7 = 4x2+8x+7
8.LHS = (2x) 2+5x = 4x2+5x ≠ 9x = RHS
The correct statement is(2x) 2+5x = 4x2+5x
9.LHS = (3x+2) 2 = (3x)2+22+2x2x3x = 9x2+4+12x ≠ RHS
The correct statement is (3x + 2) 2 = 9x2+4+12x
10.(a) Substituting x = – 3 in x2+5x+4, we have
x2+5x+4 = (– 3) 2+5(– 3)+4 = 9–15+4 = – 2. This is the correct answer.
(b) Substituting x = – 3 in x2–5x+4
x2–5x+4 = (–3) 2–5(– 3)+4 = 9+15+4 = 28. This is the correct answer
(c) Substituting x = – 3 in x2+5x
x2+5x = (– 3) 2+5(–3) = 9–15 = -6. This is the correct answer
11.LHS = (y–3)2 , which is similar to (a–b)2 identity, where (a–b) 2 = a2+b2-2ab
(y – 3)2 = y2+(3) 2–2y×3 = y2+9 –6y ≠ y2 – 9 = RHS
The correct statement is (y–3)2 = y2 + 9 – 6y
12. LHS = (z+5)2 , which is similar to (a +b)2 identity, where (a+b) 2 = a2+b2+2ab
(z+5) 2 = z2+52+2×5×z = z2+25+10z ≠ z2+25 = RHS
The correct statement is (z+5) 2 = z2+25+10z
13.LHS = (2a+3b)(a–b) = 2a(a–b)+3b(a–b)
= 2a2–2ab+3ab–3b2
= 2a2+ab–3b2
≠ 2a2–3b2 = RHS
The correct statement is (2a +3b)(a –b) = 2a2+ab–3b2
14.LHS = (a+4)(a+2) = a(a+2)+4(a+2)
= a2+2a+4a+8
= a2+6a+8
≠ a2+8 = RHS
The correct statement is (a+4)(a+2) = a2+6a+8
15.LHS = (a–4)(a–2) = a(a–2)–4(a–2)
= a2–2a–4a+8
= a2–6a+8
≠ a2-8 = RHS
The correct statement is (a–4)(a–2) = a2–6a+8
16.LHS = 3x2/3x2 = 1 ≠ 0 = RHS
The correct statement is 3x2/3x2 = 1
17.LHS = (3x2+1)/3x2 = (3x2/3x2)+(1/3x2) = 1+(1/3x2) ≠ 2 = RHS
The correct statement is (3x2+1)/3x2 = 1+(1/3x2)
18.LHS = 3x/(3x+2) ≠ 1/2 = RHS
The correct statement is 3x/(3x+2) = 3x/(3x+2)
19.LHS = 3/(4x+3) ≠ 1/4x
The correct statement is 3/(4x+3) = 3/(4x+3)
20.LHS = (4x+5)/4x = 4x/4x + 5/4x = 1 + 5/4x ≠ 5 = RHS
The correct statement is (4x+5)/4x = 1 + (5/4x)
21.LHS = (7x+5)/5 = (7x/5)+ 5/5 = (7x/5)+1 ≠ 7x = RHS
The correct statement is (7x+5)/5 = (7x/5) +1