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?

NCERT solutions for class 8 maths chapter 14 factorization for all exercises-

Solutions of NCERT 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)

CBSE NCERT solutions for class 8 maths chapter 14 factorization-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(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)

NCERT solutions for class 8 maths chapter 14 factorization-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:10(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)

Solutions of NCERT for class 8 maths chapter 14 factorization 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}

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

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,
         -36y^{3} = -1 \times 2 \times 2 \times 3 \times 3 \times y \times y \times y
           9y^{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 \Rightarrow4yz( 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}                                           usnig  (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 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}
 

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)

Happy Reading!!!

 

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