Class 8 Maths Chapter 14 Factorisation
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NCERT Solutions For Class 8 Maths Chapter 14 Factorisation
Class 8 Maths Chapter 14 Factorisation
Exercise 14.1
Page No: 208
1. 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
Solution:
(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 28p2q
14pq = 2x7xpxq
28p2q = 2x2x7xpxpxq
Common factors of 14 pq and 28 p2q 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
2.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
Solution:
(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)
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:
Exercise 14.2
Page No: 223
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
= a2+2×4×a+42
= (a+4)2
Using identity: (x+y)2 = x2+2xy+y2
(ii) p2–10p+25
= p2-2×5×p+52
= (p-5)2
Using identity: (x-y)2 = x2-2xy+y2
(iii) 25m2+30m+9
= (5m)2-2×5m×3+32
= (5m+3)2
Using identity: (x+y)2 = x2+2xy+y2
(iv) 49y2+84yz+36z2
=(7y)2+2×7y×6z+(6z)2
= (7y+6z)2
Using identity: (x+y)2 = x2+2xy+y2
(v) 4x2–8x+4
= (2x)2-2×4x+22
= (2x-2)2
Using identity: (x-y)2 = x2-2xy+y2
(vi) 121b2-88bc+16c2
= (11b)2-2×11b×4c+(4c)2
= (11b-4c)2
Using identity: (x-y)2 = x2-2xy+y2
(vii) (l+m)2-4lm (Hint: Expand (l+m)2 first)
Expand (l+m)2 using identity: (x+y)2 = x2+2xy+y2
(l+m)2-4lm = l2+m2+2lm-4lm
= l2+m2-2lm
= (l-m)2
Using identity: (x-y)2 = x2-2xy+y2
(viii) a4+2a2b2+b4
= (a2)2+2×a2×b2+(b2)2
= (a2+b2)2
Using identity: (x+y)2 = x2+2xy+y2
2. 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
Solution:
(i) 4p2–9q2
= (2p)2-(3q)2
= (2p-3q)(2p+3q)
Using identity: x2-y2 = (x+y)(x-y)
(ii) 63a2–112b2
= 7(9a2 –16b2)
= 7((3a)2–(4b)2)
= 7(3a+4b)(3a-4b)
Using identity: x2-y2 = (x+y)(x-y)
(iii) 49x2–36
= (7a)2 -62
= (7a+6)(7a–6)
Using identity: x2-y2 = (x+y)(x-y)
(iv) 16x5–144x3
= 16x3(x2–9)
= 16x3(x2–9)
= 16x3(x–3)(x+3)
Using identity: x2-y2 = (x+y)(x-y)
(v) (l+m) 2-(l-m) 2
= {(l+m)-(l–m)}{(l +m)+(l–m)}
Using 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 Identity: x2-y2 = (x+y)(x-y)
(vii) (x2–2xy+y2)–z2
= (x–y)2–z2
Using Identity: (x-y)2 = x2-2xy+y2
= {(x–y)–z}{(x–y)+z}
= (x–y–z)(x–y+z)
Using 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 Identity: x2-y2 = (x+y)(x-y) , we have
= (5a+2b-7c)(5a-2b-7c)
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+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)
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) (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
5. Factorise the following expressions.
(i) p2+6p+8
(ii) q2–10q+21
(iii) p2+6p–16
Solution:
(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: p2+6p+8 = (p+2)(p+4)
(ii) q2–10q+21
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 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)
Exercise 14.3
Page No: 227
1. Carry out the following divisions.
(i) 28x4 ÷ 56x
(ii) –36y3 ÷ 9y2
(iii) 66pq2r3 ÷ 11qr2
(iv) 34x3y3z3 ÷ 51xy2z3
(v) 12a8b8 ÷ (– 6a6b4)
Solution:
(i)28x4 = 2×2×7×x×x×x×x
56x = 2×2×2×7×x
2. 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
Solution:
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:
(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
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:
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:
(i) (y2+7y+10)÷(y+5)
First solve for 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: Factorize 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)
Factorize 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 identity.
5pq(p2–q2) ÷ 2p(p+q) = 5pq(p–q)(p+q)/2p(p+q) = 5/2q(p–q)
(vi) 12xy(9x2–16y2) ÷ 4xy(3x+4y)
Factorize 9x2–16y2 , we have
9x2–16y2 = (3x)2–(4y)2 = (3x+4y)(3x-4y) using 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)
First 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) =
Exercise 14.4
Page No: 228
Find and correct the errors in the following mathematical statements.
1. 4(x–5) = 4x–5
Solution:
4(x- 5)= 4x – 20 ≠ 4x – 5 = RHS
The correct statement is 4(x-5) = 4x–20
2. x(3x+2) = 3x2+2
Solution:
LHS = x(3x+2) = 3x2+2x ≠ 3x2+2 = RHS
The correct solution is x(3x+2) = 3x2+2x
3. 2x+3y = 5xy
Solution:
LHS= 2x+3y ≠ R. H. S
The correct statement is 2x+3y = 2x+3 y
4. x+2x+3x = 5x
Solution:
LHS = x+2x+3x = 6x ≠ RHS
The correct statement is x+2x+3x = 6x
5. 5y+2y+y–7y = 0
Solution:
LHS = 5y+2y+y–7y = y ≠ RHS
The correct statement is 5y+2y+y–7y = y
6. 3x+2x = 5x2
Solution:
LHS = 3x+2x = 5x ≠ RHS
The correct statement is 3x+2x = 5x
7. (2x) 2+4(2x)+7 = 2x2+8x+7
Solution:
LHS = (2x) 2+4(2x)+7 = 4x2+8x+7 ≠ RHS
The correct statement is (2x) 2+4(2x)+7 = 4x2+8x+7
8. (2x) 2+5x = 4x+5x = 9x
Solution:
LHS = (2x) 2+5x = 4x2+5x ≠ 9x = RHS
The correct statement is(2x) 2+5x = 4x2+5x
9. (3x + 2) 2 = 3x2+6x+4
Solution:
LHS = (3x+2) 2 = (3x)2+22+2x2x3x = 9x2+4+12x ≠ RHS
The correct statement is (3x + 2) 2 = 9x2+4+12x
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
Solution:
(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.(y–3)2 = y2–9
Solution:
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. (z+5) 2 = z2+25
Solution:
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. (2a+3b)(a–b) = 2a2–3b2
Solution:
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. (a+4)(a+2) = a2+8
Solution:
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. (a–4)(a–2) = a2–8
Solution:
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. 3x2/3x2 = 0
Solution:
LHS = 3x2/3x2 = 1 ≠ 0 = RHS
The correct statement is 3x2/3x2 = 1
17. (3x2+1)/3x2 = 1 + 1 = 2
Solution:
LHS = (3x2+1)/3x2 = (3x2/3x2)+(1/3x2) = 1+(1/3x2) ≠ 2 = RHS
The correct statement is (3x2+1)/3x2 = 1+(1/3x2)
18. 3x/(3x+2) = ½
Solution:
LHS = 3x/(3x+2) ≠ 1/2 = RHS
The correct statement is 3x/(3x+2) = 3x/(3x+2)
19. 3/(4x+3) = 1/4x
Solution:
LHS = 3/(4x+3) ≠ 1/4x
The correct statement is 3/(4x+3) = 3/(4x+3)
20. (4x+5)/4x = 5
Solution:
LHS = (4x+5)/4x = 4x/4x + 5/4x = 1 + 5/4x ≠ 5 = RHS
The correct statement is (4x+5)/4x = 1 + (5/4x)
Solution:
LHS = (7x+5)/5 = (7x/5)+ 5/5 = (7x/5)+1 ≠ 7x = RHS
The correct statement is (7x+5)/5 = (7x/5) +1
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NCERT Solutions for Class 8 contains extremely important points, and for each chapter, each concept has been simplified to make it easier to remember and increase your chances of achieving excellent exam results. Exam Preparation References Here are some tips on how these Solutions can help you prepare for the exam.
- This helps students solve many of the problems in each chapter and encourages them to make their concepts more meaningful.
- NCERT Solutions for Class 8 encourage you to update your knowledge and refine your concepts so that you can get good results in the exam.
- These NCERT Solutions For Class 8 are the best exam materials, allowing you to learn more about your week and your strengths. To get good results in the exam, it is important to overcome your weaknesses.
- Most of the questions in the exam are formulated in a similar way to NCERT textbooks. Therefore, students should review the Solutions in each chapter in order to better understand the topic.
- It is free of cost.
Tips & Strategies for Class 8 Exam Preparation
- Plan your course and syllabus and make time for revision
- Please refer to the NCERT Solutions available on the cbsestudyguru website to clarify your concepts every time you prepare for the exam.
- Use the cbsestudyguru learning app to start learning to successfully pass the exam. Provide complete teaching materials, including resolved and unresolved tasks.
- It is important to clear all your doubts before the exam with your teachers or Alex (an Al study Bot).
- When you read or study a chapter, write down algorithm formulas, theorems, etc., and review them quickly before the exam.
- Practice an ample number of question papers to make your concepts stronger.
- Take rest and a proper meal. Don’t stress too much.
Why opt for cbsestudyguru NCERT Solutions for Class 8 ?
- cbsestudyguru provide NCERT Solutions for all subjects at your fingertips.
- These Solutions are designed by subject matter experts and provide Solutions to every NCERT textbook questions.
- cbsestudyguru especially focuses on making learning interactive, effective and for all classes.
- We provide free NCERT Solutions for class 8 and all other classes.
