NCERT Solutions Class 8 Maths Chapter 14

NCERT Solutions for Class 8 Mathematics Chapter 14 - Factorisation 

Mathematics is a subject of great importance and plays a great role in our lives. It helps boost our problem-solving skills, thereby increasing students' productivity. As a result, students should focus on making Mathematics their strong subject.

NCERT Class 8 Mathematics Chapter 14 is connected with algebraic expressions covered in lower classes. Thus, students should know the basics of algebraic expressions. Some of the main topics covered in this chapter include factorisation, division of algebraic expressions, division of algebraic expressions in the case of polynomials and all the concepts related to framing equations in Algebra. 

Extramarks is a well-known online learning platform for primary and secondary education students. A team of highly experienced subject teachers and academicians design our study courses based on the CBSE syllabus and the latest NCERT guidelines.

In our NCERT Solutions for Class 8 Mathematics Chapter 14, we have included the basics of algebraic expressions and all topics and sub-topics related to this chapter. Our solutions are a comprehensive set of study resources, including chapter notes, important questions and their answers, revision notes, etc. Students can confidently rely on our study resources and use them fully to learn the chapter and prepare for their exams. All our study materials are specially designed by keeping in mind the needs of every student. 

Extramarks provides a complete suite of study materials for helping students with their Class 1 to Class 12 studies. We also have comprehensive solutions for competitive exam preparation such as JEE, NEET, etc. Students can confidently rely on the Extramarks platform for learning and scoring good marks in their exams.

Key Topics Covered In NCERT Solutions for Class 8 Mathematics Chapter 14

You have already learnt in lower classes about the role of algebraic expressions in Mathematics. In this chapter, you will learn more about the factorisation of algebraic expressions, their division, and division in the case of polynomials which is one of the most important topics of this chapter. To score well in this chapter, students should primarily focus on the division of polynomials and should try to get a conceptual understanding of this topic properly.

Students can find a variety of questions based on the division of polynomials in our NCERT Solutions for Class 8 Mathematics Chapter 14. The different sets of questions will give students the confidence to easily solve the questions they would be facing in their school and competitive examinations. 

By regularly practising and studying from our NCERT solutions, students are likely to develop a keen interest in the subject of Mathematics, which will help them easily ace their exams.

Introduction

In this section, we will learn about the factorisation of natural numbers and algebraic expressions. 

We have learnt in our earlier classes about natural numbers. We will learn how to factorise a natural number into prime factors. 

For example, if we have a natural number 30, then its prime factor is 2 * 3 * 5.

We already learnt about algebraic expressions in our earlier classes. We will learn how to factorise an algebraic expression into the prime factor.

For example, if we have an algebraic expression 5xy, its prime factor is 5 * x * y 

What is factorisation?

When we factorise an algebraic expression, it simplifies the expression as a product of its factors. These factors are in the form of natural numbers, algebraic variables or algebraic expressions.  

For example, 2xy, 5x^3y, and 5x(y+2) are factorised forms.

In other examples, the expressions; (2x+4) and 5x^2+9 are not in a factorised form. To change in factor form, we will use some methods given below:

  • Method1: method of common factors
  • Method2: factorisation by regrouping terms

Regrouping: we will rearrange the algebraic expression (Ex. 3xy+2+3+2ab), which will allow us to form the group (ex. 3xy+3+2ab+2), leading to factorisation.

  • Method3: factorisations using identities
  • Method4: factors of the form (x+a)(x+b)

The methods given above will clarify how to factorise any natural number or algebraic expressions. To learn each method's step-by-step process, please visit our Extramarks website and refer to our NCERT Solutions for Class 8 Mathematics Chapter 14.

Division of algebraic expressions 

As we learned in the earlier chapters, division is the inverse operation of multiplication.

For example, 5*6=30 gives 30÷5=6 or 30÷6=5.

  • Division of a monomial by another monomial

In this section, we will learn to divide a monomial by another monomial. 

For example, To divide 6x^2 to 2x,

Step1: factorise the monomial 2*3*x*x and 2*x

Step2: cancel the matched factor by their denominator 

(2*3*x*x ÷ 2*x) = 3*x

  • Division of a polynomial by a monomial

Here, we will learn to divide a polynomial by a monomial.

For example, divide 6x^2(2x + y) by 2x

Step1: factorise the polynomial and monomial 2*3*x*x*(2*x + y) and 2*x

Step2: cancel the matched factor by their denominator 

[2*3*x*x*(2*x + y) ÷ 2*x] = 3*x(2x + y)

Learning how to do division of polynomials is a crucial topic that will be required in advanced algebraic equations. Our Mathematics experts have explained the division of a monomial and polynomial by another monomial with many solved examples in our NCERT Solutions for Class 8 Mathematics Chapter 14. Students can register on our website and get access to these NCERT solutions. 

Division of algebraic expressions continued (polynomial ÷ polynomial)

This section will teach us to divide the polynomial by the polynomial. 

For example, To divide (7y^2 + 14x) to (y + 2)

Step1: factorise the polynomial and polynomial

7*y*y + 7*2*y and (y + 2)

7*y(y +2) and (y + 2)

Step2: cancel the matched factor by their denominator 

[7*y(y +2) ÷ (y + 2)] = 7*y

The questions related to the division of polynomials start getting harder in this section. Students can refer to our NCERT Solutions for Class 8 Mathematics Chapter 14 to get access to a lot of questions and their solved answers. A variety of questions with low to high difficulty levels are covered in our study materials. Solving them will help students clear their doubts and increase their confidence in this chapter. 

Can you find the error?

In this section, we are provided with some examples to find the error in solving different division problems. Students can register on the Extramarks website and get access to our NCERT Solutions for Class 8 Mathematics Chapter 14 to learn from study notes for this section.

NCERT Solutions for Class 8 Mathematics Chapter 14: Exercise & Answer Solutions

NCERT textbook has exercises based on each chapter, which helps develop a better understanding of the students' concepts. As a result, students must have a written solution to all these exercises. Keeping this in mind, we have designed the NCERT Solutions for Class 8 Mathematics Chapter 14, covering the detailed solutions for each exercise in the chapter. This would help them learn and test themselves while practising their problems. Students can easily get NCERT Solutions for Class 8 Mathematics Chapter 14 from the Extramarks website.

Click on the below links to view exercise-specific questions and solutions for NCERT Solutions for Class 8 Mathematics Chapter 14:

  • Chapter 14: Exercise 14.1 
  • Chapter 14: Exercise 14.2
  • Chapter 14: Exercise 14.3
  • Chapter 14: Exercise 14.4  
  • Factorisation Class 8 Solutions - Exercise and Answer Solutions

Along with Class 8 Mathematics solutions, you can explore NCERT Solutions on our Extramarks website for all primary and secondary classes.

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4 
  • NCERT Solutions Class 5
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7
  • NCERT Solutions Class 8
  • NCERT Solutions Class 9
  • NCERT Solutions Class 10
  • NCERT Solutions Class 11
  • NCERT Solutions Class 12

NCERT Exemplar Class 8 Mathematics 

NCERT exemplar covers all NCERT-related questions. The subject matter experts design it, so students trust this book during their exam preparation. It covers all the core concepts related to the chapter and formulated questions according to it. As a result, no topic of the students remains untouched.

The book has all ranges of questions right from basics to advanced, which helps in gradually building the mindset of the students, due to which they start solving a variety of questions with ease. 

NCERT Exemplar Class 8 Mathematics is available on the Extramarks website. Students can register on our website and get access to our study materials anytime. Students also learn how to manage their time efficiently during the examinations and study smartly once they refer to the NCERT Exemplar. The book develops not only the academic performance of the students but also all other aspects related to it.

Key Features of NCERT Solutions for Class 8 Mathematics Chapter 14

Extramarks is a learning platform which has proven helpful for many students in their Examination preparation. Students can start with a trial account by registering on the website. 

Students who know how to study smartly are always top performers. Lakhs of students have registered for Extramarks online learning solutions and scored above-average marks in their exams.

Academic experts smartly design our NCERT Solutions for Class 8 Mathematics Chapter 14 to make learning easy and fun for students. Some of the key features of our study solutions are: 

  • You will find all the important topics highlighted so that students know what topics to focus on the most during the preparation.
  • The NCERT solutions also have quick notes, mind maps, memorising techniques and tables for students to quickly learn every bit of the chapter.
  • After completing the NCERT Solutions for Class 8 Mathematics Chapter 14, students will understand all the aspects related to the factorisation of all algebraic expressions and will easily be able to apply in their higher classes too.

Q.1 Factorise.

(i) x2 + x y + 8x + 8y

(ii) 15xy – 6x + 5y – 2

(iii) ax + bx – ay – by

(iv) 15 pq + 15 + 9q + 25p

(v) z – 7 + 7xy – xyz

Ans

(i) x2 + x y + 8x + 8y

Rearranging the terms of the given expression as follows:

= x × x + x × y + 8 × x + 8 × y

By taking ‘x’ as common from the first two terms and ‘8’ common from the last two terms, we get

= x (x + y) + 8 (x + y)

= (x + y) (x + 8)

(ii) 15xy − 6x + 5y – 2

Rearranging the terms of the given expression as follows:

= 3 × 5 × x × y − 3 × 2 × x + 5 × y − 2

By taking ‘3x’ as common from the first two terms and ‘1’ common from the last two terms, we get

= 3x (5y − 2) + 1 (5y − 2)

= (5y − 2) (3x + 1)

(iii) ax + bx − ay − by

Rearranging the terms of the given expression as follows:

= a × x + b × x − a × y − b × y

By taking ‘x’ as common from the first two terms and ‘y’ common from the last two terms, we get

= x (a + b) − y (a + b)

= (a + b) (x − y)

(iv) 15pq + 15 + 9q + 25p = 15pq + 9q + 25p + 15

Rearranging the terms of the given expression as follows:

= 3 × 5 × p × q + 3 × 3 × q + 5 × 5 × p + 3 × 5

By taking ‘3q’ as common from the first two terms and ‘5’ common from the last two terms, we get

= 3q (5p + 3) + 5 (5p + 3)

= (5p + 3) (3q + 5)

(v) z − 7 + 7xy − xyz

Rearranging the terms of the given expression as follows:

= z − x × y × z − 7 + 7 × x × y

By taking ‘z’ as common from the first two terms and ‘−7’ common from the last two terms, we get

= z (1− xy) − 7 (1− xy)

= (1− xy) (z − 7)

Q.2

Factorise the following expressions.(i) a2 + 8a + 16 (ii) p210 p + 25 (iii) 25m2 + 30m + 9(iv) 49y2 + 84yz + 36z2 (v) 4x28x + 4(vi) 121b288bc + 16c2(vii) (l + m)24lm (Hint: Expand ( l + m)2 first)(viii) a4 + 2a2b2 + b4

Ans

(i) a2 + 8a + 16

The given expression can be written as:

= (a)2 + 2 × a × 4 + (4)2

This is of the form: x2+ y2 +2xy with x = a, y = 4 and 2xy = 8a

Therefore, by using the identity: (x + y)2 = x2 + 2xy + y2,we get

= (a + 4)2

(ii) p2 − 10p + 25

The given expression can be written as:

= (p)2 − 2 × p × 5 + (5)2

This is of the form: a2+ b2 −2ab with a= p, b= 5 and

2ab =10p

Therefore, by using the identity: (a − b)2 = a2 – 2ab + b2,we get

= (p − 5)2

(iii) 25m2 + 30m + 9

The given expression can be written as:

= (5m)2 + 2 × 5m × 3 + (3)2

This is of the form: a2+ b2 +2ab with a = 5m, b = 3 and 2ab =30m

Therefore, by using the identity: (a + b)2 = a2 + 2ab + b2,we get

= (5m + 3)2

(iv) 49y2 + 84yz + 36z2

The given expression can be written as:

= (7y)2 + 2 × (7y) × (6z) + (6z)2

This is of the form: a2 + b2 +2ab with a = 7y, b= 6z and 2ab = 84yz

Therefore, by using the identity: (a + b)2 = a2 + 2ab + b2,we get

= (7y + 6z)2

(v) 4x2 − 8x + 4

The given expression can be written as:

= (2x)2 − 2 (2x) (2) + (2)2

This is of the form: a2 + b2 − 2ab with a = 2x, b= 2 and 2ab = 8x

Therefore, by using the identity: (a − b)2 = a2 − 2ab + b2,we get

= (2x − 2)2

Again, we can take out 2 common

= [(2) (x − 1)]2

= 4(x − 1)2

(vi) 121b2 − 88bc + 16c2

The given expression can be written as:

= (11b)2 − 2 (11b) (4c) + (4c)2

This is of the form: a2 + b2 − 2ab with a = 11b, b= 4c and 2ab = 88bc

Therefore, by using the identity: (a − b)2 = a2 − 2ab + b2,we get

= (11b − 4c)2

(vii) (l + m)2 − 4lm

The given expression can be written as:

= l2 + 2lm + m2 − 4lm

= l2 − 2lm + m2

This is of the form: a2 + b2 − 2ab with a = l, b= m and 2ab = 2lm

Therefore, by using the identity: (a − b)2 = a2 − 2ab + b2,we get

= (l − m)2

(viii) a4 + 2a2b2 + b4

The given expression can be written as:

= (a2)2 + 2 (a2) (b2) + (b2)2

This is of the form: x2 + y2 + 2xy with x = a2, y= b2 and 2xy = 2a2b2

Therefore, by using the identity: (x + y)2 = x2 + 2xy + y2,we get

= (a2 + b2)2

Q.3

Factorise.(i) 4p2 9q2 (ii) 63a2 112b2 (iii) 49x236(iv) 16x5 144x3 (v) (l + m)2l m2(vi) 9x2 y2 16 (vii) (x2  2xy + y2) z2(viii) 25a2 4b2 + 28bc 49c2

Ans

(i) 4p2 − 9q2

The given expression can be written as:

= (2p)2 − (3q)2

This is of the form: a2 − b2 with a = 2p and b = 3q

Therefore, by using the identity: a2 − b2 = (a − b) (a + b), we get

= (2p + 3q) (2p − 3q)

(ii) 63a2 − 112b2

The given expression can be written as:

= 7(9a2 − 16b2)

= 7[(3a)2 − (4b)2]

This is of the form: x2 − y2 with x = 3a and y = 4b

Therefore, by using the identity: x2 − y2 = (x − y) (x + y), we get

= 7(3a + 4b) (3a − 4b)

(iii) 49x2 − 36

The given expression can be written as:

= (7x)2 − (6)2

= (7x − 6) (7x + 6)

(iv) 16x5 − 144x3

The given expression can be written as:

= 16x3(x2 − 9)

= 16 x3 [(x)2 − (3)2]

This is of the form: a2 − b2 with a = x and b = 3

Therefore, by using the identity: a2 − b2 = (a − b)(a + b), we get

= 16x3(x − 3) (x + 3)

(v) (l + m)2 − (l − m)2

This is of the form: a2 − b2 with a = (l + m) and

b = (l − m)

Therefore, by using the identity: a2 − b2 = (a − b)(a + b), we get

= [(l + m) − (l − m)] [(l + m) + (l − m)]

= (l + m − l + m) (l + m + l − m)

= 2m × 2l

= 4ml

= 4lm

(vi) 9x2y2 − 16

The given expression can be written as:

= (3xy)2 − (4)2

This is of the form: a2 − b2 with a = 3xy and b = 4

Therefore, by using the identity: a2 − b2 = (a − b)(a + b), we get

= (3xy − 4) (3xy + 4)

(vii) (x2 − 2xy + y2) − z2

By using the identity: (a − b)2 = a2 − 2ab + b2,we get

= (x − y)2 − (z)2

This is of the form: a2 − b2 with a = (x – y) and b = z

Therefore, by using the identity: a2 − b2 = (a − b)(a + b), we get

= (x − y − z) (x − y + z)

(viii) 25a2 − 4b2 + 28bc − 49c2

The given expression can be written as:

= 25a2 − (4b2 − 28bc + 49c2)

= (5a)2 − [(2b)2 − 2 × 2b × 7c + (7c)2]

By using the identity: (a − b)2 = a2 − 2ab + b2,we get

= (5a)2 − [(2b − 7c)2]

This is of the form: x2 − y2 with x = 5a and y = (2b – 7c)

Therefore, by using the identity: x2 − y2 = (x − y)(x + y), we get

= [5a + (2b − 7c)] [5a − (2b − 7c)]

= (5a + 2b − 7c) (5a − 2b + 7c)

Q.4

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

Ans

(i) ax2 + bx

= a × x × x + b × x

Here, the common factor is ‘x’.

Therefore,

ax2 + bx = x(ax + b)

(ii) 7p2 + 21q2

= 7 × p × p + 3 × 7 × q × q

Here, the common factor is ‘7’.

Therefore,

7p2 + 21q2 = 7(p2 + 3q2)

(iii) 2x3 + 2xy2 + 2xz2

Here, the common factor is ‘2x’.

Therefore,

2x3 + 2xy2 + 2xz2 = 2x(x2 + y2 + z2)

(iv) am2 + bm2 + bn2 + an2

The given expression can be written as:

= am2 + bm2 + an2 + bn2

The common factor in first two terms is m2 and the common factor in last two terms is n2.

= m2(a + b) + n2(a + b)

= (a + b) (m2 + n2)

Therefore,

am2 + bm2 + bn2 + an2 = (a + b) (m2 + n2)

(v) (lm + l) + m + 1

Rearranging the given expression as follows:

= lm + m + l + 1

Taking out common factors .

= m(l + 1) + 1(l + 1)

= (l + 1) (m + 1)

Therefore,

(lm + l) + m + 1=(l + 1) (m + 1)

(vi) y (y + z) + 9 (y + z)

= (y + z) (y + 9)

Therefore,

y (y + z) + 9 (y + z) = (y + z) (y + 9)

(vii) 5y2 − 20y − 8z + 2yz

The given expression can be written as follows:

5y2 − 20y + 2yz − 8z

Taking out common factors.

= 5y(y − 4) + 2z(y − 4)

= (y − 4) (5y + 2z)

Therefore,

5y2 − 20y − 8z + 2yz =(y − 4) (5y + 2z)

(viii) 10ab + 4a + 5b + 2

Rearranging the given expression as follows:

= 10ab + 5b + 4a + 2

Taking out the common terms

= 5b(2a + 1) + 2(2a + 1)

= (2a + 1) (5b + 2)

Therefore,

10ab + 4a + 5b + 2= (2a + 1) (5b + 2)

(ix) 6xy − 4y + 6 − 9x

Rearranging the terms of the given expression and taking out common terms.

= 6xy − 9x − 4y + 6

= 3x(2y − 3) − 2(2y − 3)

= (2y − 3) (3x − 2)

Therefore,

6xy − 4y + 6 − 9x= (2y − 3) (3x − 2)

Q.5

Factorise.(i) a4 b4 (ii) p4 81 (iii) x4 (y + z)4(iv) x4(x – z)4 (v) a42a2b2 + b4

Ans

(i) a4b4= (a2)2 (b2)2= (a2b2) (a2+b2) [By using:(a2b2)=(a+b)(ab)]= (ab) (a+b) (a2+b2)[By using:(a2b2)=(a+b)(ab)](ii)p4 81 =(p2)2 (9)2=(p2 9) (p2+ 9)[By using:(a2b2)=(a+b)(ab)]=[(p)2 (3)2] (p2+ 9)=(p 3) (p+ 3) (p2+ 9)[By using:(a2b2)=(a+b)(ab)](iii)x4(y+z)4=(x2)2[(y+z)2]2=[x2(y+z)2][x2+(y+z)2] [By using:(a2b2)=(a+b)(ab)]

=[x(y+z)][x+(y+z)][x2+(y+z)2] [Byusing:(a2b2)=(a+b)(ab)]=(xyz) (x+y+z) [x2+(y+z)2]ivx4(xz)4=(x2)2[(xz)2]2=[x2(xz)2][x2+(xz)2] [Byusing:(a2b2)=(a+b)(ab)]=[x(xz)] [x+(xz)][x2+(xz)2] [Byusing:(a2b2)=(a+b)(ab)]=z(2xz)[x2+x22xz+z2]=z(2xz) (2x22xz+z2)(v)a42a2b2+b4=(a2)22(a2)(b2)+(b2)2=(a2b2)2=[(ab)(a+b)]2[Byusing:(a2b2)=(a+b)(ab)]=(ab)2(a+b)2

Q.6

Factorise the following expressions.(i) p2 + 6p + 8 (ii) q210q + 21(iii) p2 + 6p 16

Ans

(i) p2 + 6p + 8

Here, 8 = 4 × 2 and

6 = 4 + 2

Therefore, p2 + 6p + 8 can be written as:

p2 + 2p + 4p + 8

= p(p + 2) + 4(p + 2)

= (p + 2) (p + 4)

(ii) q2 − 10q + 21

Here, 21 = (−7) × (−3) and

−10 = (−7) + (−3)

Therefore, q2 − 10q + 21 can be written as:

q2 − 7q − 3q + 21

= q(q − 7) − 3(q − 7)

= (q − 7) (q − 3)

(iii) p2 + 6p − 16

Here, 16 = (−2) × 8 and

6 = 8 + (−2)

Therefore, p2 + 6p – 16 can be written as:
= p2 + 8p − 2p − 16

= p(p + 8) − 2(p + 8)

= (p + 8) (p – 2)

Q.7

Carry out the following divisions.(i)  28x4 ÷ 56x (ii) 36y3 ÷ 9y2 (iii) 66pq2r3 ÷ 11qr2(iv) 34x3y3z3 ÷ 51xy2z3 (v) 12a8b8 ÷ (6a6b4)

Ans

(i)28x4÷56x28x4=2×2×7×x×x×x×x56x=2×2×2×7×x28x456x=2×2×7×x×x×x×x2×2×2×7×x=x32(ii)36y3÷9y236y3=2×2×3×3×y×y×y9y2 =3×3×y×y36y39y2=3×3×2×2×y×y×y3×3×y×y=4y(iii)66pq2r3÷11qr266pq2r3=2×3×11×p×q×q×r×r×r11qr2 =11×q×r×r66pq2r311qr2=11×3×2×p×q×q×r×r×r11×q×r×r=6pqr

(iv)34x3y3z3÷51xy2z334x3y3z3=2×17×x×x×x×y×y×y×z×z×z51xy2z3 =3×17×x×y×y×z×z×z34x3y3z351xy2z3=2×17×x×x×x×y×y×y×z×z×z3×17×x×y×y×z×z×z=2x2y3(v)12a8b8÷(6a6b4)12a8b8=2×2×3×a×a×a×a×a×a×a×a×b×b×b×b×b×b×b×b6a6b4 =3×2×a×a×a×a×a×a×b×b×b×b12a8b8(6a6b4)=2×2×3×a×a×a×a×a×a×a×a×b×b×b×b×b×b×b×b3×2×a×a×a×a×a×a×b×b×b×b=2a2b4

Q.8

Divide the given 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

Ans

(i)(5x26x)÷3x(5x26x)3x=x(5x6)3x=(5x6)3(ii)(3y84y6+5y4)÷y4(3y84y6+5y4)y4=y4(3y44y2+5)y4=3y44y2+5

(iii)8(x3y2z2+x2y3z2+x2y2z3)÷4x2y2z28(x3y2z2+x2y3z2+x2y2z3)4x2y2z2=8x2y2z2(x+y+z)4x2y2z2=2(x+y+z)

(iv)(x3+2x2+3x)÷2x(x3+2x2+3x)2x=x(x2+2x+3)2x=(x2+2x+3)2(v)(p3q6p6q3)÷p3q3p3q6p6q3p3q3=p3q3(q3p3)p3q3=q3p3

Q.9 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)

Ans

(i) (10x25)÷5 (10x25)5= 5(2x5)5=2x5(ii) (10x25)÷(2x5)(10x25)(2x5)=5(2x5)(2x5)=5

(iii) 10y(6y+21)÷5(2y+7)10y(6y+21)5(2y+7)=10y×3(2y+7)5(2y+7)=6y(iv) 9x2y2(3z24)÷27xy(z8)9x2y2(3z24)27xy(z8)=9x2y2×3(z8)27xy(z8)=xy

(v) 96abc(3a12)(5b 30)÷144(a4)(b6)96abc(3a12)(5b 30)144(a4)(b6)=96abc×3(a4)×5(b6)144(a4)(b6)=10abc

Q.10

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)

Ans

(i)5(2x+1)(3x+5)÷(2x+1)5(2x+1)(3x+5)(2x+1)=5(2x+1)(3x+5)(2x+1)=5(3x+5)(ii)26xy(x+5)(y4)÷13x(y4)26xy(x+5)(y4)13x(y4)=26xy(x+5)(y4)13x(y4)=2y(x+5)

(iii)52pqr(p+q)(q+r)(r+p)÷104pq(q+r)(r+p)52pqr(p+q)(q+r)(r+p)104pq(q+r)(r+p)

=2×2×13×pqr(p+q)(q+r)(r+p)2×2×2×13×pq(q+r)(r+p)=r(p+q)2

(iv)20(y+4)(y2+5y+3)÷5(y+4)20(y+4)(y2+5y+3)5(y+4)=5×4(y+4)(y2+5y+3)5(y+4)=4(y2+5y+3)

(v)x(x+1)(x+2)(x+3)÷x(x+1)x(x+1)(x+2)(x+3)x(x+1)=x(x+1)(x+2)(x+3)x(x+1)=(x+2)(x+3)

Q.11

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)

Ans

(i)(y2+7y+10)÷(y+5)First we will factorise (y2+7y+10)=y2+2y+5y+10=y(y+2)+5(y+2)=(y+2)(y+5)(y2+7y+10)(y+5)=(y+2)(y+5)(y+5)=(y+2)

(ii) (m214m32)÷(m+2)First we will factorise (m214m32)=m216m+2m32=m(m16)+2(m16)=(m16)(m+2)(m214m32)(m+2)=(m16)(m+2)(m+2)=(m16)

(iii) (5p225p+20)÷(p1)First we will factorise (5p225p+20)=5p220p5p+20=5p(p4)5(p4)=5(p1)(p4)(5p225p+20)(p1)=5(p4)(p1)(p1)=5(p4)

(iv) 4yz(z2+6z16)÷2y(z+8)First we will factorise 4yz(z2+6z16)=4yz(z2+8z2z16)=4yz[z(z+8)2(z+8)]=4yz[(z+8)(z2)]4yz(z2+6z16)2y(z+8)=4yz[(z+8)(z2)]2y(z+8)=2z(z2)

(v) 5pq(p2q2)÷2p(p+q)First we will factorise 5pq(p2q2)=5pq(pq)(p+q)[Byusing:a2b2=(a+b)(ab)]5pq(p2q2)2p(p+q)=5pq(pq)(p+q)2p(p+q)=5q(pq)2

(vi) 12xy(9x216y2)÷4xy(3x+4y)First we will factorise 12xy(9x216y2)=12xy[(3x)2(4y)2]= 12xy[(3x4y)(3x+4y)] [Byusing:a2b2=(a+b)(ab)]12xy[(3x+4y)(3x4y)] 4xy(3x+4y)=3(3x4y)

(vii) 39y3(50y298)÷26y2(5y+7)First we will factorise 39y3(50y298)=13×3×y3×2(25y249)=13×3×y3×2[(5y)272]=13×3×y3×2[(5y7)(5y+7)] [Byusing:a2b2=(a+b)(ab)]39y3(50y298)26y2(5y+7)

=13×3×y3×2[(5y7)(5y+7)] 2×13×y2(5y+7)=3y(5y7)

Q.12

Find and correct the errors in the following mathematical statements. 1. 4(x -5) = 4x- 5 2 . x(3x + 2) = 3x 2 + 2 3. 2x + 3y = 5xy 4. x + 2x + 3x = 5x 5. 5y + 2y + y- 7y = 0 6 . 3x + 2x = 5x 2 7 . (2x) 2 + 4(2x) + 7 = 2x 2 + 8x + 7 8 . (2x) 2 + 5x = 4x + 5x = 9x 9 . (3x + 2) 2 = 3x 2 + 6x + 4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@217B@

Ans

1. 4(x – 5) = 4x – 5

L.H.S.

4(x − 5)

= 4 × x − 4 × 5

= 4x − 20 ≠ R.H.S.

The correct statement is:

4(x − 5) = 4x – 20

2. x(3x + 2) = 3x2 + 2

L.H.S.

= x(3x + 2)

= x × 3x + x × 2

= 3x2 + 2x ≠ R.H.S.

The correct statement is:

x(3x + 2) = 3x2 + 2x

3. 2x + 3y = 5xy

L.H.S.

= 2x + 3y

R.H.S.

The correct statement is :

2x + 3y = 2x + 3y

4. x + 2x + 3x = 5x

L.H.S

= x + 2x + 3x

=1x + 2x + 3x

= x(1 + 2 + 3)

= 6x ≠ R.H.S.

The correct statement is :

x + 2x + 3x = 6x

5. 5y + 2y + y – 7y = 0

L.H.S.

= 5y + 2y + y − 7y

= 8y − 7y

= y ≠ R.H.S

The correct statement is:

5y + 2y + y − 7y = y

6. 3x + 2x = 5x2

L.H.S.

= 3x + 2x

= 5x ≠ R.H.S

The correct statement is:

3x + 2x = 5x

7. (2x)2 + 4(2x) + 7 = 2x2 + 8x + 7

L.H.S

= (2x)2 + 4(2x) + 7

= 4x2 + 8x + 7 ≠ R.H.S

The correct statement is:

(2x)2 + 4(2x) + 7 = 4x2 + 8x + 7

8. (2x)2 + 5x = 4x + 5x = 9x

L.H.S

= (2x)2 + 5x

= 4x2 + 5x ≠ R.H.S.

The correct statement is:

(2x)2 + 5x = 4x2 + 5x

9. (3x + 2)2 = 3x2 + 6x + 4

L.H.S.

= (3x + 2)2

= (3x)2 + 2(3x)(2) + (2)2

[By using: (a + b)2 = a2 + 2ab + b2]

= 9x2 + 12x + 4 ≠ R.H.S

The correct statement is:

(3x + 2)2 = 9x2 + 12x + 4

Q.13

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) = 915 = 24

Ans

(a) When we substitute x = −3 in x2 + 5x + 4, we get

= (−3)2 + 5 (−3) + 4

= 9 − 15 + 4

= 13 − 15

= −2

Therefore, the correct statement is:

x2 + 5x + 4 = − 2

(b) When we substitute x = −3 in x2 − 5x + 4, we get

= (−3)2 − 5 (−3) + 4

= 9 + 15 + 4

= 28

Therefore, the correct statement is:

x2 − 5x + 4 =28

(c) When we substitute x = −3 in x2 + 5x, we get

= (−3)2 + 5(−3)

= 9 − 15

= −6

Therefore, the correct statement is:

x2 + 5x = − 6

Q.14 Find and correct the errors in the below mathematical statement:

(y3)2=y29

Ans

L.H.S=(y3)2=(y)22(y)(3)+(3)2 [By using:(ab)2=a2 2ab+b2]=y26y+9 R.H.SThe correct statement is: (y 3)2=y26y+ 9

Q.15 Find and correct the errors in the below mathematical statement:

(z+5)2=z2+25

Ans

L.H.S=(z+ 5)2=(z)2+2(z)(5)+(5)2[By Using:(a+b)2=a2+ 2ab+b2]=z2+10z+25R.H.SThe correct statement is:(z+ 5)2=z2+ 10z+ 25

Q.16 Find and correct the errors in the below mathematical statement:

(2a+3b)(ab)=2a23b2

Ans

L.H.S. =(2a+ 3b) (ab) =2a×a+ 3b×a 2a×b 3b×b=2a2+ 3ab 2ab 3b2= 2a2+ab 3b2 R.H.S.The correct statement is: (2a+ 3b)(ab)=2a2+ab 3b2

Q.17 Find and correct the errors in the below mathematical statement:

(a + 4)(a + 2)=a2+8

Ans

L.H.S.=(a+ 4) (a+ 2) =(a)2+ (4 + 2) (a) + 4 × 2=a2+ 6a+ 8 R.H.SThe correct statement is:(a+ 4)(a+ 2)=a2+6a+8

Q.18 Find and correct the errors in the below mathematical statement:

(a4)(a2)=a28

Ans

L.H.S.=(a 4)(a 2)=(a)2+[( 4)+( 2)](a)+( 4)( 2)=a26a+ 8 R.H.S.The correct statement is:(a 4) (a 2) =a2 6a+ 8

Q.19 Find and correct the errors in the below mathematical statement:

3x23x2=0

Ans

L.H.S=3x23x2=3×x×x3×x×x=1R.H.SThe correct statement is:3x23x2=1

Q.20 Find and correct the errors in the below mathematical statement:

3x2+13x2=1+1=2

Ans

L.H.S=3x2+13x2=3x23x2+13x2

=1+13x2R.H.SThe correct statement is:3x2+13x2=1+13x2

Q.21 Find and correct the errors in the below mathematical statement:

3x3x+2=12

Ans

L.H.S=3x3x+2R.H.SThe correct statement is:3x3x+2=3x3x+2

Q.22 Find and correct the errors in the below mathematical statement:

34x+3=14x

Ans

L.H.S=34x+3R.H.SThe correct statement is:34x+3=34x+3

Q.23 Find and correct the errors in the below mathematical statement:

4x + 54x=5

Ans

L.H.S=4x + 54x=4x4x+54x

=1+54xR.H.SThe correct statement is:4x+54x=1+54x

Q.24 Find and correct the errors in the below mathematical statement:

7x+55=7x

Ans

L.H.S=7x+55=7x5+55

=7x5+1R.H.SThe correct statement is:7x+55=7x5+1

Q.25 Find the common factors of the given terms.

(i) 12x, 36 (ii) 2y, 22xy (iii) 14pq, 28p2q2

(iv) 2x, 3x2, 4 (v) 6abc, 24ab2, 12 a2b

(vi) 16 x3, – 4x2, 32x (vii) 10pq, 20qr, 30rp

(viii) 3x2y3, 10x3y2, 6x2y2z

Ans

(i) 12x = 2 × 2 × 3 × x

36 = 2 × 2 × 3 × 3

Therefore, a common factor of 12x and 36 is 2 × 2 × 3 = 12

(ii) 2y = 2 × y

22xy = 2 × y × 11 × x

Therefore, a common factor of 2y and 22xy is 2 × y = 2y

(iii) 14pq = 2 × 7 × p × q

28p2q2 = 2 × 7 × p × q × p × q × 2

Therefore, a common factor of 14pq and 28p2q2 is 2 × 7 × p × q = 14pq

(iv) 2x = 1× 2 × x

3x2= 1× 3 × x × x

4 = 1× 2 × 2

Therefore, a common factor of 2x, 3x2 and 4 is 1.

(v) 6abc = 2 × 3 × a × b × c

24ab2 = 2 × 3 × a × b × 2 × 2 × b

12a2b = 2 × 3 × a × b × a × 2

Therefore, a common factor of 6abc, 24ab2 and 12a2b is 2 × 3 × a × b = 6ab

(vi) 16x3 = 2 × 2 × 2 × x × 2 × x × x

−4x2 = 2 × 2 × −1 × x × x

32x = 2 × 2 × 2 × x × 2 × 2

Therefore, a common factor of 16x3, −4x2 and 32x is 2 × 2 × x = 4x

(vii) 10pq = 2 × 5 × p × q

20qr = 2 × 5 × 2 × q × r

30rp = 2 × 5 × 3 × r × p

Therefore, a common factor of 10pq, 20qr and 30rp is 2 × 5 = 10.

(viii) 3x2y3 = 3 × y × x × x × y × y

10x3y2 = 2 × 5 × x × x × y × y × x

6x2y2z = 2 × 3 × x × x × y × y × z

Therefore, a common factor of 3x2y3, 10x3y2 and 6x2y2z is x × x × y × y = x2y2

Q.26 Factorise the following expressions.

(i) 7x – 42 (ii) 6p – 12q (iii) 7a2 + 14a

(iv) – 16 z + 20 z3 (v) 20l2m + 30alm

(vi) 5x2y – 15 xy2 (vii) 10a2 – 15b2 + 20c2

(viii) – 4 a2 + 4ab – 4ca (ix) x2yz + x y2z + xyz2

(x) ax2y + bxy2 + cxyz

Ans

(i) 7x – 42

7x = 7 × x

42 = 7 × 3 × 2

Here, the common factor is 7.

Therefore,

7x − 42 = (7 × x) − (2 × 3 × 7)

= 7 (x − 6)

(ii) 6p – 12q

6p = 2 × 3 × p

12q = 2 × 3 × 2 × q

Here, the common factors are 2 and 3.

Therefore,

6p − 12q = (2 × 3 × p) − (2 × 2 × 3 × q)

= 2 × 3 [p − (2 × q)]

= 6 (p − 2q)

(iii) 7a2 + 14a

7a2 = 7 × a × a

14a = 7 × a × 2

The common factors are 7 and a.

Therefore,

7a2 + 14a = (7 × a × a) + (2 × 7 × a)

= 7 × a [a + 2]

= 7a (a + 2)

(iv) −16 z + 20 z3

16z = 2 × 2 × 2 × 2 × z

20z3 = 2 × 2 × 5 × z × z × z

Here, the common factors are 2, 2, and z.

Therefore, −16z + 20z3

= − (2 × 2 × 2 × 2 × z) + (2 × 2 × 5 × z × z × z)

= (2 × 2 × z) [− (2 × 2) + (5 × z × z)]

= 4z (− 4 + 5z2)

(v) 20l2m + 30alm

20l2m = 2 × 2 × 5 × l × l × m

30alm = 2 × 3 × 5 × a × l × m

Here, the common factors are 2, 5, l, and m.

Therefore, 20l2m + 30alm

= (2 × 2 × 5 × l × l × m) + (2 × 3 × 5 × a × l × m)

= (2 × 5 × l × m) [(2 × l) + (3 × a)]

= 10lm (2l + 3a)

(vi) 5x2y − 15 xy2

5x2y = 5 × x × x × y

15xy2 = 5 × 3 × x × y × y

Here, the common factors are 5, x, and y.

Therefore,

5x2y − 15xy2 = (5 × x × x × y) − (3 × 5 × x × y × y)

= 5 × x × y [x − (3 × y)]

= 5xy (x − 3y)

(vii) 10a2 − 15b2 + 20c2

10a2 = 2 × 5 × a × a

15b2 = 3 × 5 × b × b

20c2 = 2 × 5 × 2 × c × c

Here, the common factor is 5.

Therefore, 10a2 − 15b2 + 20c2

= (2 × 5 × a × a) − (3 × 5 × b × b) + (2 × 2 × 5 × c × c)

= 5 [(2 × a × a) − (3 × b × b) + (2 × 2 × c × c)]

= 5(2a2 − 3b2 + 4c2)

(viii) – 4 a2 + 4ab – 4ca

4a2 = 2 × 2 × a × a

4ab = 2 × 2 × a × b

4ca = 2 × 2 × a × c

Here, the common factors are 2, 2, and a.

Therefore, −4a2 + 4ab − 4ca

= − (2 × 2 × a × a) + (2 × 2 × a × b) − (2 × 2 × c × a)

= 2 × 2 × a [− (a) + b − c]

= 4a (−a + b − c)

(ix) x2yz + x y2z + xyz2

x2yz = x × y × x × z

xy2z = x × y × y × z

xyz2 = x × y × z × z

Here, the common factors are x, y, and z.

Therefore, x2yz + xy2z + xyz2

= (x × x × y × z) + (x × y × y × z) + (x × y × z × z)

= x × y × z [x + y + z]

= xyz (x + y + z)

(x) ax2y + bxy2 + cxyz

ax2y = a × x × x × y

bxy2 = b × x × y × y

cxyz = c × x × z × y

Here, the common factors are x and y.

Therefore, ax2y + bxy2 + cxyz

= (a × x × x × y) + (b × x × y × y) + (c × x × y × z)

= (x × y) [(a × x) + (b × y) + (c × z)]

= xy(ax + by + cz)

For viewing question paper please click here

FAQs (Frequently Asked Questions)
1. How can I be good at the chapter of factorisation?

The chapter factorisation is directly related with the chapter algebraic expressions covered in the lower classes. Therefore, to be good in the chapter, students should know and be able to solve algebraic expressions efficiently. Also, the chapter factorisation is based more on the knowledge of tables.

Hence, students must be thorough with tables. When students have good knowledge about algebraic expressions and tables, they just have to learn to apply the steps of factorisation. In this way, one can be good at chapter factorisation.

Students can find all the steps related to carrying out factorisation accurately in the NCERT Solutions for Class 8 Mathematics Chapter 14 available on the Extramarks website.

2. Is there any other study material on the Extramarks website besides the NCERT solutions for Class 8 Mathematics chapter 14?

Students can find a variety of questions related to the chapter and academic notes based on the NCERT textbook on the Extramarks website. Also, they can find NCERT revision notes, NCERT-based papers and mock tests on the website.