# NCERT Solutions Class 8 Maths Chapter 14 Exercise 14.4

NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4 are provided to make learning more enjoyable. Having access to NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4, Mathematics Answers, and other subject answers simplifies learning subjects like Science, Mathematics, and English. Students can also download the NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4 to revamp the entire curriculum and score more points in their exams.

Students can download PDF NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4 and all chapter exercises in one place, prepared by an experienced teacher according to the NCERT (CBSE) book guidelines. NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4 with answers to help students revise their entire syllabus and gain good scores. Students can register at Extramarks to receive all the solutions by email.

## NCERT Solutions for Class 8 Mathematics Chapter 14 Factorisation (EX 14.4) Exercise 14.4

NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4 answers are prepared in logical and simple language for quick revision. These NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4  help students create perfect answers in mathematics exams. The NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4 aims to help students master the concepts and strengthen the foundations of higher education. These NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4 were created by subject matter experts to assist students in solving all NCERT textbook problems, making them an excellent resource for exam success.Download now and start practising students’ tasks.

In NCERT Solutions for Class 8 Mathematics, Chapter 14, Exercise 14.4, calculations are ways of finding and correcting errors in mathematical statements. Finding and correcting errors in algebraic expressions involves ordering terms, multiplying them, and solving them. This skill requires a thorough understanding of factoring. Exercise 14.4 consists of 10 short answers, which are then divided into small sections to find errors and terminology. Returning the terms to their exact form is the first step in solving algebraic expressions, NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4 is also useful to find errors in expressions. NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4 teaches children this skill through practise with some examples and sample problems.

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

Factorization is the classic mathematical operation of separating components (or numbers) that are multiplied to produce a new number. A given integer has several components. Multiply 6 and 4, 8 and 3, 12 and 2, 24 and 1, and students get 24. Students can use factorization to solve various mathematics problems. When abbreviating formulas, students usually need to remove all parentheses, but sometimes fraction formulas can be shortened by factorisation, multiplication, and division. A factor of a number is any number that can be multiplied to produce that number. Students can also represent numbers by repeating division. Factoring large numbers can be difficult at first, but there are some simple methods students can use to quickly discover the factors. To find the factors of a number, students need to find all the terms that multiply together to make the number. To fully factorize, reduce the number of prime factors.

A factor tree can be used to represent the division of large integers into prime numbers. Students can put the number they want to factorise at the beginning of the statement and divide it into steps by that factor. Every time someone divides a number, students have to place the two factors of that number. Then they have to divide until all numbers are prime factors. After mastering the basics of factoring, students may be asked to find the greatest common divisor of two integers or expressions. Then they have to make a list of the divisors of both numbers and find the largest common divisor.

All the solutions to Class 8 Mathematics Exercise 14.4 are available on the website of Extramarks, and students can easily download them from Class 8 Mathematics Chapter 14 Exercise 14.4 from the website anytime they want.

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

Factoring a number or polynomial into a product of many factors of other polynomials and multiplying them to get the original number is called factorisation. To factorise, use the factorisation formula. Factorisation is the process of converting a unit (number, matrix, polynomial, etc.) into another unit or product of factors, which when multiplied together yields the original number. The factorisation formula divides a large number into smaller numbers called factors. A factor is a number that completely divides a given integer without leaving a remainder. for example, the prime factorization of 28 = 2 X 2 X 7. Before students start factoring, they should understand the important mathematical term “factor.” What is a factor? A factor is a number, an algebraic variable, or an algebraic expression that divides a number or algebraic expression without leaving a remainder. For example, the factor 9 is the 1,3,9 algebraic factorization. This is a basic algebraic technique for simplifying formulas, fractions, and solving equations. Algebraic decomposition is another name. The factors of the algebraic expression 10xyz = 2 × 5 × x × y × z. Similarly, in irreducible factor form, the algebraic expression 5(x + 1) (z + 2) is :5(x + 1) (z + 2) = 5 × (x + 1) × (z + 2)

Purpose of factoring in real life. Another typical operation based on factoring is currency exchange. According to the principle of factoring, the two factors of 100 are 4 and 25. Factoring is a student’s gateway to great things in life. If students want to be a pharmacist, astronomer, biologist, physicist, programmer, or manager and be competitive or work more than 9-5 and want to be leaders in their field and achieve great things, they need algebra-based mathematics skills, so they need to learn about input methods.

The Importance of Mathematics:

Mathematics is the study of measurements and numbers, and it became one of the first sciences to be advanced because of its significance and value. The term “arithmetic” is derived from Greek and means “mastering proclivities,” and there are numerous branches of arithmetic in technology that deal with numbers, together with geometric shapes, algebra, and others. Mathematics is utilised in nearly every step of life, from day duties like timekeeping, driving, and cooking to professions like accounting, finance, banking, engineering, and software programme improvement. These necessitate a strong mathematical foundation, and medical experiments necessitate mathematical techniques. This is a language used to explain many scientists’ accomplishments in the mentioned field. All these things make it important to study Mathematics, and therefore Extramarks suggests that students follow NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4

### NCERT Solutions for Class 8

All the solutions for Class 8 are available on the NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.4 on the Extramarks website, which can be downloaded by visiting the Class 8 Mathematics Exercise 14.4 on the website or by downloading the Extramarks apps.

Q.1

$\begin{array}{l}\text{Find and correct the errors in the following mathematical}\\ \text{statements}\text{.}\\ \text{1}\text{. 4(x -5) = 4x- 5 2}{\text{. x(3x + 2) = 3x}}^{\text{2}}\text{+ 2 3}\text{. 2x + 3y = 5xy}\\ \text{4}\text{. x + 2x + 3x = 5x 5}\text{. 5y + 2y + y- 7y = 0 6}{\text{. 3x + 2x = 5x}}^{\text{2}}\\ \text{7}{\text{. (2x)}}^{\text{2}}{\text{+ 4(2x) + 7 = 2x}}^{\text{2}}\text{+ 8x + 7}\\ \text{8}{\text{. (2x)}}^{\text{2}}\text{+ 5x = 4x + 5x = 9x}\\ \text{9}{\text{. (3x + 2)}}^{\text{2}}{\text{= 3x}}^{\text{2}}\text{+ 6x + 4}\end{array}$

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.2

$\begin{array}{l}\text{Substituting x = }-\text{3 in}\\ {\text{(a) x}}^{\text{2}}\text{+ 5x + 4 gives (}-{\text{3)}}^{\text{2}}\text{+ 5 (}-\text{3) + 4 = 9 + 2 + 4 = 15}\\ {\text{(b) x}}^{\text{2}}\text{}-\text{5x + 4 gives (}-{\text{3)}}^{\text{2}}\text{}-\text{5 (}-\text{3) + 4 = 9}-\text{15 + 4 =}-\text{2}\\ {\text{(c) x}}^{\text{2}}\text{+ 5x gives (}-{\text{3)}}^{\text{2}}\text{+ 5 (}-\text{3) =}-\text{9}-\text{15 =}-\text{24}\end{array}$

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.3 Find and correct the errors in the below mathematical statement:

${\left(\mathrm{y}-3\right)}^{2}={\mathrm{y}}^{2}-9$

Ans

$\begin{array}{l}\text{L}.\text{H}.\text{S}\\ ={\left(\mathrm{y}-\text{3}\right)}^{\text{2}}\\ ={\left(\mathrm{y}\right)}^{\text{2}}-\text{2}\left(\mathrm{y}\right)\left(\text{3}\right)+{\left(\text{3}\right)}^{\text{2}}\text{}\left[\text{By using}:{\left(\mathrm{a}-\mathrm{b}\right)}^{\text{2}}={\mathrm{a}}^{\text{2}}-\text{2}\mathrm{ab}+{\mathrm{b}}^{\text{2}}\right]\\ ={\mathrm{y}}^{\text{2}}-\text{6}\mathrm{y}+\text{9}\ne \text{R}.\text{H}.\text{S}\\ \\ \therefore \text{The correct statement is:}\\ \text{}{\left(\mathrm{y}-\text{3}\right)}^{\text{2}}={\mathrm{y}}^{\text{2}}-\text{6}\mathrm{y}+\text{9}\end{array}$

Q.4 Find and correct the errors in the below mathematical statement:

${\left(\mathrm{z}+5\right)}^{2}={\mathrm{z}}^{2}+25$

Ans

$\begin{array}{l}\text{L}.\text{H}.\text{S}\\ ={\left(\mathrm{z}+\text{5}\right)}^{\text{2}}\\ ={\left(\mathrm{z}\right)}^{\text{2}}+\text{2}\left(\mathrm{z}\right)\left(\text{5}\right)+{\left(\text{5}\right)}^{\text{2}}\text{\hspace{0.17em}}\left[\text{By Using:}{\left(\mathrm{a}+\mathrm{b}\right)}^{\text{2}}={\mathrm{a}}^{\text{2}}+\text{2}\mathrm{ab}+{\mathrm{b}}^{\text{2}}\right]\\ ={\mathrm{z}}^{\text{2}}+\text{1}0\mathrm{z}+\text{25}\ne \text{R}.\text{H}.\text{S}\\ \\ \therefore \text{The correct statement is:}\\ {\left(\mathrm{z}+\text{5}\right)}^{\text{2}}={\mathrm{z}}^{\text{2}}+\text{1}0\mathrm{z}+\text{25}\end{array}$

Q.5 Find and correct the errors in the below mathematical statement:

$\left(2\mathrm{a}+3\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)=2{\mathrm{a}}^{2}-3{\mathrm{b}}^{2}$

Ans

$\begin{array}{l}\text{L}.\text{H}.\text{S}.\text{}\\ =\left(\text{2}\mathrm{a}+\text{3}\mathrm{b}\right)\text{}\left(\mathrm{a}-\mathrm{b}\right)\text{}\\ =\text{2}\mathrm{a}×\mathrm{a}+\text{3}\mathrm{b}×\mathrm{a}-\text{2}\mathrm{a}×\mathrm{b}-\text{3}\mathrm{b}×\mathrm{b}\\ =\text{2}{\mathrm{a}}^{\text{2}}+\text{3}\mathrm{ab}-\text{2}\mathrm{ab}-\text{3}{\mathrm{b}}^{\text{2}}\\ =\text{2}{\mathrm{a}}^{\text{2}}+\mathrm{ab}-\text{3}{\mathrm{b}}^{\text{2}}\\ \ne \text{R}.\text{H}.\text{S}.\\ \\ \therefore \text{The correct statement is:}\\ \text{}\left(\text{2}\mathrm{a}+\text{3}\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)=\text{2}{\mathrm{a}}^{\text{2}}+\mathrm{ab}-\text{3}{\mathrm{b}}^{\text{2}}\end{array}$

Q.6 Find and correct the errors in the below mathematical statement:

$\left(\mathrm{a}+4\right)\left(\mathrm{a}+2\right)={\mathrm{a}}^{2}+8$

Ans

$\begin{array}{l}\text{L}.\text{H}.\text{S}.\\ =\left(\mathrm{a}+\text{4}\right)\text{}\left(\mathrm{a}+\text{2}\right)\text{}\\ ={\left(\mathrm{a}\right)}^{\text{2}}+\text{}\left(\text{4}+\text{2}\right)\text{}\left(\mathrm{a}\right)\text{}+\text{4}×\text{2}\\ ={\mathrm{a}}^{\text{2}}+\text{6}\mathrm{a}+\text{8}\\ \ne \text{R}.\text{H}.\text{S}\\ \\ \therefore \text{The correct statement is:}\\ \left(\mathrm{a}+\text{4}\right)\left(\mathrm{a}+\text{2}\right)={\mathrm{a}}^{\text{2}}+\text{6}\mathrm{a}+\text{8}\end{array}$

Q.7 Find and correct the errors in the below mathematical statement:

$\left(\mathrm{a}-4\right)\left(\mathrm{a}-2\right)={\mathrm{a}}^{2}-8$

Ans

$\begin{array}{l}\text{L}.\text{H}.\text{S}.\\ =\left(\mathrm{a}-\text{4}\right)\left(\mathrm{a}-\text{2}\right)\\ ={\left(\mathrm{a}\right)}^{\text{2}}+\left[\left(-\text{4}\right)+\left(-\text{2}\right)\right]\left(\mathrm{a}\right)+\left(-\text{4}\right)\left(-\text{2}\right)\\ ={\mathrm{a}}^{\text{2}}-\text{6}\mathrm{a}+\text{8}\\ \ne \text{R}.\text{H}.\text{S}.\\ \\ \therefore \text{The correct statement is:}\\ \left(\mathrm{a}-\text{4}\right)\text{}\left(\mathrm{a}-\text{2}\right)\text{}={\mathrm{a}}^{\text{2}}-\text{6}\mathrm{a}+\text{8}\end{array}$

Q.8 Find and correct the errors in the below mathematical statement:

$\frac{3{\mathrm{x}}^{2}}{3{\mathrm{x}}^{2}}=0$

Ans

$\begin{array}{l}\text{L.H.S}\\ =\frac{3{\mathrm{x}}^{2}}{3{\mathrm{x}}^{2}}\\ =\frac{3×\mathrm{x}×\mathrm{x}}{3×\mathrm{x}×\mathrm{x}}\\ =1\\ \ne \text{R.H.S}\\ \\ \therefore \text{The correct statement is}:\\ \frac{3{\mathrm{x}}^{2}}{3{\mathrm{x}}^{2}}=1\\ \end{array}$

Q.9 Find and correct the errors in the below mathematical statement:

$\frac{3{\mathrm{x}}^{2}+1}{3{\mathrm{x}}^{2}}=1+1=2$

Ans

$\begin{array}{l}\text{L.H.S}\\ =\frac{3{\mathrm{x}}^{2}+1}{3{\mathrm{x}}^{2}}\\ =\frac{3{\mathrm{x}}^{2}}{3{\mathrm{x}}^{2}}+\frac{1}{3{\mathrm{x}}^{2}}\end{array}$

$\begin{array}{l}=1+\frac{1}{3{\mathrm{x}}^{2}}\\ \ne \text{R.H.S}\\ \\ \therefore \text{The correct statement is:}\\ \frac{3{\mathrm{x}}^{2}+1}{3{\mathrm{x}}^{2}}=1+\frac{1}{3{\mathrm{x}}^{2}}\end{array}$

Q.10 Find and correct the errors in the below mathematical statement:

$\frac{3\mathrm{x}}{3\mathrm{x}+2}=\frac{1}{2}$

Ans

$\begin{array}{l}\text{L.H.S}\\ =\frac{3\mathrm{x}}{3\mathrm{x}+2}\\ \ne \text{R.H.S}\\ \\ \therefore \text{The correct statement is:}\\ \frac{3\mathrm{x}}{3\mathrm{x}+2}=\frac{3\mathrm{x}}{3\mathrm{x}+2}\end{array}$

Q.11 Find and correct the errors in the below mathematical statement:

$\frac{3}{4\mathrm{x}+3}=\frac{1}{4\mathrm{x}}$

Ans

$\begin{array}{l}\text{L.H.S}\\ =\frac{3}{4\mathrm{x}+3}\\ \ne \text{R.H.S}\\ \\ \therefore \text{The correct statement is:}\\ \frac{3}{4\mathrm{x}+3}=\frac{3}{4\mathrm{x}+3}\end{array}$

Q.12 Find and correct the errors in the below mathematical statement:

$\frac{4\mathrm{x}+5}{4\mathrm{x}}=5$

Ans

$\begin{array}{l}\text{L.H.S}\\ =\frac{4\mathrm{x}+5}{4\mathrm{x}}\\ =\frac{4\mathrm{x}}{4\mathrm{x}}+\frac{5}{4\mathrm{x}}\end{array}$

$\begin{array}{l}=1+\frac{5}{4\mathrm{x}}\\ \ne \text{R.H.S}\\ \\ \therefore \text{The correct statement is:}\\ \frac{4\mathrm{x}+5}{4\mathrm{x}}=1+\frac{5}{4\mathrm{x}}\end{array}$

Q.13 Find and correct the errors in the below mathematical statement:

$\frac{7\mathrm{x}+5}{5}=7\mathrm{x}$

Ans

$\begin{array}{l}\text{L.H.S}\\ =\frac{7\mathrm{x}+5}{5}\\ =\frac{7\mathrm{x}}{5}+\frac{5}{5}\end{array}$

$\begin{array}{l}=\frac{7\mathrm{x}}{5}+1\\ \ne \text{R.H.S}\\ \\ \therefore \text{The correct statement is:}\\ \\ \frac{7\mathrm{x}+5}{5}=\frac{7\mathrm{x}}{5}+1\end{array}$

## 1. What does factor in Mathematics mean?

In Mathematics NCERT Solutions For Class 8 Mathematics Chapter 14 Exercise 14.4, a factor is a number or algebraic expression that divides another number or expression evenly, leaving no remainder or residue. A number that students multiply to get another number is called a factor. There are multiple factorizations for a given integer (multiple kinds of factorizations). Factors are important because they tell students something about the properties of various numbers, such as whether they are even or odd, or whether they are perfect squares and cubes. Students can refer to NCERT Solutions For Class 8 Mathematics Chapter 14 Exercise 14.4 .

## 2. What is meant by factorization?

Factorization is the process of dividing a large number into smaller numbers, which when multiplied together yield the original value. These components or coefficients can be integers, variables, or even algebraic expressions. The students should practice all factoring problems on NCERT Solutions For Class 8 Mathematics Chapter 14 Exercise 14.4 to fully understand the basics of factorisation and perform well on all tests with the help of Extramarks.

## 3. What is the significance of the NCERT Class 8 Mathematics Chapter 14 Exercise 14.4?

NCERT Solutions Chapter 14 Factoring for Class 8 Mathematics is a reliable resource formulated according to the latest CBSE syllabus to help students learn mathematics. By regularly practising the questions and examples included in these solutions, students will develop the conceptual fluency and correct answering skills necessary to study mathematics. The ready-to-use practical examples in these solutions make it easy for students to understand difficult procedures and solutions.