# NCERT Solutions Class 8 Maths Chapter 14 Exercise 14.2﻿

The area of Mathematics known as algebra aids in the representation of circumstances or problems as mathematical expressions. To create a meaningful mathematical expression, it takes variables like x, y, and z together with mathematical operations like addition, subtraction, multiplication, and division. The students can understand its subtopics with the help of tools like the NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2 .

Algebra is used in all areas of Mathematics, including trigonometry, calculus, and coordinate geometry. 2x + 4 = 8 is a straightforward algebraic expression.

In algebra, symbols are used, and operators are used to connect the symbols to one another. It is more than just a mathematical idea; it is a skill that students utilise on a regular basis without even being aware of it.

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A branch of mathematics known as algebra deals with symbols and the mathematical operations performed on them. Variables are the name given to these symbols because they lack set values. Humans frequently observe a constant change in specific values in their day-to-day situations. But the need to depict these shifting values is ongoing.

These values are usually represented in algebra by variables, which are symbols like x, y, z, p, or q. These symbols are also subjected to various arithmetic operations—like addition, subtraction, multiplication, and division—in order to ascertain their values.

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Elementary algebra or Algebra 1:

The conventional subjects covered in a contemporary primary algebra course are covered in Elementary Algebra. Arithmetic includes both numbers and mathematical operations such as +, -, and x. But in algebra, variables—such as x, a, n, and y—are frequently used to represent the integers. It is the first step that demonstrates the systematic exploration of all the properties of a system of real numbers. It also permits the common formulation of the laws of arithmetic, such as “a + b” = “b + a.” Different chapters in NCERT books cover this topic, and the students can access tools similar to those in the NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2, that are targeted towards this topic to understand it better.

Variables, evaluating expressions and equations, the characteristics of equalities and inequalities, solving algebraic equations, solving linear equations with one or two variables, etc. are some of the subjects covered in elementary algebra.

Advanced algebra or Algebra 2 :

This is algebra at the intermediate level. Compared to pre-algebra, this algebra has a higher level of equations to answer. Students can go through the additional algebraic concepts by using advanced algebra, such as:

• Equations involving inequality
• Matrices
• Solution to a system of linear equations
• Graphing linear equations and functions
• Curved portions
• Equation of a Polynomial
• Series and sequences
• Reasonable statements
• Trigonometry

Abstract Algebra :

One area of algebra known as “abstract algebra” studies principles about algebraic systems, irrespective of the particulars of some operations. In particular situations, these operations have particular characteristics. As a result, certain conclusions can be drawn about these qualities. Hence, the name of this area of mathematics is “abstract algebra.”

Algebraic structures, including fields, groups, modules, rings, lattices, vector spaces, etc., are dealt with in abstract algebra.

Linear Algebra :

A division of algebra known as linear algebra has applications in both pure and applied mathematics. It deals with the vector spaces’ linear mappings. It also has to do with studying lines and planes. It is the research of linear sets of equations having properties of transformation. Almost all branches of mathematics use it. It deals with linear equations for linear functions and how they are represented in matrices and vector spaces. The following are some key areas that linear algebra covers:

• Linear equations
• Vector Spaces
• Relations
• Matrices and matrix decomposition
• Relations and Computations

Commutative Algebra :

One of the branches of algebra that analyses commutative rings and their ideals is called commutative algebra. Commutative algebra is a prerequisite for both algebraic geometry and algebraic number theory. Polynomial rings, rings of algebraic integers, and other rings are included. Commutative algebra is used in many other branches of mathematics, including differential topology, invariant theory, order theory, and general topology. It has played a noteworthy part in contemporary pure mathematics. The students can better understand this topic with the help of tools that were created by the experts in mathematics, like the NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2, but particularly those that were created for this topic.

## NCERT Class 8 Maths Solutions, Chapter 14 Factorisation (EX 14.2) Exercise 14.2

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### Access NCERT Solutions for Class 8 Maths Chapter 14-Factorisation

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### NCERT Class 8 Maths Solutions, Chapter 14 Factorisation Exercise 14.2

There is a beauty in taking a seemingly simple problem and repeatedly combining and simplifying it until there is only one value for each variable. The process can be fun, and the outcome can be quite fulfilling. That’s what Algebra is, with all its parts. That can be better understood with the help of resources like the NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2, and many more that are available on the Extramarks website.

A valuable life skill worth mastering is algebra. It takes students beyond elementary mathematics and gets them ready for calculus and statistics. It is helpful for a wide variety of vocations, some of which a student might pursue as a second career. In the home and when evaluating news articles, algebra is helpful. It also strengthens logical reasoning.

1) Algebra is more accurate and faster than “Basic” Mathematics.

Similar to how multiplication by 12 is quicker than counting to 24 or adding 2 by 12, algebra makes problem-solving quicker and simpler than it would be without it. Additionally, algebra enables the solution of previously unsolvable life problems like charting curves that go beyond the scope of basic math abilities. Chapter 14 teaches factorization, which is a part of Algebra, and the students can understand it better with the help of the NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2

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Calculus and statistics are two types of mathematics that many people utilise in their careers, even though the idea of learning one type of mathematics in order to acquire more types of arithmetic may not be immediately rewarding. Statistics are generally employed in a variety of positions in the business world, the media, the health and wellness industry, politics, the social sciences, and many other sectors. Understanding statistics helps students become better workers, citizens, and information consumers. This can be achieved with a tool similar to the NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2 .

3) Numerous intricate processes, such as how an object’s speed changes over time, are better understood because of calculus. Calculus is a tool that engineers and scientists use to create consumer goods, medicinal procedures, and new technologies. Any student interested in a job in science, medicine, computer modelling, or engineering must learn calculus. They can start to form the base by making use of all the related resources provided on the Extramarks website, like the NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2 .

4) Later on, knowing algebra could be a useful skill.

Even if a student is certain that they will not be entering a profession that requires statistics or calculus, many people change professions and entire careers throughout their working lives, and factorization and other parts of algebra may play an important role at some point or another. The students are advised to gain a strong foundation in this topic with the help of tools like the NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2. and more.

Factorization is the process of breaking down a number or polynomial into a product of numerous factors of other polynomials that, when multiplied, produce the original number. Use the factorization formula to divide an integer into its factors. A number, matrix, or polynomial can be converted into the product of another entity or factors, which, when multiplied together, produce the original number. This process is known as “factorization.” A large number is split up into smaller numbers, or factors, using the factorization formula. A factor is a number that evenly divides an integer by itself with no remainder. Prime Factorization of 28 is, for instance, 2 X 2 X 7, and before we begin factorization, let’s first go through the mathematical concept of “factor.” The students are advised to make the most of resources like the NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2, if they wish to master this topic. The NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2, can prove to be very beneficial in helping the students gain in-depth knowledge of the topic. The Solutions for Class 8 Maths Exercise 14.2  are prepared by the experts in the topic. The NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2, are therefore essential for the students to better understand this particular chapter.

Finding the factors of an algebraic expression, also known as locating two or more expressions whose product is the given expression, is the process of factorization. The factorization of algebraic expressions is the process of identifying two or more expressions whose product is the given expression. A factor is a number that evenly divides the inputted number. It merely refers to representing a number as the product of two other numbers. Similar to this, algebraic expressions are written as the product of their factors. The only exception is that an algebraic expression here combines addition or subtraction with arithmetic operations like numbers and variables. The students can find this to be a challenging topic, but they are advised by the experts to make use of the NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2, to get a better understanding of the topic.

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### NCERT Solutions for Class 8

A number is expressed as the product of any two numbers using the term “factor.” Finding factors for any mathematical object, such as an integer, polynomial, or algebraic expression, is done through the process of factorization. Finding the factors of the provided algebraic expression is what is meant when an algebraic expression is factorised. The students can practise this topic and understand it better with the help of the NCERT Solutions for Class 8 Maths, Chapter 14, Exercise 14.2, and other tools on the Extramarks website that are targeted towards this topic.

There are numerous ways to factorise algebraic expressions. The most typical techniques for factoring algebraic expressions include:

• Using common factors to factorise.
• Factorization by term regrouping.
• Utilizing identities to factorise.

Mathematicians frequently employ the procedure of factoring. This aids in resolving various mathematical issues or in the simplification of formulas. A polynomial is divided into two or more factors through factoring. Students have to keep in mind that the original phrase should still be obtained when they multiply these components.

Numerous daily tasks include the use of factors. Students are aware that factoring makes it possible to divide something into several equal parts, so the concept of factoring is present in any division into equal parts.

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Q.1

$\begin{array}{l}\text{Factorise the following expressions.}\\ {\text{(i) a}}^{\text{2}}\text{+ 8a + 16}\\ {\text{(ii) p}}^{\text{2}}-\text{10 p + 25}\\ {\text{(iii) 25m}}^{\text{2}}\text{+ 30m + 9}\\ {\text{(iv) 49y}}^{\text{2}}{\text{+ 84yz + 36z}}^{\text{2}}\text{}\\ {\text{(v) 4x}}^{\text{2}}-\text{8x + 4}\\ {\text{(vi) 121b}}^{\text{2}}-{\text{88bc + 16c}}^{\text{2}}\\ {\text{(vii) (l + m)}}^{\text{2}}-{\text{4lm (Hint: Expand ( l + m)}}^{\text{2}}\text{first)}\\ {\text{(viii) a}}^{\text{4}}{\text{+ 2a}}^{\text{2}}{\text{b}}^{\text{2}}{\text{+ b}}^{\text{4}}\end{array}$

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.2

$\begin{array}{l}\text{Factorise.}\\ {\text{(i) 4p}}^{\text{2}}\text{ }-{\text{ 9q}}^{\text{2}}\text{}\\ {\text{(ii) 63a}}^{\text{2}}-{\text{ 112b}}^{\text{2}}\text{}\\ {\text{(iii) 49x}}^{\text{2}}\text{ }-\text{36}\\ {\text{(iv) 16x}}^{\text{5}}\text{ }-{\text{ 144x}}^{\text{3}}\text{}\\ {\text{(v) (l + m)}}^{\text{2}}\text{ }-\text{ }{\left(\text{l}-\text{ m}\right)}^{\text{2}}\\ {\text{(vi) 9x}}^{\text{2}}{\text{y}}^{\text{2}}\text{ }-\text{ 16}\\ {\text{(vii) (x}}^{\text{2}}-{\text{ 2xy + y}}^{\text{2}}\text{)}-{\text{ z}}^{\text{2}}\\ {\text{(viii) 25a}}^{\text{2}}-{\text{ 4b}}^{\text{2}}\text{+ 28bc}-{\text{ 49c}}^{\text{2}}\end{array}$

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.3

$\begin{array}{l}\text{Factorise the expressions.}\\ {\text{(i) ax}}^{\text{2}}\text{+ bx}\\ {\text{(ii) 7p}}^{\text{2}}{\text{+ 21q}}^{\text{2}}\text{}\\ {\text{(iii) 2x}}^{\text{3}}{\text{+ 2xy}}^{\text{2}}{\text{+ 2xz}}^{\text{2}}\\ {\text{(iv) am}}^{\text{2}}{\text{+ bm}}^{\text{2}}{\text{+ bn}}^{\text{2}}{\text{+ an}}^{\text{2}}\text{}\\ \text{(v) (lm + l) + m + 1}\\ \text{(vi) y (y + z) + 9 (y + z)}\\ {\text{(vii) 5y}}^{\text{2}}\text{ }-\text{ 20y }-\text{ 8z + 2yz}\\ \text{(viii) 10ab + 4a + 5b + 2}\\ \text{(ix) 6xy }-\text{ 4y + 6 }-\text{ 9x}\end{array}$

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.4

$\begin{array}{l}\text{Factorise.}\\ {\text{(i) a}}^{\text{4}}-{\text{ b}}^{\text{4}}\text{}\\ {\text{(ii) p}}^{\text{4}}-\text{ 81}\\ {\text{(iii) x}}^{\text{4}}-{\text{ (y + z)}}^{\text{4}}\\ {\text{(iv) x}}^{\text{4}}-{\text{(x – z)}}^{\text{4}}\text{}\\ {\text{(v) a}}^{\text{4}}-{\text{2a}}^{\text{2}}{\text{b}}^{\text{2}}\text{+ b4}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }\mathrm{}{\mathrm{a}}^{4}-{\mathrm{b}}^{4}\\ =\text{}{\left({\mathrm{a}}^{2}\right)}^{2}-\text{}{\left({\mathrm{b}}^{2}\right)}^{2}\\ =\text{}\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)\text{}\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\text{}\left[\mathrm{By}\text{}\mathrm{using}:\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\\ =\text{}\left(\mathrm{a}-\mathrm{b}\right)\text{}\left(\mathrm{a}+\mathrm{b}\right)\text{}\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\left[\mathrm{By}\text{}\mathrm{using}:\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\\ \\ \left(\mathrm{ii}\right)\text{ }{\mathrm{p}}^{4}-\text{}81\text{}\\ ={\left({\mathrm{p}}^{2}\right)}^{2}-\text{}{\left(9\right)}^{2}\\ =\left({\mathrm{p}}^{2}-\text{}9\right)\text{}\left({\mathrm{p}}^{2}+\text{}9\right)\left[\mathrm{By}\text{}\mathrm{using}:\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\\ =\left[{\left(\mathrm{p}\right)}^{2}-\text{}{\left(3\right)}^{2}\right]\text{}\left({\mathrm{p}}^{2}+\text{}9\right)\\ =\left(\mathrm{p}-\text{}3\right)\text{}\left(\mathrm{p}+\text{}3\right)\text{}\left({\mathrm{p}}^{2}+\text{}9\right)\left[\mathrm{By}\text{}\mathrm{using}:\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\\ \\ \left(\mathrm{iii}\right)\text{ }{\mathrm{x}}^{4}-{\left(\mathrm{y}+\mathrm{z}\right)}^{4}\\ ={\left({\mathrm{x}}^{2}\right)}^{2}-{\left[{\left(\mathrm{y}+\mathrm{z}\right)}^{2}\right]}^{2}\\ =\left[{\mathrm{x}}^{2}-{\left(\mathrm{y}+\mathrm{z}\right)}^{2}\right]\left[{\mathrm{x}}^{2}+{\left(\mathrm{y}+\mathrm{z}\right)}^{2}\right]\text{}\left[\mathrm{By}\text{}\mathrm{using}:\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\end{array}$ $\begin{array}{l}=\left[\mathrm{x}-\left(\mathrm{y}+\mathrm{z}\right)\right]\left[\mathrm{x}+\left(\mathrm{y}+\mathrm{z}\right)\right]\left[{\mathrm{x}}^{2}+{\left(\mathrm{y}+\mathrm{z}\right)}^{2}\right]\\ \text{}\left[\mathrm{By}\text{ }\mathrm{using}:\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\\ =\left(\mathrm{x}-\mathrm{y}-\mathrm{z}\right)\text{}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\text{}\left[{\mathrm{x}}^{2}+{\left(\mathrm{y}+\mathrm{z}\right)}^{2}\right]\\ \\ \left(\mathrm{iv}\right)\text{ }{\mathrm{x}}^{\text{4}}-{\left(\mathrm{x}-\mathrm{z}\right)}^{\text{4}}\\ ={\left({\mathrm{x}}^{\text{2}}\right)}^{\text{2}}-{\left[{\left(\mathrm{x}-\mathrm{z}\right)}^{\text{2}}\right]}^{\text{2}}\\ =\left[{\mathrm{x}}^{\text{2}}-{\left(\mathrm{x}-\mathrm{z}\right)}^{\text{2}}\right]\left[{\mathrm{x}}^{\text{2}}+{\left(\mathrm{x}-\mathrm{z}\right)}^{\text{2}}\right]\text{}\left[\mathrm{By}\text{ }\mathrm{using}:\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\\ =\left[\mathrm{x}-\left(\mathrm{x}-\mathrm{z}\right)\right]\text{}\left[\mathrm{x}+\left(\mathrm{x}-\mathrm{z}\right)\right]\left[{\mathrm{x}}^{\text{2}}+{\left(\mathrm{x}-\mathrm{z}\right)}^{\text{2}}\right]\\ \text{}\left[\mathrm{By}\text{ }\mathrm{using}:\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\\ =\mathrm{z}\left(\text{2}\mathrm{x}-\mathrm{z}\right)\left[{\mathrm{x}}^{\text{2}}+{\mathrm{x}}^{\text{2}}-\text{2}\mathrm{xz}+{\mathrm{z}}^{\text{2}}\right]\\ =\mathrm{z}\left(\text{2}\mathrm{x}-\mathrm{z}\right)\text{}\left(\text{2}{\mathrm{x}}^{\text{2}}-\text{2}\mathrm{xz}+{\mathrm{z}}^{\text{2}}\right)\\ \\ \left(\mathrm{v}\right)\text{ }{\mathrm{a}}^{\text{4}}-\text{2}{\mathrm{a}}^{\text{2}}{\mathrm{b}}^{\text{2}}+{\mathrm{b}}^{\text{4}}\\ ={\left({\mathrm{a}}^{\text{2}}\right)}^{\text{2}}-\text{2}\left({\mathrm{a}}^{\text{2}}\right)\left({\mathrm{b}}^{\text{2}}\right)+{\left({\mathrm{b}}^{\text{2}}\right)}^{\text{2}}\\ ={\left({\mathrm{a}}^{\text{2}}-{\mathrm{b}}^{\text{2}}\right)}^{\text{2}}\\ ={\left[\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)\right]}^{\text{2}}\text{\hspace{0.17em}}\left[\mathrm{By}\text{ }\mathrm{using}:\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\\ ={\left(\mathrm{a}-\mathrm{b}\right)}^{\text{2}}{\left(\mathrm{a}+\mathrm{b}\right)}^{\text{2}}\end{array}$

Q.5

$\begin{array}{l}\text{Factorise the following expressions.}\\ {\text{(i) p}}^{\text{2}}\text{+ 6p + 8}\\ {\text{(ii) q}}^{\text{2}}-\text{10q + 21}\\ {\text{(iii) p}}^{\text{2}}\text{+ 6p}-\text{16}\end{array}$

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)