# NCERT Solutions for Class 8 Maths Chapter 14 Factorization

NCERT Solutions for Class 8 Maths Chapter 14 Factorization: Decomposition of expression or mathematical object as a product of several factors is known as factorization. It is a process to factorize algebraic expressions and write this expression as a product of its factors. These are usually smaller or simpler objects of the same kind. It makes easy to multiply and find the least common multiple and greatest common factor. 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. 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 CBSE NCERT solutions for class 8 maths chapter 14 factorization. Check NCERT solutions from class 6 to 12 to learn science and maths.

## Important Topics of NCERT Class 8 Maths Chapter 14 Factorization:

• 14.1 Introduction
• 14.2 What is Factorization?
• 14.3 Division of Algebraic Expressions
• 14.4 Division of Algebraic Expressions Continued(Polynomial divide; Polynomial)
• 14.5 Can you Find the Error?

## Solutions of NCERT for class 8 maths chapter 14 factorization topic 14.2.1 method of common factor

Question:(i) Factorise:

$12 x +36$

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

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 :

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

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,

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

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

$(i) 12 x , 36$

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

$(ii) 2y , 22xy$

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

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

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.

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

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

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

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

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

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

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

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

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

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

$(i)7x -42$

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

$(ii)6 p - 12 q$

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

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

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

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

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

$20 l^2 m + 30 alm$

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

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

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

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

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

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

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

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

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

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

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,
$ax^2y+bxy^2+cxyz = ( a \times x \times x \times y)+( b \times x \times y \times y)+(c \times x \times y \times z)$                                           $= (x\times y)( a \times x+ b \times y+c \times z)$

$= xy(ax+by+cz)$

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

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

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

$ax + bx - ay - by$

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

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

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

$z-7 + 7 xy - xyz$

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

NCERT solutions for class 8 maths chapter 14 factorization-Exercise: 14.2

Question:1(i) Factorise the following expressions

$a ^ 2 + 8a + 16$

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

$p^2 -10 p + 25$

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

$25 m ^2 + 30 m + 9$

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

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

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

$4 x^2 - 8x + 4$

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

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

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

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

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

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

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 :

$4 p^2 - 9 q ^2$

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

$63 a ^2 - 112 b ^ 2$

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

$49 x^2 - 36$

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

$16 x^5 - 144 x ^ 3$

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

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

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

$9 x ^2 y^2 - 16$

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

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

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

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

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

$ax ^2 + bx$

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

$7p^2 + 21 q ^2$

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

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

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

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

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

$( lm + l ) + m + 1$

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

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

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

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

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

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

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

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

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)$

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

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

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

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}$$)$

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

$p^ 2 + 6 p + 8$

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

$q ^ 2 - 10 q + 21$

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

$p^2 + 6 p - 16$

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)$

## Solutions of NCERT for class 8 maths chapter 14 factorization topic 14.3.1 division of a monomial by another monomial

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

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

## CBSE NCERT solutions for class 8 maths chapter 14 factorization-Exercise: 14.3

Question:1(i) Carry out the following divisions

$28 x ^ 4 \div 56 x$

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

This is done using factorization.

Question:1(ii) Carry out the following divisions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$( 10 x - 25) \div 5$

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

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

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

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

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

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

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

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

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

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

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

Question:4(ii) Divide as directed

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$2 x + 3y = 5 xy$

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
$2x + 3y = 2x + 3y$

$(4)x + 2x + 3x = 5x$

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
$x + 2x + 3x = 6x$

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

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
$5y + 2y + y - 7y = y$

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

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
$3x + 2x = 5x$

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

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

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

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
$(2x)^{2}+5x = 4x^2+5x$

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

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

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

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   $(-3)^{2}+5(-3)+4= -2$

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

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

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

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

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$

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

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

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

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

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) = $2a^{2} +ab - 3b^{2}$

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

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)  = $a^{2} + 6a + 8$

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

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)  =  $a^{2} - 6a+ 8$

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

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

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

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

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

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,

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

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

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

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

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

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 $\frac{4x+5}{4x} = 1 + \frac{5}{4x} = \frac{4x+5}{4x}$

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

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  $\frac{7x+5}{5} = \frac{7x}{5} + 1 = \frac{7x+5}{5}$

## NCERT solutions for class 8 maths: Chapter-wise

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

## 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 solutions of NCERT for class 8 maths chapter 14 factorizations you will come across some applications like simplifying expressions and solving quadratic equations. Some important expressions from CBSE NCERT solutions for class 8 maths chapter 14 factorizations are given below which you should remember.

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