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# NCERT Solutions for Class 8 Maths Chapter 14: Factorisation - Exercise 14.4

Last updated date: 14th Aug 2024
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## NCERT Solutions for Class 8 Maths Chapter 14 (EX 14.4)

Free PDF download of NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.4 and all chapter exercises at one place prepared by an expert teacher as per NCERT (CBSE) books guidelines. Class 8 Maths Chapter 14 Factorisation Exercise 14.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

 Class: NCERT Solutions for Class 8 Subject: Class 8 Maths Chapter Name: Chapter 14 - Factorisation Exercise: Exercise - 14.4 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

Every NCERT Solution is provided to make the study simple and interesting. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 8 Science , Maths solutions and solutions of other subjects. You can also download NCERT Solutions for Class 8 Maths to help you to revise complete syllabus and score more marks in your examinations.

## Access NCERT Solutions for Class 7 Mathematics Chapter 14- Factorisation

Exercise 14.4

Refer the page 15 to 20 for the exercise 14.4 in the PDF

1. Find and correct the errors in the statement${\mathbf{4}}\left( {{\mathbf{x}} - {\mathbf{5}}} \right) = {\mathbf{4x}} - {\mathbf{5}}$.

Ans:  From the above question we get,

$4(x - 5) = 4x - 20$

Thus, $4x - 20 \ne 4x - 5$

So, the correct statement is $4x - 20$.

2. Find and correct the errors in the statement${\mathbf{x}}\left( {{\mathbf{3x}} + {\mathbf{2}}} \right) = {\mathbf{3}}{{\mathbf{x}}^{\mathbf{2}}} + {\mathbf{2}}$

Ans: From the above question we get,

$x(3x + 2) = 3{x^2} + 2x$

Again, $3{x^2} + 2x \ne 3{x^2} + 2$

So, the correct answer is $3{x^2} + 2x$

3. Find and correct the errors in the statement: ${\mathbf{2x}} + {\mathbf{3y}} = {\mathbf{5xy}}$

Ans: $L.H.S = 2x + 3y \ne R.H.S$

The correct statement is $2x + 3y = 2x + 3y$

4. Find and correct the errors in the statement: x + 2x + 3x = 5x

Ans: $L.H.S = x + 2x + 3x = 1x + 2x + 3x = x\left( {1 + 2 + 3} \right) = 6x \ne {\text{R}}.H.S.$

The correct statement is x + 2x + 3x = 6x

5. Find and correct the errors in the statement: ${\mathbf{5y}} + {\mathbf{2y}} + {\mathbf{y}} - {\mathbf{7y}} = {\mathbf{0}}$

Ans: $L.H.S. = 5y + 2y + y - {\text{7}}y = 8y - 7y = y \ne R.H.S$

The correct statement is 5y + 2y + y − 7y = y

6. Find and correct the errors in the statement: 3x + 2x = 5x 2

Ans: $L.H.S. = 3x + 2x = 5x \ne {\text{R}}.H.S$

The correct statement is $3x + 2x = 5x$

7. Find and correct the errors in the statement: ${\left( {{\mathbf{2x}}} \right)^{\mathbf{2}}} + {\mathbf{4}}\left( {{\mathbf{2x}}} \right) + {\mathbf{7}} = {\mathbf{2}}{{\mathbf{x}}^{\mathbf{2}}} + {\mathbf{8x}} + {\mathbf{7}}$

Answer:$\;L.H.S = {\left( {2x} \right)^2} + 4\left( {2x} \right) + 7 = 4{x^2} + 8x + 7 \ne R.H.S$

The correct statement is ${\left( {2x} \right)^2} + 4\left( {2x} \right) + 7 = 4{x^2} + 8x + 7$

8. Find and correct the errors in the statement:

Ans:$L.H.S = {\left( {2x} \right)^2} + 5x = 4{x^2} + 5x \ne R.H.S.$The correct statement is ${\left( {2x} \right)^2} + 5x = 4{x^2} + 5x$

9. Find and correct the errors in the statement: (3x + 2)2 = 3x 2 + 6x + 4

Ans: $L.H.S. = {\left( {3x + 2} \right)^2} = {\left( {3x} \right)^2} + 2\left( {3x} \right)\left( 2 \right) + {\left( 2 \right)^2}$

$\left[ {{{\left( {a + b} \right)}^2} = {a^2} + 2ab + {b^2}} \right] = 9{x^2} + 12x + 4 \ne R.H.S$

The correct statement is ${\left( {3x + 2} \right)^2} = 9{x^2} + {\text{1}}2x + 4$

10. Find and correct the errors in the following mathematical statement. Substituting ${\mathbf{x}} = - {\mathbf{3}}$in

a) ${{\mathbf{x}}^{\mathbf{2}}} + {\mathbf{5x}} + {\mathbf{4}}$ gives ${\left( { - {\mathbf{3}}} \right)^{\mathbf{2}}} + {\mathbf{5}}\left( { - {\mathbf{3}}} \right) + {\mathbf{4}} = {\mathbf{9}} + {\mathbf{2}} + {\mathbf{4}} = {\mathbf{15}}$

Ans: For$x = - 3$,

${x^2} + 5x + 4$

$= {\left( { - 3} \right)^2} + 5\left( { - 3} \right) + 4$

$= 9 - 15 + 4$

$= 13 - 15$

$= - 2$

b) ${{\mathbf{x}}^{\mathbf{2}}} - {\mathbf{5x}} + {\mathbf{4}}$ gives ${\left( { - {\mathbf{3}}} \right)^{\mathbf{2}}} - {\mathbf{5}}\left( { - {\mathbf{3}}} \right) + {\mathbf{4}} = {\mathbf{9}} - {\mathbf{15}} + {\mathbf{4}} = - {\mathbf{2}}$

Ans: For $x = - 3$,

$\;{x^2} - 5x + 4$

$= {\left( { - 3} \right)^2} - 5\left( { - 3} \right) + 4$

$= 9 + 15 + 4$

$= 28$

c) ${{\mathbf{x}}^{\mathbf{2}}} + {\mathbf{5x}}$ gives ${\left( { - {\mathbf{3}}} \right)^{\mathbf{2}}} + {\mathbf{5}}\left( { - {\mathbf{3}}} \right) = - {\mathbf{9}} - {\mathbf{15}} = - {\mathbf{24}}$

Ans:For $x = - 3$,

${x^2} + 5x$

$= {\left( { - 3} \right)^2} + 5\left( { - 3} \right)$

$= 9 - 15$

$= - 6$

11. Find and correct the errors in the statement: ${\left( {{\mathbf{y}} - {\mathbf{3}}} \right)^{\mathbf{2}}} = {\mathbf{y}} - {\mathbf{9}}$

Ans: $L.H.S = \left( {y - 3} \right)2 = {\left( y \right)^2} - 2\left( y \right)\left( 3 \right) + {\left( 3 \right)^2}\left[ {{{\left( {a - b} \right)}^2} = {a^2} - 2ab + {b^2}} \right] = {y^2} - 6y + 9 \ne R.H.S\,$The correct statement is ${\left( {y - 3} \right)^2} = {y^2} - 6y + 9$

12. Find and correct the errors in the statement: ${\left( {{\mathbf{z}} + {\mathbf{5}}} \right)^{\mathbf{2}}} = {{\mathbf{z}}^{\mathbf{2}}} + {\mathbf{25}}$

Ans: $L.H.S = {\left( {z + 5} \right)^2} = {\left( z \right)^2} + 2\left( z \right)\left( 5 \right) + {\left( 5 \right)^2}\left[ {{{\left( {a{\text{ }} + {\text{ }}b} \right)}^2} = {a^2} + 2ab + {b^2}} \right] = {z^2} + 10z + 25 \ne R.H.S$The correct statement is ${\left( {z + 5} \right)^2} = {z^2} + {\text{1}}0z + 25$

13. Find and correct the errors in the statement: $\left( {{\mathbf{2a}} + {\mathbf{3b}}} \right)\left( {{\mathbf{a}} - {\mathbf{b}}} \right) = {\mathbf{2}}{{\mathbf{a}}^{\mathbf{2}}} - {\mathbf{3}}{{\mathbf{b}}^{\mathbf{2}}}$

Ans: $L.H.S. = \left( {2a + 3b} \right)\left( {a - b} \right) = 2a \times a + 3b \times a - 2a \times b - 3b \times b = 2{a^2} + 3ab - 2ab - 3{b^2} = 2{a^2} + {\text{a}}b - 3{b^2} \ne R.H.S.$The correct statement is $\left( {2a + 3b} \right)\left( {a - b} \right) = 2a2 + {\text{a}}b - 3{b^2}$

14. Find and correct the errors in the statement: $\left( {{\mathbf{a}} + {\mathbf{4}}} \right)\left( {{\mathbf{a}} + {\mathbf{2}}} \right) = {{\mathbf{a}}^{\mathbf{2}}} + {\mathbf{8}}$

Ans:$L.H.S. = \left( {a + 4} \right)\left( {a + 2} \right) = {\left( a \right)^2} + \left( {4 + 2} \right)\left( a \right) + 4 \times 2 = {a^2} + 6a + 8 \ne R.H.S$correct statement is $\left( {a + 4} \right)\left( {a + 2} \right) = {a^2} + 6a + 8$

15. Find and correct the errors in the statement: $\left( {{\mathbf{a}} - {\mathbf{4}}} \right)\left( {{\mathbf{a}} - {\mathbf{2}}} \right) = {{\mathbf{a}}^{\mathbf{2}}} - {\mathbf{8}}$

Ans: $L.H.S. = \left( {a - 4} \right)\left( {a - 2} \right) = {\left( a \right)^2} + \left[ {\left( { - 4} \right) + \left( { - 2} \right)} \right]\left( a \right) + \left( { - 4} \right)\left( { - 2} \right) = {a^2} - 6a + 8 \ne R.H.S.$

The correct statement is $\left( {a - 4} \right)\left( {a - 2} \right) = {a^2} - {\text{6}}a + 8$

16. Find and correct the errors in the statement $\dfrac{{3{x^2}}}{{3{x^2}}} = 0$

Ans: L.H.S = The correct statement is

LHS

$\dfrac{{3{x^2}}}{{3{x^2}}} = \dfrac{{3 \times x \times x}}{{3 \times x \times x}} = 1 \ne R.H.S.$

The correct statement is$\dfrac{{3{x^2}}}{{3{x^2}}} = 1$

17. Find and correct the errors in the statement: $\dfrac{{3{x^2} + 1}}{{3{x^2}}} = 1 + 1 = 2$

Ans: The correct statement is

$\dfrac{{3{x^2} + 1}}{{3{x^2}}}$

$= \dfrac{{3{x^2}}}{{3{x^2}}} + \dfrac{1}{{3{x^2}}}$

$= 1 + \dfrac{1}{{3{x^2}}}$

$\ne R.H.S$

The correct statement is $\dfrac{{3{x^2} + 1}}{{3{x^2}}} = 1 + \dfrac{1}{{3{x^2}}}$

18. Find and correct the errors in the statement: $\dfrac{{3x}}{{3x + 2}} = \dfrac{1}{2}$

Ans: L.H.S =$\dfrac{{3x}}{{3x + 2}} \ne R.H.S.$

The correct statement is $\dfrac{{3x}}{{3x + 2}} = \dfrac{{3x}}{{3x + 2}}$

19. Find and correct the errors in the statement: $\dfrac{3}{{4x + 3}} = \dfrac{1}{{4x}}$

Ans: L.H.S. =$\dfrac{3}{{4x + 3}}$ ≠ R.H.S.

The correct statementis $\dfrac{3}{{4x + 3}} = \dfrac{3}{{4x + 3}}$

20. Find and correct the errors in the statement: $\dfrac{{4x + 5}}{{4x}} = 5$

Ans: L.H.S. $\dfrac{{4x + 5}}{{4x}} = \dfrac{{4x}}{{4x}} + \dfrac{5}{{4x}} = 1 + \dfrac{5}{{4x}} \ne R.H.S$

The correct statement is$\dfrac{{4x + 5}}{{4x}} = 1 + \dfrac{5}{{4x}}$

21. Find and correct the errors in the statement: $\dfrac{{7x + 5}}{5} = 7x$

Ans: L.H.S. =$\dfrac{{7x + 5}}{5} = \dfrac{{7x}}{5} + \dfrac{5}{5} = \dfrac{{7x}}{5} + 1 \ne R.H.S$

The correct statement is$\dfrac{{7x + 5}}{5} = \dfrac{{7x}}{5} + 1$

## NCERT Solutions for Class 8 Maths Chapter 14 Factorisation (Ex 14.4) Exercise 14.4

Opting for the NCERT solutions for Ex 14.4 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 14.4 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 14 Exercise 14.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 14 Exercise 14.4, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 8 Maths Chapter 14 Exercise 14.4 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

## FAQs on NCERT Solutions for Class 8 Maths Chapter 14: Factorisation - Exercise 14.4

1. What is the Importance of NCERT Solutions for Chapter 14 of Maths Class 8?

Chapter 14 of Maths Class 8 deals in factorization. The definition of factorisation is explained in this PDF. The process of factorisation is the representation of a given statement as the product of two factors. These elements might be integers, variables, or algebraic expressions themselves. The PDF of solutions also states that it uses several systematic techniques to discover factors for the provided phrases.

2. What are the important topics covered in Chapter 14 of Maths Class 8?

All key subjects of the factorization are covered in Chapter 14 of Maths Class 8 are factors of natural numbers, factors of algebraic expressions, factorization techniques, and division of algebraic expressions. The solutions of this chapter are available on Vedantu’s website(vedantu.com) as well as Vedantu’s mobile app. The solutions are free of cost.

3. How many questions are there in Chapter 14 Exercise 14.4 of Maths Class 8?

There are 10 in Class 8 Maths Chapter 14 Factorization Exercise 14.4, most of which are short answer questions with subparts. Factoring algebraic equations and dividing polynomials by polynomials are the most common techniques used in these issues.

4. What is the meaning of factorisation according to Chapter 14 of Maths Class 8?

Algebraic expression factorization is a crucial ability for advanced math courses. It means breaking down an equation into smaller numbers. Factorization of expressions is a difficult procedure that necessitates a thorough grasp of factors and the methods for listing them. Students may swiftly gain this information by thoroughly practising NCERT Solutions of Chapter 14 of Maths Class 8 in order to prepare effectively for examinations. These solutions also help pupils build confidence in preparation for a variety of competitive studies.

5. What is the formula for factorisation for solving sums in Exercise 14.4?

The factorization formula breaks down a large number into smaller numbers or factors. A factor is a number that divides a given integer completely without leaving any residual. The factorization formula for a given value is as follows N = Xa × Yb × Zc

N = Any number

X, Y, and Z = Factors of number N

a, b, and c = exponents of factors X, Y, and Z respectively.

6. What is factorisation algebra according to Chapter 14 of Maths Class 8?

The term "algebraic factorization" refers to the process of identifying the factors of a given expression, which entails identifying two or more expressions whose product is the given expression. Factorization of algebraic expressions is the process of finding two or more expressions whose product is the given expression.