NCERT Solutions for Class 8 Maths Chapter 14 Factorization (Updated For 2023-24)

NCERT Solutions for Class 8 Maths Chapter 14 Factorization

Edited By Ramraj Saini | Updated on Mar 01, 2024 06:47 PM IST

Factorization Class 8 Questions And Answers provided here. These NCERT Solutions are created by expert team at craeers360 keeping in mind of the latest syllabus and pattern of CBSE 2023-23. In NCERT solutions for Class 8 Maths chapter 14 Factorization, you will be dealing with questions related to algebraic expressions and natural numbers. Important topics like methods of common factors, factorization using identities, factorization by regrouping terms, factors of the form (x + a) ( x + b), and division of algebraic expressions are covered in this chapter.

New: ALLEN TALLENTEX. Get Scholarships & Cash prizes. Register Now

This Story also Contains

Factorization Class 8 Questions And Answers PDF Free Download
Factorization Class 8 Solutions - Important Formulae
Factorization Class 8 NCERT Solutions (Intext Questions and Exercise)
Factorization class 8 NCERT solutions - Topics
NCERT Solutions for Class 8 Maths - Chapter Wise
Key Features of Factorization Class 8 Solutions
NCERT Solutions for Class 8 - Subject Wise
NCERT Books and NCERT Syllabus

NCERT Solutions for Class 8 Maths Chapter 14 Factorization

There are 4 exercises with 34 questions given in the NCERT textbook. All these questions are prepared in the solutions of NCERT for Class 8 Maths chapter 14 Factorization in a step-by-step manner. It will be easy for you to understand the concept. For a better understanding of the concept, there are some practice questions given after every topic. You will find solutions to these practice questions also in NCERT Solutions for Class 8 maths by clicking on the link.

Factorization Class 8 Questions And Answers PDF Free Download

Download PDF

Factorization Class 8 Solutions - Important Formulae

Factorization: Factorization is the process of expressing an algebraic equation as a product of its components. These components can be numbers, variables, or algebraic expressions.

Irreducible Factor: An irreducible factor is a component that cannot be further factored into a product of factors.

Method to Do Factorization:

The common factor approach involves three steps:

Write each term of the statement as a product of irreducible elements.

Look for and separate the similar components.

Combine the remaining elements in each term using the distributive law.

The regrouping approach involves grouping terms in a way that brings out a common factor across the groups.

Common Factor Identity: Certain factorable expressions take the form of:

a² + 2ab + b² = (a + b)²
a² - 2ab + b² = (a - b)²
a² - b² = (a + b)(a - b)
x² + (a + b)x + ab = (x + a)(x + b)

Dividing a Polynomial by a Monomial: When dividing a polynomial by a monomial, you can divide each term of the polynomial by the monomial or use the common factor technique.

Division of Algebraic Expressions: In division of algebraic expressions, you factor both the dividend and the divisor, then cancel common factors.

Division Formula: Dividend = Divisor × Quotient or Dividend = Divisor × Quotient + Remainder.

Free download NCERT Solutions for Class 8 Maths Chapter 14 Factorization for CBSE Exam.

Factorization Class 8 NCERT Solutions (Intext Questions and Exercise)

NCERT Solutions for Class 8 Maths Chapter 14 Factorization - Topic 14.2.1 Method Of Common Factor

Question:(i) Factorise:

$12 x +36$

Answer:

We have
$12x = 2 \times 2 \times 3 \times x$
$36 = 2 \times 2 \times 3 \times 3$

So, we have $2 \times 2 \times 3$ common in both
Therefore,

$12x + 36 =$ $2 \times 2 \times 3 (x + 3)$

$12x + 36 = 12(x + 3)$

Question:(ii) Factorise : 22y-32z

Answer:

We have,
$22y=$ $2 \times 11 \times y$
$33z =$ $3 \times 11 \times z$
So, we have 11 common in both
Therefore,

$22y - 33z = 11(2y - 3z)$

Question:(iii) Factorise :

$( iii) \: \: 14 pq + 35 pqr$

Answer:

We have
$14pq =$ $2 \times 7 \times p \times q$
$35pqr =$ $5 \times 7 \times p \times q \times r$
So, we have

$7 \times p \times q$ common in both
Therefore,

$14pq + 35pqr =7pq (2 + 5r)$

Class 8 maths chapter 14 question answer - Exercise: 14.1

Question:1(i) Find the common factors of the given terms.

$(i) 12 x , 36$

Answer:

We have
$12x ={\color{Red} 2 \times 2 \times 3}\times x$
$36 = {\color{Red} 2 \times 2 \times 3}\times 3$
So, the common factors between the two are

$2\times2\times3=12$

Question:1(ii) Find the common factors of the given terms

$(ii) 2y , 22xy$

Answer:

We have,
$2y = {\color{Red} 2 \times y}$
$22xy = {\color{Red} 2} \times 11 \times x {\color{Red} \times y}$
Therefore, the common factor between these two is 2y

Question:1(iii) Find the common factors of the given terms

$(iii) 14 pq, 28 p^2 q^2$

Answer:

We have,
$14pq = {\color{Red} 2 \times 7 \times p \times q}$
$28p^2q^2 = 2 \times {\color{Red} 2 \times 7 \times p} \times p{\color{Red} \times q} \times q$
Therefore, the common factor is

$2\times7\times p\times q=14pq$

Question:1(iv) Find the common factors of the given terms.

$(iv) 2x , 3x ^2 , 4$

Answer:

We have,
$2x = 2 \times x$
$3x^2 = 3 \times x \times x$
$4 = 2 \times 2$
Therefore, the common factor between these three is 1

Question:1(v) Find the common factors of the given terms

$( v ) 6 abc , 24 ab^2 , 12 a^2 b$

Answer:

We have,
$6abc ={\color{Red} 2 \times 3 \times a \times b }\times c$
$24ab^2 = 2 \times 2\times {\color{Red} 2\times 3 \times a \times b} \times b$
$12a^2b = 2 \times {\color{Red} 2\times 3 \times a} \times a{\color{Red} \times b}$
Therefore, the common factors is

$2 \times 3 \times a \times b = 6ab$

Question:1(vi) Find the common factors of the given terms

$(vi)16 x ^ 3 , -4 x ^ 2 , 32 x$

Answer:

We have,
$16x^3 = 2 \times 2 \times {\color{Red} 2 \times 2 \times x} \times x \times x$
$4x^2 = {\color{Red} 2 \times 2 \times x} \times x$
$32x = 2 \times 2 \times 2 \times{\color{Red} 2 \times 2 \times x}$
Therefore, the common factors is

$2 \times 2 \times x = 4x$

Question:1(vii) Find the common factors of the given terms

$(vii) 10 pq , 20 qr , 30 rp$

Answer:

We have,
$10pq ={\color{DarkRed} 2 \times 5} \times p \times q$
$20qr = 2\times{\color{DarkRed} 2 \times 5 }\times q \times r$
$30rp ={\color{DarkRed} 2}\times 3{\color{DarkRed} \times 5} \times r \times p$
Therefore, the common factors between these three is

$2 \times 5 =10$

Question:1(viii) Find the common factors of the given terms

$(viii)3 x ^2 y^3 , 10 x ^3 y ^ 2 , 6 x^ 2 y^2 z$

Answer:

We have,
$3x^{2}y^{2}$ $= 3 \times {\color{Red} x \times x \times y \times y}$
$10x^{3}y^{2}$ $=2 \times 5 \times x \times {\color{Red} x\times x \times y \times y}$
$6x^{2}y^{2}z$ $=2 \times 3 \times{\color{Red} x \times x \times y \times y} \times z$
Therefore, the common factors between these three are $x\times x\times y \times y =$ $x^{2}y^{2}$

Question:2(i) Factorise the following expressions

$(i)7x -42$

Answer:

We have,
$7x = 7 \times x \\ 42=7\times 2 \times 3=7\times 6\\ 7x-42=7x-7\times 6=7(x-6)$

Therefore, 7 is a common factor

Question:2(ii) Factorise the following expressions

$(ii)6 p - 12 q$

Answer:

We have,
$6p = 2 \times 3 \times p$
$12q = 2 \times 2 \times 3 \times q$
$\therefore$ on factorization

$6p -12q = (2\times 3 \times p) - (2\times 2 \times 3 \times q) = (2\times 3)(p-2q) = 6(p-2q)$

Question:2(iii) Factorise the following expressions

$(iii)7 a ^2 + 14 a$

Answer:

We have,
$7a^2 = 7 \times a \times a$
$14a = 2 \times 7 \times a$
$\therefore$ $7a^2+14a = (7\times a \times a)+(2 \times 7 \times a) = (7 \times a)(a+2)$
$= 7a(a+2)$

Question:2(iv) Factorise the following expressions

$(iv)-16 z + 20 z^3$

Answer:

We have,
$-16z = -1 \times 2 \times 2 \times 2 \times 2 \times z$
$20z^3 = 2 \times 2 \times 5 \times z \times z \times z$
$\therefore$ on factorization we get,
$-16z+20z^3 = (-1 \times 2 \times 2 \times 2 \times 2 \times z)+(2 \times 2 \times 5 \times z \times z \times z )$
$= (2\times 2 \times z)(-1 \times 2 \times 2+ 5 \times z \times z )$
$= 4z(-4+5z^2 )$

Question:2(v) Factorise the following expressions

$20 l^2 m + 30 alm$

Answer:

We have,
$20l^2m = 2 \times 2 \times 5 \times l \times l \times m$
$30alm = 2 \times 3 \times 5 \times a \times l \times m$
$\therefore$ on factorization we get,
$20l^2m+30alm =(2\times 2 \times 5 \times l \times l \times m) + (2 \times 3 \times 5 \times a \times l \times m)$
$=(2\times 5 \times l \times m)(2\times l + 3 \times a )$
$=10lm(2l+3a)$

Question:2(vi) Factorise the following expressions

$5 x^2 y - 15 xy^2$

Answer:

We have,
$5x^2y = 5 \times x\times x \times y$
$15xy^2 =3\times 5 \times x\times y \times y$
$\therefore$ on factorization we get,
$5x^2y - 15xy^2 = (5 \times x \times x \times y ) - (3\times 5 \times x \times y \times y )$
$=(5\times x \times y) ( x - 3\times y )$
$=5xy (x-3y)$

Question:2(vii) Factorise the following expressions

$10 a ^2 - 15 b^2 +20 c^2$

Answer:

We have,
$10a^2 = 2 \times 5 \times a \times a$
$15b^2 = 3 \times 5 \times b \times b$
$20c^2 = 2\times 2 \times 5 \times c \times c$
$\therefore$ on factorization we get,
$10a^2-15b^2+20c^2 = (2\times 5 \times a \times a)-(3\times 5 \times b \times b)+(2\times 2 \times 5 \times c \times c)$ $=5 (2 \times a \times a-3 \times b \times b+2\times 2 \times c \times c)$
$=5(2a^2-3b^2+4c^2)$

Question:2(viii) Factorise the following expressions

$- 4 a ^2 + 4 ab - 4ca$

Answer:

We have,
$-4a^2 = -1\times 2 \times 2 \times a\times a$
$4ab = 2 \times 2 \times a\times b$
$4ca = 2 \times 2 \times c\times a$
$\therefore$ on factorization we get,
$-4a^2+4ab-4ca = (-1 \times 2 \times 2 \times a\times a )+( 2 \times 2 \times a\times b )- (2 \times 2 \times c\times a)$

$=(2 \times 2 \times a) (-1 \times a + b - c)$
$= 4a(-a+b-c)$

Question:2(ix) Factorise the following expressions

$x^2 yz + xy^2 z + xyz^2$

Answer:

We have,
$x^2yz =x \times x \times y \times z$
$xy^2z =x \times y \times y \times z$
$xyz^2 =x \times y \times z \times z$
Therefore, on factorization we get,
$x^2yz+xy^2z+xyz^2 =(x \times x \times y \times z)+(x \times y \times y \times z)+(x \times y \times z \times z)$

$=( x \times y \times z)(x + y + z)$
$=xyz(x+y+z)$

Question:2(x) Factorise the following expressions

$a x^2 y + bxy^2 + cxyz$

Answer:

We have,
$ax^2y = a \times x \times x \times y$
$bxy^2 = b \times x \times y \times y$
$cxyz = c \times x \times y \times z$
Therefore, on factorization we get,
$ax^2y+bxy^2+cxyz = ( a \times x \times x \times y)+( b \times x \times y \times y)+(c \times x \times y \times z)$ $= (x\times y)( a \times x+ b \times y+c \times z)$

$= xy(ax+by+cz)$

Question:3(i) Factorise $x ^ 2 + xy + 8 x + 8y$

Answer:

We have,
$x^2 = x \times x$
$xy = x \times y$
$8x = 8 \times x$
$8y = 8 \times y$
Therefore, on factorization we get,
$x^2+xy+8x+8y = (x \times x)+(x\times y )+(8 \times x)+(8 \times y)$
$= x(x +y )+8(x+ y)$
$= (x+8)(x+y)$

Question:3(ii) Factorise

$15 xy -6 x +5 y -2$

Answer:

We have,
$15xy = 3 \times 5 \times x \times y$
$6x = 2 \times 3 \times x$
$5y = 5 \times y$
$2 = 2$
Therefore, on factorization we get,
$15xy - 6x +5y-2 = (3\times 5 \times x \times y)-(2 \times 3 \times x)+(5\times y)-2$
$=(5 \times y)(3\times x + 1)-2(3\times x + 1)$
$=(5y-2)(3x+1)$

Question:3(iii) Factorise

$ax + bx - ay - by$

Answer:

We have,
$ax+bx-ay-by = a(x-y)-b(x-y)$
$=(a-b)(x-y)$
Therefore, on factorization we get,
$(a-b)(x-y)$

Question:3(iv) Factorise

$15 pq + 15 + 9q + 25p$

Answer:

We have,
$15pq + 15 + 9q + 25p = 5 p(3q + 5) + 3 (3q + 5)$
$= (3q + 5)(5p + 3)$
Therefore, on factorization we get,
$(3q + 5)(5p + 3)$

Question:3(v) Factorise

$z-7 + 7 xy - xyz$

Answer:

We have,
$z - 7 + 7xy - xyz = z(1 - xy) -7(1 - xy)$
$= (1 - xy)(z - 7)$
Therefore, on factorization we get,
$(1 - xy)(z - 7)$

Class 8 maths chapter 14 NCERT solutions - Exercise: 14.2

Question:1(i) Factorise the following expressions

$a ^ 2 + 8a + 16$

Answer:

We have,
$a^2 + 8a + 16 = a^2+ 4a + 4a + 16$
$= a(a + 4) + 4 (a+4)$
$= (a+4)(a+4) =$ $(a+4)^{2}$
Therefore,
$a^2+8a+16 = (a+4)^2$

Question:1(ii) Factorise the following expressions

$p^2 -10 p + 25$

Answer:

We have,
$p^2 - 10p + 25 = p^2 - 5p - 5p + 25$
$= p(p - 5) -5 (p -5)$
$= (p - 5)(p - 5) =$ $(p-5)^{2}$
Therefore,
$p^2-10p+25 =(p-5)^2$

Question:1(iii) Factorise the following expressions

$25 m ^2 + 30 m + 9$

Answer:

We have,
$25m^2 + 30m + 9 = 25m^2 + 15m + 15m + 9$
$= 5m (5m + 3) +3(5m + 3)$
$= (5m + 3) (5m + 3) =$ $(5m+3)^{2}$
Therefore,
$25m^2+30m+9 = (5m+3)^2$

Question:1(iv) Factorise the following expressions

$49 y^2 + 84 yz + 36 z^2$

Answer:

We have,
$49 y^2 + 84 yz + 36 z^2$ $= 49y^2 + 42yz + 42yz + 36z^2$
$= 7y(7y + 6z) + 6z(7y + 6z)$
$= (7y + 6z)(7y + 6z) =$ $(7y+ 6z)^{2}$
Therefore,
$49y^2+84yz+36z^2=(7y+6z)^2$

Question:1(v) Factorise the following expressions

$4 x^2 - 8x + 4$

Answer:

We have,
$4 x^2 - 8x + 4$ $= 4x^2 - 4x - 4x + 4$
$= 4x(x - 1) -4(x - 1)$
$= 4(x-1)(x-1) \\\ \ \ = 4(x-1)^{2}$

Question:1(vi) Factorise the following expressions

$121 b^2 - 88 bc + 16 c^2$

Answer:

We have,
$121 b^2 - 88 bc + 16 c^2$ $= 121b^2 - 44bc - 44bc + 16c^2$
$= 11b(11b - 4c) - 4c(11b - 4c)$
$= (11b-4c)(11b-4c) =$ $(11b -4c)^{2}$
Therefore,
$121 b^2 - 88 bc + 16 c^2$ $=$ $(11b -4c)^{2}$

Question:1(vii) Factorise the following expressions

$( l+m ) ^2 - 4lm$

Answer:

We have,
$( l+m ) ^2 - 4lm$ = $l^{2} + 2ml + m^{2} - 4lm$ $(using \ (a+b)^{2} = a^{2} + 2ab + b^{2})$
= $l^{2} - 2lm + m^{2}$
= $(l-m)^{2}$ $(using \ (a-b)^{2} = a^{_2} -2ab + b^{2})$

Question:1(viii) Factorise the following expressions

$a ^4 +2 a ^2 b ^ 2 + b ^ 4$

Answer:

We have,
$a ^4 +2 a ^2 b ^ 2 + b ^ 4$ = $a^{4}$ + $a^{2}b^{2}$ + $a^{2}b^{2}$ + $b^{4}$
= $a^{2}(a^{2 }+ b^{2}) + b^{2}(a^{2}+b^{2})$ = $(a^{2}+b^{2})(a^{2}+b^{2})$ = $(a^{2}+b^{2})^{2}$

Question:2(i) Factorise :

$4 p^2 - 9 q ^2$

Answer:

This can be factorized as follows
$4 p^2 - 9 q ^2$ = $(2p)^{2} - (3q)^{2}$ $= (2p - 3q)(2p + 3q)$ $(using \ (a)^{2} - (b)^{2} = (a-b)(a+b))$

Question:2(ii) Factorise the following expressions

$63 a ^2 - 112 b ^ 2$

Answer:

We have,
$63 a ^2 - 112 b ^ 2$ $= 7$ $(9a^{2} - 16b^{2})$ $= 7$ $((3a)^{2} - (4b)^{2})$ $=7 (3a - 4b)(3a + 4b)$
$(using \ (a)^{2} - (b)^{2} = (a-b)(a+b))$

Question:2(iii) Factorise

$49 x^2 - 36$

Answer:

This can be factorised as follows
$49 x^2 - 36$ = $(7x)^{2} - (6)^{2}$ $= (7x - 6)(7x + 6)$ $(using \ (a)^{2} - (b)^{2} = (a-b)(a+b) )$

Question:2(iv) Factorise

$16 x^5 - 144 x ^ 3$

Answer:

The given question can be factorised as follows
$16 x^5 - 144 x ^ 3$ $= 16x^3(x^{2}- 9)$
$= 16x^3((x)^{2}- (3)^{2})$ $= 16x^3(x-3)(x+3)$ $(using \ (a)^{2}- (b)^{2} = (a-b)(a+b))$

Question:2(v) Factorise

$(l+m) ^ 2 - ( l- m ) ^2$

Answer:

We have,
$(l+m) ^ 2 - ( l- m ) ^2$ $= [(l + m) - (l - m)][(l + m) + (l - m)]$ (using $a^{2} - b^{2} = (a-b)(a+b)$ )
$= (l + m - l + m)(l + m + l - m)$
$= (2m)(2l) = 4ml$

Question:2(vi) Factorise

$9 x ^2 y^2 - 16$

Answer:

We have,
$9 x ^2 y^2 - 16$ = $(3xy)^{2} -(4)^{2}$ (using $(a)^{2} -(b)^{2} = (a-b) (a+b)$ )
$= (3xy - 4 )(3xy + 4)$

Question:2(vii) Factorise

$( x ^2 -2xy + y^2 ) - z ^2$

Answer:

We have,
$( x ^2 -2xy + y^2 ) - z ^2$ = $(x-y)^{2} - z^{2}$ $(using \ (a-b)^{2} = a^{2} -2ab + b^{2})$
$= (x - y - z)(x - y + z)$ $(using \ (a)^{2} - (b)^{2} = (a -b ) (a+b))$

Question:2(viii) Factorise

$25 a ^2 -4 b ^2 + 28 bc - 49 c ^2$

Answer:

We have,
$25 a ^2 -4 b ^2 + 28 bc - 49 c ^2$ = $25a^{2} - (2b-7c)^{2}$ $(using \ (a-b)^{2} = a^{2} -2ab + b^{2})$
= $(5a)^{2} - (2b-7c)^{2}$ $(using \ (a)^{2} - (b)^{2} = (a -b ) (a+b))$
$=(5a - (2b - 7c))(5a + (2b - 7c)$ )
$= (5a - 2b + 7c)(5a + 2b - 7c )$

Question:3(i) Factorise the following expressions

$ax ^2 + bx$

Answer:

We have,
$ax^2 = a \times x \times x$
$bx = b \times x$
Therefore,
$ax ^2 + bx$ $= (a \times x \times x) + (b \times x)$
$= x(a \times x + b)$
$= x(ax + b)$

Question:3(ii) Factorise the following expressions

$7p^2 + 21 q ^2$

Answer:

We have,
$7p^2 = 7 \times p \times p$
$21q^3 = 3 \times 7 \times q \times q$
Therefore,
$7p^2 + 21 q ^2$ $= (7 \times p \times p) + (3 \times 7 \times q \times q)$
$=7$ $(p^{2}+ 3q^{2})$

Question:3(iii) Factorise the following expressions

$2 x^3 + 2xy^2 + 2 xz ^2$

Answer:

We have,
$2x^3 = 2 \times x \times x \times x$
$2xy^2 = 2 \times x \times y \times y$
$2xz^2 = 2 \times x \times z \times z$
Therefore,
$2 x^3 + 2xy^2 + 2 xz ^2$ $= (2 \times x \times x \times x) + ( 2 \times x \times y \times y) + ( 2 \times x \times z \times z)$
$= (2 \times x) [(x \times x) + (y \times y ) + (z \times z)]$
$= 2x(x^2+y^2+z^2)$

Question:3(iv) Factorise the following expressions

$am^2 + bm ^2 + bn ^2 + an^2$

Answer:

We have,
$am^2 + bm ^2 + bn ^2 + an^2$ $= m^2(a + b) + n^2(a + b)$
$= (a + b)$ $(m^{2 }+n^{2})$

Question:3(v) Factorise the following expressions

$( lm + l ) + m + 1$

Answer:

We have,
$( lm + l ) + m + 1$ $= lm + l + m + 1$
$= l(m + 1) +1(m + 1)$
$= (m + 1)(l + 1)$

Question:3(vi) Factorise the following expressions

$y ( y + z ) + 9 ( y + z )$

Answer:

We have,
$y ( y + z ) + 9 ( y + z )$
Take ( y+z) common from this
Therefore,
$y ( y + z ) + 9 ( y + z )$ $= (y + z)(y + 9)$

Question:3(vii) Factorise the following expressions

$5 y ^ 2 - 20 y - 8z + 2yz$

Answer:

We have,
$5 y ^ 2 - 20 y - 8z + 2yz$ $= 5y(y - 4) + 2z(y - 4)$
$= (y - 4)(5y + 2z)$
Therefore,
$5 y ^ 2 - 20 y - 8z + 2yz$ $= (y - 4)(5y + 2z)$

Question:3(viii) Factorise

$10 ab + 4a + 5b + 2$

Answer:

We have,
$10 ab + 4a + 5b + 2$ $= 2a(5b + 2) + 1(5b + 2)$
$= (5b + 2)(2a + 1)$
Therefore,
$10 ab + 4a + 5b + 2$ $= (5b + 2)(2a + 1)$

Question:3(ix) Factorise the following expressions

$6 xy - 4 y + 6 - 9 x$

Answer:

We have,
$6 xy - 4 y + 6 - 9 x$ $= 2y(3x - 2) - 3 (3x - 2)$
$= (3x - 2)(2y - 3)$
Therefore,
$6 xy - 4 y + 6 - 9 x$ $= (3x - 2)(2y - 3)$

Question:4(i) Factorise $a ^ 4 - b ^ 4$

Answer:

We have,
$a ^ 4 - b ^ 4$ = $(a^{2})^{2} - (b^{2})^{2} = (a^{2} - b^{2})(a^{2} + b^{2}) = (a-b)(a+b)(a^{2} + b^{2})$
$using \ (x^{2} - y^{2}) = (x-y)(x+y)$

Question:4(ii) Factorise $p ^ 4 - 81$

Answer:

We have,
$p ^ 4 - 81$ =
$(p^{2})^{2} - (9)^{2} = (p^{2} - 9)(p^{2}+9) \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (p^{2}-(3)^{2})(p^{2}+9)\\ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (p-3)(p+3)(p^{2}+9)$ $using \ a^{2} - b^{2} = (a-b)(a+b)$

Question:4(iii) Factorise $x ^4 - ( y + z )^4$

Answer:

We have,
$x ^4 - ( y + z )^4$ =
$(x^{2})^{2} -((y+z)^{2})^{2} = (x^{2} - (y+z)^{2})(x^{2} +(y+z)^{2})\\ \Rightarrow (x-(y+z))(x+(y+z))(x^{2} +(y+z)^{2})$
$(using \ a^{2} -b^{2} = (a-b)(a+b))$

Question:4(iv) Factorise $x ^ 4 - ( x-z ) ^ 4$

Answer:

We have,
$x ^ 4 - ( x-z ) ^ 4$ = $(x^{2})^{2} - ((x-z)^{2})^{2}$ $using \ a^{2}-b^{2} = (a-b)(a+b)$
= $(x^{2} - (x-z)^{2})(x^{2}+(x-z)^{2})$
= $(x+(x-z))(x - (x-z))(x^{2}+(x-z)^{2})$
= $(2x - z)(z)$ $($ $x^{2}+(x-z)^{2}$ $)$

Question:4(v) Factorise $a ^ 4 - 2 a^2 b^2 + b ^ 4$

Answer:

We have,
$a ^ 4 - 2 a^2 b^2 + b ^ 4$ = $a^{4} - a^{2}b^{2} - a^{2}b^{2} + b^{4}$
= $a^{2}(a^{2} - b^{2}) - b^{2}(a^{2} - b^{2})$
= $(a^{2} - b^{2}) (a^{2}-b^{2})$ $using \ a^{2}-b^{2} = (a-b)(a+b)$
= $(a^{2} - b^{2})^{2}$
= $((a - b)(a+b))^{2}$
= $(a - b)^{2}(a+b)^{2}$

Question:5(i) Factorise the following expression

$p^ 2 + 6 p + 8$

Answer:

We have,
$p^ 2 + 6 p + 8$ = $p^{2} + 2p + 4p + 8$
$= p(p + 2) + 4(p + 2)$
$=(p + 2)(p + 4)$
Therefore,
$p^ 2 + 6 p + 8$ $=(p + 2)(p + 4)$

Question:5(ii) Factorise the following expression

$q ^ 2 - 10 q + 21$

Answer:

We have,
$q ^ 2 - 10 q + 21$ = $q^{2} - 7q -3q + 21$
$= q(q - 7) -3(q - 7)$
$=(q - 7)(q - 3)$
Therefore,
$q ^ 2 - 10 q + 21$ $=(q - 7)(q - 3)$

Question:5(iii) Factorise the following expression

$p^2 + 6 p - 16$

Answer:

We have,
$p^2 + 6 p - 16$ = $p^{2} + 8p - 2p - 16$
$= p(p + 8) -2(p + 8)$
$=(p - 2)(p + 8)$
Therefore,
$p^2 + 6 p - 16$ $=(p - 2)(p + 8)$

Class 8 factorization NCERT solutions - Topic 14.3.1 Division Of A Monomial By Another Monomial

Question:(i) Divide $24 xy^2 z^3 \: \: by \: \: 6 yz^2$

Answer:

We have,
$\frac{24xy^{2}z^{3}}{6yz^{2}} =\frac{2\times 2\times 2\times3\times y \times y \times z\times z\times z}{2\times 3 \times y \times z \times z}= 4xyz$

Question:(ii) Divide $63 a ^ 2 b^ 4 c ^6 \: \: by \: \: 7 a ^2 b^ 2 c ^3$

Answer:

We have,
$\frac{63a^{2}b^{4}c^{6}}{7a^{2}b^{2}c^{3}}=\frac{3\times 3 \times 7 \times a \times a \times b \times b\times b^2 \times c \times c \times c \times c^3}{7a^{2}b^{2}c^{3}} = 9b^{2}c^{3}$

NCERT Solutions for Class 8 Maths Chapter 14 Factorization-Exercise: 14.3

Question:1(i) Carry out the following divisions

$28 x ^ 4 \div 56 x$

Answer:

, $\frac{28x^{4}}{56x} = \frac{2 \times 2 \times 7 \times x \times x \times x \times x}{2 \times 2 \times 2 \times 7 \times x} = \frac{x^{3}}{2}$

This is done using factorization.

Question:1(ii) Carry out the following divisions

$-36 y^3 \div 9 y^2$

Answer:

We have,
$-36$ $y^{3}$ $= -1 \times 2 \times 2 \times 3 \times 3 \times y \times y \times y$
$9$ $y^{2 }$ $= 3 \times 3 \times y \times y$
Therefore,

$\frac{-36y^{3}}{9y^{2}} = \frac {-1 \times 2 \times 2 \times 3 \times 3 \times y \times y \times y}{3 \times 3 \times y \times y} = -4y$

Question:1(iii) Carry out the following divisions

$66 pq^2 r ^ 3 \div 11 q r ^2$

Answer:

We have,
$66pq^2r^3 = 2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r$
$11qr^2 = 11 \times q \times r \times r$
Therefore,
$\frac{66pq^{2}r^{3}}{11qr^{2}} = \frac{2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r}{11 \times q \times r \times r} = 6pqr$

Question:1(iv) Carry out the following divisions

$34 x^ 3 y^3 z ^ 3 \div 51 x y^2 z ^ 3$

Answer:

We have,

$\therefore \frac{34x^{3}y^{3}z^{3}}{51xy^{2}z^{3}} = \frac{2 \times 17\times \ x \times x \times x \times y \times y \times y \times z\times z \times z}{3 \times 17 \times x \times y \times y \times z \times z\times z} = \frac{2x^{2}y}{3}$

Question:1(v) Carry out the following divisions

$12 a ^ 8 b^ 8 \div ( -6 a ^ 6 b ^ 4 )$

Answer:

We have,

$\frac{12a^8b^8}{-6a^4b^4}= \frac{2 \times 2 \times 3 \times a \times a \times a^{6} \times b \times b \times b \times b \times b^{4}}{-1 \times 2 \times 3 \times a^{6} \times b^{4}} = -2a^{2}b^{4}$

Question:2(i) Divide the given polynomial by the given monomial

$( 5x ^2 -6x ) \div 3x$

Answer:

We have,
$5x^2 - 6x = x(5x - 6)$

$\therefore \frac{5x^{2}-6}{3x} = \frac{x(5x-6)}{3x} = \frac{5x-6}{3}$

Question:2(ii) Divide the given polynomial by the given monomial

$( 3 y ^8 - 4 y^6 + 5 y ^4 )\div y ^ 4$

Answer:

We have,
$3y^{8} - 4y^{6} + 5y^{4} = y^{4}(3y^{4}-4y^{2} + 5)$
$\therefore \frac{y^{4}(3y^{4}-4y^{2}+5)}{y^{4}} = (3y^{4}-4y^{2}+5)$

Question:2(iii) Divide the given polynomial by the given monomial

$8 ( x ^3 y^2 z ^2 + x^2 y^3 z^2 + x ^2 y^2 z^3 ) \div 4 x ^2 y ^2 z ^2$

Answer:

We have,
$8(x^{3}y^{2}z^{2} + x^{2}y^{3}z^{2} + x^{2} y^{2}z^{3}) = 8x^{2}y^{2}z^{2}(x+y+z)$
$\therefore \frac{8(x^{3}y^{2}z^{2} + x^{2}y^{3}z^{2} + x^{2} y^{2}z^{3})}{4x^{2}y^{2}z^{2}} =\frac{ 8x^{2}y^{2}z^{2}(x+y+z)}{4x^{2}y^{2}z^{2}} =2(x+y+z)$

Question:2(iv) Divide the given polynomial by the given monomial

$( x^3 +2 x ^2 + 3 x ) \div 2x$

Answer:

We have,
$x^{3} + 2x^{2} + 3x = x(x^{2} + 2x + 3)$

$\therefore \frac{x^{3} + 2x^{2} + 3x}{2x} = \frac{x(x^{2} + 2x + 3)}{2x} = \frac{x^{2} + 2x + 3}{2}$

Question:2(v) Divide the given polynomial by the given monomial

$( p ^ 3 q ^6 - p ^ 6 q ^ 3 ) \div p ^3 q ^3$

Answer:

We have,
$(p^{3}q^{6} - p^{6}q^{3}) = p^{3}q^{3}(q^{3} - p^{3})$
$\therefore \frac{(p^{3}q^{6} - p^{6}q^{3})}{p^{3}q^{3}} = \frac{p^{3}q^{3}(q^{3} - p^{3})}{p^{3}q^{3}} = (q^{3} - p^{3})$

Question:3(i) workout the following divisions

$( 10 x - 25) \div 5$

Answer:

We have,
$10x -25 = 5(2x - 5)$
Therefore,
$\frac{10x-25}{5}= \frac{5(2x-5)}{5} = 2x - 5$

Question:3(ii) workout the following divisions

$( 10 x -25 ) \div ( 2x -5 )$

Answer:

We have,
$10x-25 = 5(2x - 5 )$
Therefore,
$\frac{10x-25}{2x-5} = \frac{5(2x-5)}{2x-5} = 5$

Question:3(iii) workout the following divisions

$10 y ( 6y +21 ) \div 5 ( 2y + 7 )$

Answer:

We have,
$10y(6y + 21) = 2 \times y \times 5 \times 3(2y + 7)$
Therefore,
$\frac{10y(6y+21)}{5(2y+7)} = \frac{2 \times 5 \times y \times 3(2y+7)}{5(2y+7)} = 6y$

Question:3(iv) workout the following divisions

$9 x ^2 y^2 ( 3z -24 ) \div 27 xy ( z-8 )$

Answer:

We have,
$9x^{2}y^{2}(3z-24) = 9x^{2}y^{2} \times 3(z-8) = 27x^{2}y^{2}(z-8)$

$\therefore \frac{9x^{2}y^{2}(3z-24)}{27xy(z-8)} = \frac{27x^{2}y^{2}(z-8)}{27xy(z-8)} = xy$

Question:3(v) workout the following divisions

$96 abc ( 3a -12 ) ( 5 b -30 ) \div 144 (a-4 ) ( b- 6 )$

Answer:

We have,
$96abc(3a - 12)(5b - 30) = 2 \times 48abc \times 3(a - 4) \times 5(b - 6)$
$= 2 \times144abc (a - 4) \times 5(b - 6)$
Therefore,
$\frac{96abc(3a-12)(5b-30)}{144(a-4)(b-6)} = \frac{2 \times 144abc (a-4) \times 5 (b - 6)}{144(a-4)(b-6)} = 10abc$

Question:4(i) Divide as directed

$5 ( 2x +1 ) ( 3x +5 ) \div ( 2x +1)$

Answer:

We have,
$\frac{5(2x+1)(3x+5)}{2x+1} = 5(3x+5)$

Question:4(ii) Divide as directed

$26 xy ( x+5 ) ( y-4) \div 13 x ( y-4 )$

Answer:

We have,
$\frac{26xy(x+5)(y-4)}{13x(y-4)} = \frac{2 \times 13xy(x+5)(y-4)}{13x(y-4)} =2y(x+5)$

Question:4(iii) Divide as directed

$52 pqr ( p+ q ) ( q+ r ) ( r +p)\div 104 pq ( q+r ) ( r + p )$

Answer:

We have,
$\frac{52pqr(p+q)(q+r)(r+p)}{104pq(q+r)(r+p)} = \frac{r(p+q)}{2}$

Question:4(iv) Divide as directed

$20 ( y+4 ) ( y^2 + 5 y + 3 ) \div 5 (y +4 )$

Answer:

We have,
$\frac{20(y+4)(y^{2}+5y+3)}{5(y+4)} =\frac{4 \times 5(y+4)(y^{2}+5y+3)}{5(y+4)} = 4(y^{2}+5y+3)$

Question:4(v) Divide as directed

$x ( x+1 ) ( x+2 ) ( x+3 ) \div x ( x+1 )$

Answer:

We have,
$\frac{x(x+1)(x+2)(x+3)}{x(x+1)} = (x+2)(x+3)$

Question:5(i) Factorise the expression and divide then as directed

$( y ^ 2 + 7 y + 1 0 ) \div ( y + 5 )$

Answer:

We have,
$\frac{y^{2}+7y+10}{y+5} = \frac{y^{2}+2y +5y +10}{y+5} =\frac{y(y+2)+5(y+2)}{y+5}\\ \\ \Rightarrow \frac{(y+5)(y+2)}{(y+5)} = (y+2)$

Question:5(ii) Factorise the expression and divide then as directed

$( m^2 - 14 m -32 ) \div ( m +2 )$

Answer:

We have,
$\frac{m^{2}-14m-32}{m+2} = \frac{m^{2}+2m-16m-32}{m+2} = \frac{m(m+2)-16(m+2)}{m+2}\\ \\\Rightarrow \frac{(m-16)(m+2)}{m+2} = m-16$

Question:5(iii) Factorise the expression and divide then as directed

$( 5 p^2 -25p + 20 ) \div ( p-1 )$

Answer:

We have,
$\frac{5p^{2}-25p+20}{p-1} = \frac{5p^{2} -5p -20p +20}{p-1} = \frac{5p(p-1)-20(p-1)}{p-1}\\ \\ \frac{(5p-20)(p-1)}{p-1} = 5p-20$

Question:5(iv) Factorise the expression and divide then as directed

$4 yz ( z^2 + 6z -16 ) \div 2y ( z+8 )$

Answer:

We first simplify our numerator
So,
$4yz( z^2+ 6z - 16)$
Add and subtract 64 $\Rightarrow$ $4yz( z^2- 64 + 6z - 16 + 64)$
$= 4yz(z^2-8^2 + 6z + 48)$
$= 4yz((z + 8)(z - 8) + 6(z + 8))$ $using \ a^{2} -b^{2} = (a - b)(a + b)$
$= 4yz (z + 8)(z - 8 + 6)$
$= 4yz(z + 8)(z - 2)$
Now,
$\frac{4yz(z^{2}+6z-16)}{2y(z+8)} = \frac{4yz(z+8)(z-2)}{2y(z+8)}= 2z(z-2)$

Question:5(v) Factorise the expression and divide then as directed

$5 pq ( p^2 - q ^ 2 ) \div 2 p ( p + q )$

Answer:

We have,
$\frac{5pq(p^{2} - q^{2})}{2p(p+q)} = \frac{5pq(p-q)(p+q)}{2p(p+q)} \ \ \ \ \ \ \ \ \ \ \ \ \ using \ a^{2}-b^{2} = (a-b)(a+b) \ \ \\. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{5q(p-q)}{2}$

Question:5(vi) Factorise the expression and divide then as directed

$12 xy ( 9 x^2 - 16 y^2 ) \div 4 xy ( 3 x + 4 y )$

Answer:

We first simplify our numerator,
$12xy$ ( $9x^{2} -16y^{2}$ ) = $12xy$ $(3x)^{2} -(4y)^{2}$

using $(a)^{2} -(b)^{2} = (a-b)(a+b)$
$= 12xy((3x - 4y)(3x + 4y))$
Now,
$\frac{12xy(9x^{2} - 16y^{2})}{4xy(3x + 4y)} = \frac{12xy(3x+4y)(3x-4y)}{4xy(3x+4y)} = 3(3x-4y)$

Question:5(vii) Factorise the expression and divide then as directed

$39 y^2 ( 50 y^2 - 98 ) \div 26 y^2 ( 5y +7 )$

Answer:

We first simplify our numerator,
$39y^{2}(50y^{2} -98) = 39y^{2} \times 2(25y^{2} - 49)$ using $(a)^{2} -(b)^{2} = (a-b)(a+b)$
= $78y^{2} ((5y)^{2} - (7)^{2})$
= $78y^{2} (5y - 7)(5y+7)$
Now,
$\frac{39y^{2}(50y^{2}-98)}{26y^{2}(5y +7)} = \frac{78y^{2}(5y-7)(5y+7)}{26y^{2}(5y+7)} = 3(5y-7)$

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

Question:1 Find and correct the errors in the following mathematical statements

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

Answer:

Our L.H.S.
$= 4(x - 5) = 4x - 20$
R.H.S. $= 4x -5$
It is clear from the above that L.H.S. is not equal to R.H.S.
So, correct statement is
$4(x - 5) = 4x - 20$

Question:2 Find and correct the errors in the following mathematical statements

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

Answer:

Our L.H.S.
$= x(3x + 2) = 3x^2 + 2x$
R.H.S.= $3x^2 + 2$
It is clear from the above that L.H.S. is not equal to R.H.S.
So, correct statement is
$= x(3x + 2) = 3x^2 + 2x$

Question:3 Find and correct the errors in the following mathematical statements

$2 x + 3y = 5 xy$

Answer:

Our L.H.S. $= 2x + 3y$
R.H.S. = $5xy$
It is clear from the above that L.H.S. is not equal to R.H.S.
SO, correct statement is
$2x + 3y = 2x + 3y$

Question:4 Find and correct the errors in the following mathematical statements

$x + 2x + 3x = 5x$

Answer:

Our L.H.S. $= x + 2x + 3x = 6x$
R.H.S. $= 5x$
It is clear from the above that L.H.S. is not equal to R.H.S.
So, correct statement is
$x + 2x + 3x = 6x$

Question:5 Find and correct the errors in the following mathematical statements

$(Q5)\ 5 y + 2y + y - 7y = 0$

Answer:

Our L.H.S. is
$5y + 2y + y - 7y = y$
R.H.S. = 0
IT is clear from the above that L.H.S. is not equal to R.H.S.
So, Correct statement is
$5y + 2y + y - 7y = y$

Question:6 Find and correct the errors in the following mathematical statements

$3 x +2 x = 5 x ^2$

Answer:

Our L.H.S. is
$3x + 2x = 5x$
R.H.S. = $5x^2$
It is clear from the above that L.H.S. is not equal to R.H.S.
So, correct statement is
$3x + 2x = 5x$

Question:7 Find and correct the errors in the following mathematical statements

$( 2x )^2 + 4 ( 2x ) + 7 = 2 x^2 + 8 x +7$

Answer:

Our L.H.S. is
$(2x)^2 + 4(2x) + 7 = 4x^2 + 8x + 7$
R.H.S. $= 2x^2+8x+7$
It is clear from the above that L.H.S. is not equal to R.H.S.
So, correct statement is
$(2x)^2 + 4(2x) + 7 = 4x^2 + 8x + 7$

Question:8 Find and correct the errors in the following mathematical statements

$( 2 x)^2 + 5 x = 4 x +5x = 9 x$

Answer:

Our L.H.S. is
$\Rightarrow (2x)^{2}+5x = 4x^2+5x$
R.H.S. = 9x
It is clear from the above that L.H.S. is not equal to R.H.S.
So, the correct statement is
$(2x)^{2}+5x = 4x^2+5x$

Question:9 Find and correct the errors in the following mathematical statements

$( 3x +2 )^2 = 3 x ^2 + 6x + 4$

Answer:

LHS IS

$(3x + 2)^{2 } = (3x)^{2} + 2(3x)(2) +(2)^{2}$ using $(a + b)^{2 } = (a)^{2} + 2(a)(b) +(b)^{2}$
$= 9x^2 + 12x + 4$

RHS IS

$3 x ^2 + 6x + 4$

$\boldsymbol{LHS} \neq \boldsymbol{RHS}$

Correct statement is

$(3x + 2)^{2 } = (3x)^{2} + 2(3x)(2) +(2)^{2}$ $= 9x^2 + 12x + 4$

Question:10(a) Find and correct the errors in the following mathematical statements

Substituting $x = -3$ in $x ^ 2 + 5 x + 4 \: \: gives \: \: ( -3 ) ^ 2 + 5 ( -3 ) + 4 = 9 + 2 + 4 = 15$

Answer:

We need to substitute x = -3 in

$x^{2}+5x+4$
$=(-3)^{2}+5(-3)+4$
$= 9 - 15 + 4$
$= -2 \neq 15$

so the given statement is wrong
Correct statement is $(-3)^{2}+5(-3)+4= -2$

Question:10(b) find and correct the errors in the following mathematical statements

Substituiting x = -3 in $x ^2 -5 x + 4 \: \: gives \: \: ( -3)^2 - 5 ( -3 ) + 4 = 9 - 15 + 4 = -2$

Answer:

We need to substitute x = -3 in $x^2 - 5x + 4$
$= (-3)^2-5(-3) + 4$
$= 9 + 15 + 4=28$
so the given statement is wrong
Correct statement is

$x^2 - 5x + 4=28$

Question:10(c) find and correct the errors in the following mathematical statements

Substituting x = - 3 in $x ^2 + 5 x \: \: gives \: \: ( -3 ) ^ 2 + 5 ( -3 ) = -9-15 = -24$

Answer:

We need to Substitute x = - 3 in $x^{2} + 5x$
= $(-3)^{2} + 5(-3)$
= 9 - 15
= - 6 $\neq$ R.H.S
Correct statement is Substitute x = - 3 in $x^{2} + 5x$ gives -6

Question:11 Find and correct the errors in the following mathematical statements

$( y - 3 ) ^ 2 = y ^ 2 -9$

Answer:

Our L.H.S. is $(y - 3 )^{2}$
= $(y )^{2} + 2(y)(-3) + (-3)^{2}$ using $(a-b)^{2} = (a )^{2} + 2(a)(-b) + (-b)^{2}$
= $y^{2}$ $- 6x + 9$ $\neq$ R.H.S.

Correct statement is

$(y - 3 )^{2}$ = $y^{2}$ $- 6x + 9$

Question:12 Find and correct the errors in the following mathematical statements

$( z+5 ) ^2 = z^2 + 25$

Answer:

Our L.H.S. is $(z+5)^{2}$
= $(z)^{2} + 2(z)(5) + (5)^{2}$ using $(a+b)^{2} = (a)^{2} + 2(a)(b) + (b)^{2}$
= $(z)^{2}$ $+ 10z + 25$ $\neq$ R.H.S.
Correct statement is

$(z+5)^{2}$ = $(z)^{2}$ $+ 10z + 25$

Question:13 Find and correct the errors in the following mathematical statements.

$( 2a +3b ) ( a-b) = 2 a ^2 - 3 b^2$

Answer:

Our L.H.S. is (2a + 3b)(a -b)
= $2a^{2} -2ab + 3ab - 3b^{2}$
= $2a^{2} +ab - 3b^{2}$ $\neq$ R.H.S.
Correct statement is (2a + 3b)(a -b) = $2a^{2} +ab - 3b^{2}$

Question:14 Find and correct the errors in the following mathematical statements.

$( a + 4 ) ( a +2 ) = a ^ 2 + 8$

Answer:

Oue L.H.S. is (a + 4)(a + 2)
= $a^{2} + 2a + 4a + 8$
= $a^{2} + 6a + 8$ $\neq$ R.H.S.
Correct statement is (a + 4)(a + 2) = $a^{2} + 6a + 8$

Question:15 Find and correct the errors in the following mathematical statements.

$(a - 4 ) ( a - 2 )= a ^2 - 8$

Answer:

Our L.H.S. is (a - 2) (a - 4)
= $a^{2} - 4a - 2a + 8$
= $a^{2} - 6a+ 8$ $\neq$ R.H.S.
Correct statement is (a - 2) (a - 4) = $a^{2} - 6a+ 8$

Question:16 Find and correct the errors in the following mathematical statements.

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

Answer:

Our L.H.S. is
$\Rightarrow \frac{3x^{2}}{3x^{2}}$
R.H.S. = 0
It is clear from the above that L.H.S. is not equal to R.H.S.
So, correct statement is
$\frac{3x^{2}}{3x^{2}} = 1$

Question:17 Find and correct the errors in the following mathematical statements.

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

Answer:

Our L.H.S. is
$\Rightarrow \frac{3x^2+1}{3x^2}$
R.H.S. = 2
It is clear from the above stattement that L.H.S. is not equal to R.H.S.
So, correct statement is
$\frac{3x^{2}+1}{3x^{2}} = 1 + \frac{1}{3x^{2}} = \frac{3x^{2}+1}{3x^{2}}$

Question:18 find and correct the errors in the following mathematical statements.

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

Answer:

Our L.H.S.

$\Rightarrow \frac{3x}{3x+2}$

R.H.S. = 1/2

It can be clearly observed that L.H.S is not equal to R.H.S

So, the correct statement is,

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

Question:19 find and correct the errors in the following mathematical statements

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

Answer:

Our L.H.S. is $\Rightarrow \frac{3}{4x+3} = \frac{3}{4x+3} \neq$ R.H.S.

Correct statement is $\frac{3}{4x+3} = \frac{3}{4x+3}$

Question:20 find and correct the errors in the following mathematical statements

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

Answer:

Our L.H.S. is $\Rightarrow \frac{4x+5}{4x} = \frac{4x}{4x} + \frac{5}{4x} = 1 + \frac{5}{4x} \neq$ R.H.S.

Correct statement is $\frac{4x+5}{4x} = 1 + \frac{5}{4x} = \frac{4x+5}{4x}$

Question:21 find and correct the errors in the following mathematical statements

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

Answer:

Our L.H.S. is $\Rightarrow \frac{7x+5}{5} = \frac{7x}{5} + \frac{5}{5} = \frac{7x}{5} + 1 \neq$ R.H.S.

Correct statement is $\frac{7x+5}{5} = \frac{7x}{5} + 1 = \frac{7x+5}{5}$

Factorization class 8 NCERT solutions - Topics

What is Factorization?
Division of Algebraic Expressions
Division of Algebraic Expressions Continued(Polynomial divide; Polynomial)
Can you Find the Error?

NCERT Solutions for Class 8 Maths - Chapter Wise

Chapter -1	Rational Numbers
Chapter -2	Linear Equations in One Variable
Chapter-3	Understanding Quadrilaterals
Chapter-4	Practical Geometry
Chapter-5	Data Handling
Chapter-6	Squares and Square Roots
Chapter-7	Cubes and Cube Roots
Chapter-8	Comparing Quantities
Chapter-9	Algebraic Expressions and Identities
Chapter-10	Visualizing Solid Shapes
Chapter-11	Mensuration
Chapter-12	Exponents and Powers
Chapter-13	Direct and Inverse Proportions
Chapter-14	Factorization
Chapter-15	Introduction to Graphs
Chapter-16	Playing with Numbers

Key Features of Factorization Class 8 Solutions

Comprehensive Coverage: Maths chapter 14 class 8 solutions cover all topics and concepts related to factorization as per the Class 8 syllabus.

Step-by-Step Solutions: Class 8 maths ch 14 question answer are detailed, step-by-step explanations for each problem, making it easy for students to understand and apply mathematical concepts related to factorization.

Variety of Problems: A wide range of problems, including exercises and additional questions, to help students practice and test their understanding of factorization methods are discussed in ch 14 maths class 8.

NCERT Solutions for Class 8 - Subject Wise

Factorization is a key skill to solve a problem where you need to find the value of x. It will strengthen your foundations of algebra, trigonometry, calculus, and higher class maths. It has a lot of applications like calculation, make multiplication easy, prime factorization, finding LCM and HCF, solving polynomial equations, quadratic equations, and simplifying expression, etc. In NCERT solutions for Class 8 Maths chapter 14 Factorizations, you will come across some applications like simplifying expressions and solving quadratic equations. Some important expressions from NCERT solutions for Class 8 Maths chapter 14 Factorizations are given below which you should remember.

$a^{2}+2 a b+b^{2}=(a+b)^{2}$
$a^{2}-2 a b+b^{2}=(a-b)^{2}$
$a^{2}-b^{2}=(a+b)(a-b)$
$x^{2}+(a+b) x+a b=(x+a)(x+b)$

NCERT Books and NCERT Syllabus

Frequently Asked Questions (FAQs)

1. What are the important topics of Factorization ?

Factorization of algebraic expression, division of a monomial by another monomial, division of a polynomial by a monomial, and division of a polynomial by polynomial are covered in this chapter.

2. How many chapters are there in the CBSE class 8 maths ?

There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths.

3. Does CBSE provide NCERT solution for class 8 ?

No, CBSE doesn't provide NCERT solutions for any class and subject.

4. Where can I find the complete solutions of NCERT for class 8 ?

Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

5. Where can I find the complete solutions of NCERT for class 8 maths ?

Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

6. Which is the official website of NCERT ?

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

Also Read

NCERT Solutions for Class 8 Science Chapter 11 Force and Pressure

NCERT Solutions for Class 8 Science Chapter 10 Reaching The age of adolescence

NCERT Solutions for Class 8 Science Chapter 9 Reproduction in Animals

NCERT Solutions for Class 8 Science Chapter 7 Conservation of Plants and Animals

NCERT Solutions for Class 8 Science Chapter 3 Synthetic Fibres and Plastics

NCERT Solutions for Class 8 Science Chapter 2 Microorganisms Friend and Foe

NCERT Syllabus for Class 8 2024-25 - Download All Subjects PDF

Articles

MPSOS 10th Result 2024 - Check MPSOS Class 10 December 2024 Exam Result @mpsos.nic.in

Oct 29, 2024

Education Loan for School Students in India – Eligibility, Benefits & Application Process

Oct 29, 2024

How to Track NSP Payment Status – Step-by-Step Guide for Scholarship Updates

Oct 29, 2024

TOSS SSC Time Table 2025, Telangana Open School Class 10 Exam Dates @telanganaopenschool.org

Oct 29, 2024

IB Primary Years Programme (PYP) – Curriculum Overview and Key Benefits

Oct 28, 2024

Importance of Internships in Student Life - 10 Points, 100 Words, 200 Words, 500 Words

Oct 28, 2024

Cambridge Advanced Curriculum: AS and A levels (Grade 11 & 12)

Oct 28, 2024

Cambridge School Fees 2024-25 | Admission, Tuition Fee for Cambridge IGCSE Schools

Oct 25, 2024

Upcoming School Exams

Himachal Pradesh Board of Secondary Education 10th Examination

Application Date:09 September,2024 - 14 November,2024

Exam Pattern

Admit Card

Result

Himachal Pradesh Board of Secondary Education 12th Examination

Application Date:09 September,2024 - 14 November,2024

Result

Exam Pattern

National Institute of Open Schooling 12th Examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Admit Card

National Institute of Open Schooling 10th examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Mock Test

National Means Cum-Merit Scholarship

Application Date:05 October,2024 - 04 November,2024

Eligibility Criteria

Application Process

Exam Pattern

View All School Exams

Get answers from students and experts

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

Exams

Top Schools

Products & Resources

NCERT Study Material

Exams

Colleges

Predictors

Resources

Exams

Colleges & Courses

Predictors

Resources

Exams

Colleges

Predictors

Resources

Exams

Colleges

Predictors & E-Books

Resources

Exams

Predictors & Articles

Colleges

Resources

Exams

Colleges

Resources

Exams

Resources

Top Courses & Careers

Colleges

Exams

Colleges

Resources

Quick Links

Exams

Colleges

Resources

Exams

Colleges

Resources

Diploma Colleges

Exams

Resources

Upcoming Events

Other Exams

Exams

Colleges

Top Countries

Student Visas

Exams

Colleges

Upcoming Events

Resources

Engineering Preparation

Medical Preparation

Online Courses

Products

Top Streams

Specializations

Resources

Top Providers

NCERT Solutions for Class 8 Maths Chapter 14 Factorization

Factorization Class 8 Questions And Answers PDF Free Download

Factorization Class 8 Solutions - Important Formulae

Factorization Class 8 NCERT Solutions (Intext Questions and Exercise)

Factorization class 8 NCERT solutions - Topics

NCERT Solutions for Class 8 Maths - Chapter Wise

Key Features of Factorization Class 8 Solutions

NCERT Solutions for Class 8 - Subject Wise

NCERT Books and NCERT Syllabus

Frequently Asked Questions (FAQs)

Also Read

Articles

Upcoming School Exams

Himachal Pradesh Board of Secondary Education 10th Examination

Himachal Pradesh Board of Secondary Education 12th Examination

National Institute of Open Schooling 12th Examination

National Institute of Open Schooling 10th examination

National Means Cum-Merit Scholarship