NCERT Solutions for Class 8 Maths Chapter 14 Factorisation
The NCERT Solutions are aimed at mastering the concepts and acquiring comprehensive knowledge about the various types of questions asked in CBSE Class 8 Mathematics Examinations. All the questions are provided with appropriate solutions which will come in handy to revise the CBSE class 8 Mathematics syllabus. Solutions for Maths Class 8 are designed by highly knowledgeable subject experts, as per the CBSE curriculum. Class 8 Maths NCERT Solutions are helpful for the students to streamline their last-minute revision. Students can download pdf and study offline.
Download PDF of NCERT Solutions for Class 8 Maths Chapter 14 Factorisation
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Access Answers to NCERT 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, โ 4x2 , 32 x
(vii) 10 pq, 20qr, 30 rp
(viii) 3x2y3 , 10x3y2 , 6x2y2z
Solution:
(i) Factors of 12x and 36
12x = 2ร2ร3รx
36 = 2ร2ร3ร3
Common factors of 12x and 36 are 2, 2, 3
and , 2ร2ร3 = 12
(ii) Factors of 2y and 22xy
2y = 2รy
22xy = 2ร11รxรy
Common factors of 2y and 22xy are 2, y
and ,2รy = 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 = 2รx
3x2= 3รxรx
4 = 2ร2
Common factors of 2x, 3x2 and 4 is 1.
(v) Factors of 6abc, 24ab2 and 12a2b
6abc = 2ร3รaรbรc
24ab2 = 2ร2ร2ร3รaรbรb
12 a2 b = 2ร2ร3รaรaรb
Common factors of 6 abc, 24ab2 and 12a2b are 2, 3, a, b
and, 2ร3รaรb = 6ab
(vi) Factors of 16x3 , -4x2and 32x
16 x3 = 2ร2ร2ร2รxรxรx
โ 4x2 = -1ร2ร2รxรx
32x = 2ร2ร2ร2ร2รx
Common factors of 16 x3 , โ 4x2 and 32x are 2,2, x
and, 2ร2รx = 4x
(vii) Factors of 10 pq, 20qr and 30rp
10 pq = 2ร5รpรq
20qr = 2ร2ร5รqรr
30rp= 2ร3ร5รrรp
Common factors of 10 pq, 20qr and 30rp are 2, 5
and, 2ร5 = 10
(viii) Factors of 3x2y3 , 10x3y2 and 6x2y2z
3x2y3 = 3รxรxรyรyรy
10x3 y2 = 2ร5รxรxรxรyรy
6x2y2z = 3ร2รxรxรyรyรz
Common factors of 3x2y3, 10x3y2 and 6x2y2z are x2, y2
and, x2รy2 = x2y2
2.Factorise the following expressions
(i) 7xโ42
(ii) 6pโ12q
(iii) 7a2+ 14a
(iv) -16z+20 z3
(v) 20l2m+30alm
(vi) 5x2y-15xy2
(vii) 10a2-15b2+20c2
(viii) -4a2+4abโ4 ca
(ix) x2yz+xy2z +xyz2
(x) ax2y+bxy2+cxyz
Solution:
![NCERT Solution For Class 8 Maths Chapter 14 Ncert solutions class 8 chapter 14-1](https://cdn1.byjus.com/wp-content/uploads/2020/08/ncert-solutions-class-8-chapter-14-1.png)
![NCERT Solution For Class 8 Maths Chapter 14 Ncert solutions class 8 chapter 14-2](https://cdn1.byjus.com/wp-content/uploads/2020/08/ncert-solutions-class-8-chapter-14-2.jpeg)
(vii) 10a2-15b2+20c2
10a2 = 2ร5รaรa
โ 15b2 = -1ร3ร5รbรb
20c2 = 2ร2ร5รcรc
Common factor of 10 a2 , 15b2 and 20c2 is 5
10a2-15b2+20c2 = 5(2a2-3b2+4c2 )
(viii) โ 4a2+4ab-4ca
โ 4a2 = -1ร2ร2รaรa
4ab = 2ร2รaรb
โ 4ca = -1ร2ร2รcรa
Common factor of โ 4a2 , 4ab , โ 4ca are 2, 2, a i.e. 4a
So,
โ 4a2+4 ab-4 ca = 4a(-a+b-c)
(ix) x2yz+xy2z+xyz2
x2yz = xรxรyรz
xy2z = xรyรyรz
xyz2 = xรyรzรz
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 = aรxรxรy
bxy2 = bรxรyรy
cxyz = cรxรyรz
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) 15xyโ6x+5yโ2
(iii) ax+bxโayโby
(iv) 15pq+15+9q+25p
(v) zโ7+7xyโxyz
Solution:
![NCERT Solution For Class 8 Maths Chapter 14-3 Ncert solutions class 8 chapter 14-3](https://cdn1.byjus.com/wp-content/uploads/2020/08/ncert-solutions-class-8-chapter-14-3.png)
Exercise 14.2 Page No: 223
1. Factorise the following expressions.
(i) a2+8a+16
(ii) p2โ10p+25
(iii) 25m2+30m+9
(iv) 49y2+84yz+36z2
(v) 4x2โ8x+4
(vi) 121b2โ88bc+16c2
(vii) (l+m)2โ4lm (Hint: Expand (l+m)2 first)
(viii) a4+2a2b2+b4
Solution:
(i) a2+8a+16
= a2+2ร4รa+42
= (a+4)2
Using identity: (x+y)2 = x2+2xy+y2
(ii) p2โ10p+25
= p2-2ร5รp+52
= (p-5)2
Using identity: (x-y)2 = x2-2xy+y2
(iii) 25m2+30m+9
= (5m)2-2ร5mร3+32
= (5m+3)2
Using identity: (x+y)2 = x2+2xy+y2
(iv) 49y2+84yz+36z2
=(7y)2+2ร7yร6z+(6z)2
= (7y+6z)2
Using identity: (x+y)2 = x2+2xy+y2
(v) 4x2โ8x+4
= (2x)2-2ร4x+22
= (2x-2)2
Using identity: (x-y)2 = x2-2xy+y2
(vi) 121b2-88bc+16c2
= (11b)2-2ร11bร4c+(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+2รa2รb2+(b2)2
= (a2+b2)2
Using identity: (x+y)2 = x2+2xy+y2
2. Factorise.
(i) 4p2โ9q2
(ii) 63a2โ112b2
(iii) 49x2โ36
(iv) 16x5โ144x3 differ
(v) (l+m)2-(l-m) 2
(vi) 9x2y2โ16
(vii) (x2โ2xy+y2)โz2
(viii) 25a2โ4b2+28bcโ49c2
Solution:
(i) 4p2โ9q2
= (2p)2-(3q)2
= (2p-3q)(2p+3q)
Using identity: x2-y2 = (x+y)(x-y)
(ii) 63a2โ112b2
= 7(9a2 โ16b2)
= 7((3a)2โ(4b)2)
= 7(3a+4b)(3a-4b)
Using identity: x2-y2 = (x+y)(x-y)
(iii) 49x2โ36
= (7a)2 -62
= (7a+6)(7aโ6)
Using identity: x2-y2 = (x+y)(x-y)
(iv) 16x5โ144x3
= 16x3(x2โ9)
= 16x3(x2โ9)
= 16x3(xโ3)(x+3)
Using identity: x2-y2 = (x+y)(x-y)
(v) (l+m) 2-(l-m) 2
= {(l+m)-(lโm)}{(l +m)+(lโm)}
Using Identity: x2-y2 = (x+y)(x-y)
= (l+mโl+m)(l+m+lโm)
= (2m)(2l)
= 4 ml
(vi) 9x2y2โ16
= (3xy)2-42
= (3xyโ4)(3xy+4)
Using Identity: x2-y2 = (x+y)(x-y)
(vii) (x2โ2xy+y2)โz2
= (xโy)2โz2
Using Identity: (x-y)2 = x2-2xy+y2
= {(xโy)โz}{(xโy)+z}
= (xโyโz)(xโy+z)
Using Identity: x2-y2 = (x+y)(x-y)
(viii) 25a2โ4b2+28bcโ49c2
= 25a2โ(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) 5y2โ20yโ8z+2yz
(viii) 10ab+4a+5b+2
(ix)6xyโ4y+6โ9x
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) 5y2โ20yโ8z+2yz = 5y(yโ4)+2z(yโ4) = (yโ4)(5y+2z)
(viii) 10ab+4a+5b+2 = 5b(2a+1)+2(2a+1) = (2a+1)(5b+2)
(ix) 6xyโ4y+6โ9x = 6xyโ9xโ4y+6 = 3x(2yโ3)โ2(2yโ3) = (2yโ3)(3xโ2)
4.Factorise.
(i) a4โb4
(ii) p4โ81
(iii) x4โ(y+z) 4
(iv) x4โ(xโz) 4
(v) a4โ2a2b2+b4
Solution:
(i) a4โb4
= (a2)2-(b2)2
= (a2-b2) (a2+b2)
= (a โ b)(a + b)(a2+b2)
(ii) p4โ81
= (p2)2-(9)2
= (p2-9)(p2+9)
= (p2-32)(p2+9)
=(p-3)(p+3)(p2+9)
(iii) x4โ(y+z) 4 = (x2)2-[(y+z)2]2
= {x2-(y+z)2}{ x2+(y+z)2}
= {(x โ(y+z)(x+(y+z)}{x2+(y+z)2}
= (xโyโz)(x+y+z) {x2+(y+z)2}
(iv) x4โ(xโz) 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) a4โ2a2b2+b4 = (a2)2-2a2b2+(b2)2
= (a2-b2)2
= ((aโb)(a+b))2
5. Factorise the following expressions.
(i) p2+6p+8
(ii) q2โ10q+21
(iii) p2+6pโ16
Solution:
(i) p2+6p+8
We observed that, 8 = 4ร2 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) q2โ10q+21
Observed that, 21 = -7ร-3 and -7+(-3) = -10
q2โ10q+21 = q2โ3q-7q+21
= q(qโ3)โ7(qโ3)
= (qโ7)(qโ3)
This implies q2โ10q+21 = (qโ7)(qโ3)
(iii) p2+6pโ16
We observed that, -16 = -2ร8 and 8+(-2) = 6
p2+6pโ16 = p2โ2p+8pโ16
= p(pโ2)+8(pโ2)
= (p+8)(pโ2)
So, p2+6pโ16 = (p+8)(pโ2)
Exercise 14.3 Page No: 227
1. Carry out the following divisions.
(i) 28x4 รท 56x
(ii) โ36y3 รท 9y2
(iii) 66pq2r3 รท 11qr2
(iv) 34x3y3z3 รท 51xy2z3
(v) 12a8b8 รท (โ 6a6b4)
Solution:
(i)28x4 = 2ร2ร7รxรxรxรx
56x = 2ร2ร2ร7รx
![NCERT Solution For Class 8 Maths Chapter 14 Ncert solutions class 8 chapter 14-4](https://cdn1.byjus.com/wp-content/uploads/2020/08/ncert-solutions-class-8-chapter-14-4.png)
![NCERT Solution For Class 8 Maths Chapter 14 Ncert solutions class 8 chapter 14-5](https://cdn1.byjus.com/wp-content/uploads/2020/08/ncert-solutions-class-8-chapter-14-5.png)
![NCERT Solution For Class 8 Maths Chapter 14 Ncert solutions class 8 chapter 14-6](https://cdn1.byjus.com/wp-content/uploads/2020/08/ncert-solutions-class-8-chapter-14-6.png)
![NCERT Solution For Class 8 Maths Chapter 14 Ncert solutions class 8 chapter 14-7](https://cdn1.byjus.com/wp-content/uploads/2020/08/ncert-solutions-class-8-chapter-14-7.png)
![NCERT Solution For Class 8 Maths Chapter 14 Ncert solutions class 8 chapter 14-8](https://cdn1.byjus.com/wp-content/uploads/2020/08/ncert-solutions-class-8-chapter-14-8.png)
2. Divide the given polynomial by the given monomial.
(i)(5x2โ6x) รท 3x
(ii)(3y8โ4y6+5y4) รท y4
(iii) 8(x3y2z2+x2y3z2+x2y2z3)รท 4x2 y2 z2
(iv)(x3+2x2+3x) รท2x
(v) (p3q6โp6q3) รท p3q3
Solution:
![NCERT Solution For Class 8 Maths Chapter 14 Ncert solutions class 8 chapter 14-9](https://cdn1.byjus.com/wp-content/uploads/2020/08/ncert-solutions-class-8-chapter-14-9.png)
3. Work out the following divisions.
(i) (10xโ25) รท 5
(ii) (10xโ25) รท (2xโ5)
(iii) 10y(6y+21) รท 5(2y+7)
(iv) 9x2y2(3zโ24) รท 27xy(zโ8)
(v) 96abc(3aโ12)(5bโ30) รท 144(aโ4)(bโ6)
Solution:
(i) (10xโ25) รท 5 = 5(2x-5)/5 = 2x-5
(ii) (10xโ25) รท (2xโ5) = 5(2x-5)/( 2x-5) = 5
(iii) 10y(6y+21) รท 5(2y+7) = 10yร3(2y+7)/5(2y+7) = 6y
(iv) 9x2y2(3zโ24) รท 27xy(zโ8) = 9x2y2ร3(z-8)/27xy(z-8) = xy
![NCERT Solution For Class 8 Maths Chapter 14 Ncert solutions class 8 chapter 14-10](https://cdn1.byjus.com/wp-content/uploads/2020/08/ncert-solutions-class-8-chapter-14-10.png)
4. Divide as directed.
(i) 5(2x+1)(3x+5)รท (2x+1)
(ii) 26xy(x+5)(yโ4)รท13x(yโ4)
(iii) 52pqr(p+q)(q+r)(r+p) รท 104pq(q+r)(r+p)
(iv) 20(y+4) (y2+5y+3) รท 5(y+4)
(v) x(x+1) (x+2)(x+3) รท x(x+1)
Solution:
![NCERT Solution For Class 8 Maths Chapter 14 Ncert solutions class 8 chapter 14-11](https://cdn1.byjus.com/wp-content/uploads/2020/08/ncert-solutions-class-8-chapter-14-11.png)
5. Factorise the expressions and divide them as directed.
(i) (y2+7y+10)รท(y+5)
(ii) (m2โ14mโ32)รท(m+2)
(iii) (5p2โ25p+20)รท(pโ1)
(iv) 4yz(z2+6zโ16)รท2y(z+8)
(v) 5pq(p2โq2)รท2p(p+q)
(vi) 12xy(9x2โ16y2)รท4xy(3x+4y)
(vii) 39y3(50y2โ98) รท 26y2(5y+7)
Solution:
(i) (y2+7y+10)รท(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)รท(y+5) = (y+2)(y+5)/(y+5) = y+2
(ii) (m2โ14mโ32)รท (m+2)
Solve for m2โ14mโ32, we have
m2โ14mโ32 = m2+2m-16mโ32 = m(m+2)โ16(m+2) = (mโ16)(m+2)
Now, (m2โ14mโ32)รท(m+2) = (mโ16)(m+2)/(m+2) = m-16
(iii) (5p2โ25p+20)รท(pโ1)
Step 1: Take 5 common from the equation, 5p2โ25p+20, we get
5p2โ25p+20 = 5(p2โ5p+4)
Step 2: Factorize p2โ5p+4
p2โ5p+4 = p2โp-4p+4 = (pโ1)(pโ4)
Step 3: Solve original equation
(5p2โ25p+20)รท(pโ1) = 5(pโ1)(pโ4)/(p-1) = 5(pโ4)
(iv) 4yz(z2 + 6zโ16)รท 2y(z+8)
Factorize z2+6zโ16,
z2+6zโ16 = z2-2z+8zโ16 = (zโ2)(z+8)
Now, 4yz(z2+6zโ16) รท 2y(z+8) = 4yz(zโ2)(z+8)/2y(z+8) = 2z(z-2)
(v) 5pq(p2โq2) รท 2p(p+q)
p2โq2 can be written as (pโq)(p+q) using identity.
5pq(p2โq2) รท 2p(p+q) = 5pq(pโq)(p+q)/2p(p+q) = 5/2q(pโq)
(vi) 12xy(9x2โ16y2) รท 4xy(3x+4y)
Factorize 9x2โ16y2 , we have
9x2โ16y2 = (3x)2โ(4y)2 = (3x+4y)(3x-4y) using identity: p2โq2 = (pโq)(p+q)
Now, 12xy(9x2โ16y2) รท 4xy(3x+4y) = 12xy(3x+4y)(3x-4y) /4xy(3x+4y) = 3(3x-4y)
(vii) 39y3(50y2โ98) รท 26y2(5y+7)
st solve for 50y2โ98, we have
50y2โ98 = 2(25y2โ49) = 2((5y)2โ72) = 2(5yโ7)(5y+7)
Now, 39y3(50y2โ98) รท 26y2(5y+7) =
![NCERT Solution For Class 8 Maths Chapter 14-12 Ncert solutions class 8 chapter 14-12](https://cdn1.byjus.com/wp-content/uploads/2020/08/ncert-solutions-class-8-chapter-14-12.png)
Exercise 14.4 Page No: 228
1. 4(xโ5) = 4xโ5
Solution:
4(x- 5)= 4x โ 20 โ 4x โ 5 = RHS
The correct statement is 4(x-5) = 4xโ20
2. x(3x+2) = 3x2+2
Solution:
LHS = x(3x+2) = 3x2+2x โ 3x2+2 = RHS
The correct solution is x(3x+2) = 3x2+2x
3. 2x+3y = 5xy
Solution:
LHS= 2x+3y โ R. H. S
The correct statement is 2x+3y = 2x+3 y
4. x+2x+3x = 5x
Solution:
LHS = x+2x+3x = 6x โ RHS
The correct statement is x+2x+3x = 6x
5. 5y+2y+yโ7y = 0
Solution:
LHS = 5y+2y+yโ7y = y โ RHS
The correct statement is 5y+2y+yโ7y = y
6. 3x+2x = 5x2
Solution:
LHS = 3x+2x = 5x โ 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 โ 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 โ 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 โ RHS
The correct statement is (3x + 2) 2 = 9x2+4+12x
10. Substituting x = โ 3 in
(a) x2 + 5x + 4 gives (โ 3) 2+5(โ 3)+4 = 9+2+4 = 15
(b) x2 โ 5x + 4 gives (โ 3) 2โ 5( โ 3)+4 = 9โ15+4 = โ 2
(c) x2 + 5x gives (โ 3) 2+5(โ3) = โ 9โ15 = โ 24
Solution:
(a) Substituting x = โ 3 in x2+5x+4, we have
x2+5x+4 = (โ 3) 2+5(โ 3)+4 = 9โ15+4 = โ 2. This is the correct answer.
(b) Substituting x = โ 3 in x2โ5x+4
x2โ5x+4 = (โ3) 2โ5(โ 3)+4 = 9+15+4 = 28. This is the correct answer
(c)Substituting x = โ 3 in x2+5x
x2+5x = (โ 3) 2+5(โ3) = 9โ15 = -6. This is the correct answer
11.(yโ3)2 = y2โ9
Solution:
LHS = (yโ3)2 , which is similar to (aโb)2 identity, where (aโb) 2 = a2+b2-2ab.
(y โ 3)2 = y2+(3) 2โ2yร3 = y2+9 โ6y โ y2 โ 9 = RHS
The correct statement is (yโ3)2 = y2 + 9 โ 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+2ร5รz = z2+25+10z โ z2+25 = RHS
The correct statement is (z+5) 2 = z2+25+10z
13. (2a+3b)(aโb) = 2a2โ3b2
Solution:
LHS = (2a+3b)(aโb) = 2a(aโb)+3b(aโb)
= 2a2โ2ab+3abโ3b2
= 2a2+abโ3b2
โ 2a2โ3b2 = RHS
The correct statement is (2a +3b)(a โb) = 2a2+abโ3b2
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
โ a2+8 = RHS
The correct statement is (a+4)(a+2) = a2+6a+8
15. (aโ4)(aโ2) = a2โ8
Solution:
LHS = (aโ4)(aโ2) = a(aโ2)โ4(aโ2)
= a2โ2aโ4a+8
= a2โ6a+8
โ a2-8 = RHS
The correct statement is (aโ4)(aโ2) = a2โ6a+8
16. 3x2/3x2 = 0
Solution:
LHS = 3x2/3x2 = 1 โ 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) โ 2 = RHS
The correct statement is (3x2+1)/3x2 = 1+(1/3x2)
18. 3x/(3x+2) = ยฝ
Solution:
LHS = 3x/(3x+2) โ 1/2 = RHS
The correct statement is 3x/(3x+2) = 3x/(3x+2)
19. 3/(4x+3) = 1/4x
Solution:
LHS = 3/(4x+3) โ 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 โ 5 = RHS
The correct statement is (4x+5)/4x = 1 + (5/4x)
Solution:
LHS = (7x+5)/5 = (7x/5)+ 5/5 = (7x/5)+1 โ 7x = RHS
The correct statement is (7x+5)/5 = (7x/5) +1
Also Access |
CBSE Notes for Class 8 Maths Chapter 14 |
NCERT Solutions for Class 8 Maths Chapter 14 Factorisation
Class 8 NCERT exercise wise questions and answers will help students frame a perfect solution in the Maths Exam. These exercise questions can provide students with a topic wise preparation strategy. Some of the important topics introduced in Class 8 NCERT Solutions Maths are Factorisation, Factors of natural numbers, Factors of algebraic expressions and Division of Algebraic Expressions.
NCERT Solutions For Class 8 Maths Chapter 14 Exercises:
Get detailed solutions for all the questions listed under the below exercises:
Exercise 14.1 Solutions: 3 Questions (Short answer type)
Exercise 14.2 Solutions: 5 Questions (Short answer type)
Exercise 14.3 Solutions: 5 Questions (Short answer type)
Exercise 14.4 Solutions: 21 Questions (Short answer type)
NCERT Solutions for Class 8 Maths Chapter 14 Factorisation
NCERT Class 8 Maths Chapter 14, deals primarily with the factorisation of the numbers and algebraic expressions using algebraic identities. Students will also learn about, Division of Algebraic Expressions, Division of a monomial by another monomial, Division of a polynomial by a monomial and Division of Polynomial by Polynomial.
The main topics covered in this chapter include:
Exercise | Topic |
14.1 | Introduction |
14.2 | What is Factorisation? |
14.3 | Division of Algebraic Expressions |
14.4 | Division of Algebraic Expressions |
14.5 | Can you Find the Error? |
Key Features of NCERT Solutions for Class 8 Maths Chapter 14 Factorisation
- These NCERT solutions help students in understanding the concepts clearly.
- Simple and precise language is used to explain the topics.
- All concepts have been explained in detail.
- Subject experts have consolidated all exercise questions at one place for practice.
- NCERT Solutions are helpful for the preparation of competitive exams.
Frequently Asked Questions on NCERT Solutions for Class 8 Maths Chapter 14
Are NCERT Solutions for Class 8 Maths Chapter 14 important from the exam point of view?
Yes, the 14th chapter Factorization of NCERT Solutions for Class 8 is important from an exam perspective. It contains short answer questions as well as long answer questions to increase the ability of problem-solving. These solutions are created by the BYJUโS expert faculty to help students in the preparation of their examinations.
How many exercises are there in NCERT Solutions for Class 8 Maths Chapter 14?
There are four exercises in the NCERT Solutions for Class 8 Maths Chapter 14. These Solutions of NCERT Maths Class 8 Chapter 14 help the students in solving the problems adroitly and efficiently. They also focus on cracking the Solutions of Maths in such a way that it is easy for the students to understand.
What are the main topics covered in the NCERT Solutions for Class 8 Maths Chapter 14?
The main topics covered in the NCERT Solutions for Class 8 Maths Chapter 14 are
14.1 โ Introduction
14.2 โ Definition of Factorisation
14.3 โ Division of Algebraic Expressions
14.4 โ Division of Algebraic Expressions
14.5 โ Can you Find the Error?