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Class 8 Exam  >  Class 8 Notes  >  NCERT Textbooks & Solutions for Class 8  >  NCERT Solutions: Factorisation- 1

NCERT Solutions for Class 8 Maths - Factorisation- 1

Question: Factorise 

(i) 12x + 36           
(ii) 22y – 33z          
(iii) 14pq + 35pqr
Solution:
(i) We have 12x = 3 * 2 * 2 * x = (2 * 2 * 3) * x
∴NCERT Solutions for Class 8 Maths - Factorisation- 1
NCERT Solutions for Class 8 Maths - Factorisation- 1
NCERT Solutions for Class 8 Maths - Factorisation- 1
 

Exercise 14.1

Q1: Find the common factors of the given terms.

(i) 12x, 36      
(ii) 2y, 22xy      
(iii) 14pq, 28p2q2
(iv) 2x, 3x2, 4
(v) 6abc, 24ab2, 12a2b
(vi) 16x3, –4x2, 32x
(vii) 10pq, 20qr, 30rp      
(viii) 3x2y3, 10x3y2, 6x2y2xz
Solution: 
NCERT Solutions for Class 8 Maths - Factorisation- 1
(ii) ∵                         2y = 2 * y = (2 * y)
and                        22y = 2 * 11 X y = (2 X y) X 11
∴ the common factor   = 2 X y
                                   = 2y
NCERT Solutions for Class 8 Maths - Factorisation- 1
                                       3x2 = 1 * 3 * x * x = 1 * 3 * x
and                                  4 = 1 * 2 * 2 = 1 * 2 * 2
∴ the common factor = 1
Note: 1 is a factor of every term.

(v) ∵            6abc = 2 * 3 * a * b * c = (2 * 3 * a * b) * c
                   24 ab2 = 2 * 2 * 2 * 3 * a * b * b = (2 * 3 * a * b) * 2 * 2 * b
                  12 a2b = 2 * 2 * 3 * a * a * b = (2 * 3 * a * b) * 2 * a
∴ the common factor = 2 * 3 * a * b
                                  = 6ab

(vi) ∵                   16x3 = 2 * 2 * 2 * 2 * x * x * x = (2 * x * x
                           –4x2 = –1 * 2 * 2 * x * x = (2 * 2 * x) * (–1) * x
                            32x = 2 * 2 * 2 * 2 * 2 * x = (2 * 2 * x) * 2 * 2 * 2
∴ the common factor = 2 * 2 * x
                                  = 4x

(vii) ∵                10pq  = 2 * 5 * p * q = (2 * 5) * p * q
                           20qr = 2 * 2 * 5 * q * r = (2 * 5) * 2 * q * r
                           30rp = 2 * 3 * 5 * r * p = (2 * 5) * 3 * r * p
∴ the common factor  = 2 * 5
                                  = 10

(viii) ∵                3x2y3 = 3 * x * x * y * y * y
                                    = (x * x * y * y) * 3 * y
                       10x3y2 = 2 * 5 * x * x * x * y * y
                                    = (x * x * y * y) * 2 * 5 * x
                        6x2y2z = 2 * 3 * x * x * y * y * z
                                    = (x * x * y * y) * 2 * 3 * z
∴ the common factor   = (x * x * y * y)
                                    = x2y2

Q2: Factorise the following expressions.


(i) 7x – 42      
(ii) 6p – 12q            
(iii) 7a2  + 14a
(iv) –16z + 20z3 (v) 20 l2m + 30alm
(vi) 5x2y – 15xy2
(vii) 10a2 – 15 b2 + 20c2
(viii) –4a2 + 4ab – 4ca
(ix) x2yz + xy2z + xyz2
(x) ax2y + bxy2 + cxyz
Solution:
(i) ∵           7x = 7 * x = (7) * x
                                  42 = 2 * 7 * 3 = (7) * 2 * 3
∴                        7x – 42 = 7[(x) – (2 * 3)]
                                       = 7[x – 6]

(ii) ∵          6p = 2 * 3 * p = (2) * (3) * p = (2 * 3) * p
               12q = 2 * 2 * 3 * q = (2) * 2 * (3) * q = (2 * 3) * 2 * q
∴     6p – 12q = (2 * 3) [(p) – (2 * q)]
                      = 6[p – 2q]

(iii) ∵           7a2 = 7 * a * a = (7 * a) * a
                   14a = 2 * 7 * a = (7 * a) * 2
∴       7a2 – 14a = (7 * a)[a + 2]
                          = 7a(a + 2)

(iv) ∵       –16z = (–1) * 2 * 2 * 2 * 2 * z = (2 * 2 * z) * (–1) * 2
               20z3 = 2 * 2 * 5 * z * z * z = (2 * 2 * z) * 5 * z * z
∴  –16z + 20z3 = (2 * 2 * z) [(–1) * 4 + 5 * z * z]
                        = 4z[–4 + 5z2]

(v) ∵         20l2m = 2 * 2 * 5 * l * l * m = (2 * 5 * l * m) * 2 * l
                 30alm = 2 * 3 * 5 * a * l * m = (2 * 5 * l * m) * 3 * a
∴ 20l2m + 30alm = (2 * 5 * l * m)[ 2 * l + 3 * a]
                            = 10lm[2l + 3a]

(vi) ∵           5x2y = 5 * x * x * y = (5 * x * y)[x]
                 15xy2 = 5 * 3 * x * y * y = (5 * x * y)[3 * y]
∴    5x2y – 15xy2 = (5 * x * y)[x – 3 * y]
                            = 5xy(x – 3y)

(vii) ∵                10a2 = 2 * 5 * a * a = (5)[2 * a * a]
                         15b2 = 3 * 5 * b * b = (5)[3 * b * b]
                          20c2 = 2 * 2 * 5 c * c = (5)[2 * 2 * c * c]
∴ 10a2 – 15b2 + 20c2 = (5)[2 * a * a + 3b * b + 2 * 2 * c * c]
                                  = 5[2a2 + 3b2 + 4c2]

(viii) ∵              –4a2 = (–1) * 2 * 2 * a * a = (2 * 2 * a)[(–1) * a]
                         4ab = 2 * 2 * a * b = (2 * 2 * a)[b]
                       –4ca = (–1) * 2 * 2 * c * a = (2 * 2 * a)[(–1) * c]
∴ –4a2 + 4ab – 4ca = (2 * 2 * a)[(–1) * a + b – c]
                                = 4a[–a + b –c]

(ix) ∵                  x2yz = x * x * y * z = (xyz)[x]
                          xy2z = x * y * y * z = (xyz)[y]
                          xyz2 = x * y * z * z = (xyz)[z]
∴ x2yz + xy2z + xyz2 = (xyz)[x + y + z]
                                  = xyz(x + y + z)

(x) ∵                  ax2y = a * x * x * y = (x * y)[a * x]
                         bxy2 = b * x * y * y = (x * y)[b * y]
                         cxyz = c * x * y * z = (x * y)[c * z]
∴ ax2y + bxy2 + cxyz = (x * y)[a * x + b * y + c * z]
                                  = xy(ax + by + cz)


Q3: Factorise:

(i) x2 + xy + 8x + 8y
(ii) 15xy – 6x + 5y – 2
(iii) ax + bx – ay – by
(iv) 15pq + 15 + 9q + 25p
(v) z – 7 + 7xy – xyz
Solution: 

(i) x2 + xy + 8x + 8y = x[x + y] + 8[x + y]

                                = (x + y)(x + 8)


(ii) 15xy – 6x + 5y – 2 = 3x[5y – 2] + 1[5y – 2]

                                   = (5y – 2)(3x +1)


(iii) ax + bx – ay – by = x[a + b] + (–y)[a + b]

                                  = (x – y)[a + b]


(iv) Regrouping the terms, we have

15pq + 15 + 9q + 25p = 15pq + 9q + 25 p + 15

                                   = 3q[5p + 3] + 5[5p + 3]

                                   = (5p + 3)[3q + 5]


(v) Regrouping the terms, we have

z – 7 + 7xy – xyz = z – 7 – xyz + 7xy

                            = 1[z – 7] – xy[z – 7]

                            = (z – 7)(1 – xy)



Factorisation Using Identities

We know that

         (a + b)2 = a2 + 2ab + b2

         (a – b)2 = a2– 2ab + b2

(a + b)(a – b) = a2 – b2

Using these identities, we can say that

a2 + 2ab + b2 = (a + b)2

                      = (a + b)(a + b)

a2 – 2ab + b2 = (a – b)2

                      = (a – b)(a – b)

          a2 – b2 = (a + b)(a – b)

Note: For factorising an expression of the type x2 + px + q [where p = (a + b) and q = (ab)], we have

x2 + px + q = x2 + (a + b)x + ab

                  = x2 + ax + bx + ab

                  = x(x + a) + b(x + a)

                  = (x + a)(x + b)


Exercise

 14.2


Q1: Factorise the following expressions.

(i) a2 + 8a + 16

(ii) P2 – 10p + 25

(iii) 25m2 + 30 m + 9

(iv) 49y2 + 84yz + 36z2

(v) 4x2 – 8x + 4

(vi) 121b2 – 88bc + 16c2

(vii) (l + m)2– 4lm

(viii) a4 + 2a2b2 + b4

[Hint: Expand (l + m)2 first.]

Solution:

 (i) We have   a2 + 8a + 16 = (a)2 + 2(a)(4) + (4)2

                                          = (a + 4)2 = (a + 4)(a + 4)

∴                   a2 + 8a + 16 = (a + 4)2 = (a + 4)(a + 4)


(ii) We have p2 – 10p + 25 = P2 – 2(p)(5) + (5)2

                                           = (p – 5)2 = (p – 5)(p – 5)

∴                  P2 – 10p + 25 = (p – 5)2 = (p – 5)(p – 5)


(iii) We have     25m2 + 30m + 9 = (5m)2 + 2(5m)(3) + (3)2

                                                    = (5m + 3)2 = (5m + 3)(5m + 3)

∴                       25m2 + 30m + 9 = (5m + 3)2 = (5m + 3)(5m + 3)


(iv) We have 49y2 + 84yz + 36z = (7y)2 + 2(7y)(6z) + (6z)2

                                                   = (7y + 6z)2

                                                              = (7y + 6z)(7y + 6z)

 ∴                   49y2 + 84yx + 36 z2 = (7y + 6z)2 = (7y + 6z)(7y + 6z)


(v) We have 4x2 – 8x + 4 = (2x)2 – 2(2x)(2) + (2)2

                                        = (2x – 2)2 = (2x – 2)(2x – 2)

                                        = 2(x – 1)2(x – 1)

                                        = 4(x – 1)(x – 1)

∴                 4x2 – 8x + 4 = 4(x – 1)(x – 1) = (4(x – 1)2


(vi) We have 121b2 – 88bc + 16c2 = (11b)2 – 2(11b)(4c) + (4c)2

                                                       = (11b – 4c)2 = (11b – 4c)(11b – 4c)

∴                  121b2 – 88bc + 16c2 = (11b – 4c)2 = (11b – 4c)(11b – 4c)


(vii) We have   (l + m)2 – 4lm = (l2 + 2lm + m2) – 4lm

                                              [Collecting the like terms 2lm and –4lm]

                                             = l2 + (2lm – 4lm) + m2

                                             = l2 + 2lm + m2

                                             = (l)2 – 2(l)(m) + (m)2

                                            = (l – m)2 = (l – m)(l – m)

∴                  (l + m)2 – 4lm = (l – m)2 = (l – m)(l – m)


(viii) We have a4 + 2a2b2 + b4 = (a2)2 + 2(a2)(b2) + (b2)2

                                                = (a2 + b2)2 = (a2 + b2)(a2 + b2)

∴                   a4 + 2a2b2 + b4 = (a2 + b2)2 = (a2 + b2)(a2 + b2)



Q2: Factorise:

(i) 4p2 – 9q2

(ii) 63a2 – 112b2

(iii) 49x2 – 36

(iv) 16x5 – 144x3

(v) (l + m)2 – (l – m)2

(vi) 9x2y2 – 16

(vii) (x2 – 2xy + y2) – z2

(viii) 25a2 – 4b2 + 28bc – 49c2

Solution:

 (i) ∵ 4p2 – 9q2 = (2p)2 – (3q)2

                       = (2p – 3q)(2p + 3q)        [Using a2 – b2 = (a + b)(a – b)]

∴    4p2 – 9q2 = (2p – 3q)(2p + 3q)


(ii) We have

       63a2 – 112b2 = 7 * 9a2 – 7 * 16b2

                            = 7[9a2 – 16b2)

∵       9a2 – 16b2 = (3a)2 – (4b)2

                           = (3a + 4b)(3a – 4b)          [Using a2 – b2 = (a + b)(a – b)]

∴ 63a2 – 112b2 = 7[(3a + 4b)(3a – 4b)]



(iii) ∵  49 x2 – 36 = (7x)2 – (6)2

                           = (7x – 6)(7x + 6)              [Using a2 – b2 = (a – b)(a + b)]

∴        49x2 – 36 = (7x – 6)(7x + 6)


(iv) We have 16x5 = 16x3 * x2 and 144x3 = 16x3 * 9

∴     16x5 – 144x3 = (16x3) * x2 – (16x3) * 9

                             = 16x3[x2 – 9]

                             = 16x3[(x)2 – (3)2]

                             = 16x3[(x + 3)(x – 3)]          [(Using a2 – b2 = (a – b)(a + b)]

∴     16x5 – 144x3 = 16x3(x + 3)(x – 3)


(v) Using the identity a2 – b2 = (a + b)(a – b), we have

                  (l + m)2 – (l – m)2 = [(l + m) + (l – m)][(l + m) – (l – m)]

                                               = [l + m + l – m][l + m – l + m]

                                               = (2l)(2m) = 2 * 2(l * m)

                                               = 4lm


(vi) We have    9x2y2 – 16 = (3xy)2 – (4)2

                                          = (3xy + 4)(3xy – 4)        [Using a2 – b2 = (a + b)(a – b)]

∴                      9x2y2 – 16 = (3xy + 4)(3xy – 4)


(vii) We have x2 – 2xy + y2= (x – y)2

∴         (x2 – 2xy + y2) – z2 = (x – y)2 – (z)2

                                          = [(x – y) + z][(x – y) – z]

                                                 [Using a2 – b2 = (a + b)(a – b)]

                                          = (x – y + z)(x – y – z)

∴       (x2 – 2xy + y2) – z2 = (x – y + z)(x – y – z)


(viii) We have       –4b2 + 28bc – 49c2 = (–1)[4b2 – 28bc + 49c2]

                                                             = –1[(2b)2 – 2(2b)(7c) + (7c)2]

                                                             = –[2b – 7c]2

∴                25a2 – 4b2 + 28bc – 49c2  = 25a2 – (2b – 7c)2

Now, using                              a2 – b2 = (a + b)(a – b), we have

                                [5a]2 – [2b – 7c]2 = (5a + 2b – 7c)(5a – 2b + 7c)

∴               25a2 – 4a2 + 28bc – 49c2 = (5a + 2b – 7c)(5a – 2b + 7c)



Q3: Factorise the expressions.

(i) ax2 + bx
(ii) 7p2 + 21q2
(iii) 2x3 + 2xy2 + 2xz2
(iv) am2 + bm2 + bn2 + an2
(v) (lm + l) + m + 1
(vi) y(y + z) + 9(y + z)
(vii) 5y2 – 20y – 8z + 2yz
(viii) 10ab + 4a + 5b + 2
(ix) 6xy – 4y + 6 – 9x
Solution:
(i) ∵       ax2 = a * x * x = (x)[ax]

               bx = b * x = (x)[b]

∴   ax2 + bx = x[ax + b]


(ii) We have 7p2 + 21q2 = 7 * p * p + 7 * 3 * q * q

= 7[p * p + 3 * q * q]

= 7[p2 + 3q2]


(iii) Taking out 2x as common from each term, we have

      2x3 + 2xy2 + 2xz2 = 2x[x2 + y2 + z2]


(iv) We can take out m2 as common from the first two terms and n2 as common from the last two terms,

∴ am2 + bm2 + bn2 + an2 = m2(a + b) + n2(b + a)

                                         = m2(a + b) + n2(a + b)

                                         = (a + b)[m2 + n2]


(v) We have     lm + l = l(m + 1)

∴    (lm + l) + (m + 1) = l(m + 1) + (m + 1)

                                  = (m + 1)[l + 1]


(vi) We have (y + z) as a common factor to both terms.

∴     y(y + z) + 9(y + z) = (y + z)(y + 9)


(vii) We have 5y2 – 20y = 5y(y – 4)

and                –8z + 2yz = 2z(–4 + y)

                                      = 2z(y – 4)

∴ 5y2 – 20y – 8z + 2yz = 5y(y – 4) + 2z(y – 4)

                                     = (y – 4)[5y + 2z]


(viii) We have,    10ab + 4a = 2a(5b + 2)

and                         (5b + 2) = 1(5b + 2)

∴           10ab + 4a + 5b + 2 = 2a(5b + 2) + 1(5b + 2)

                                            = (5b + 2)[2a + 1]


(ix) Regrouping the terms, we have

6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6

                            = 2y[3x – 2] – 3[3x – 2]

                            = (3x – 2)[2y – 3]


Q4: Factorise:

(i) a4 – b4
(ii) p4 – 81
(iii) x4 – (y + z)4
(iv) x4 – (x – z)4
(v) a4 – 2a2b2 + b4
Solution:
(i) Using a2 – b2 = (a – b)(a + b), we have

              a4 – b4 = (a2)2 – (b2)2

                          = (a2 + b2)(a2 – b2)

                          = (a2 + b2)[(a + b)(a – b)]

                          = (a2 + b2)(a + b)(a – b)


(ii) We have    P4 – 81 = (p2)2 – (92)2

Now using      a2 – b2 = (a + b)(a – b), we have

                (p2)2 – (9)2 = (p2 + 9)(p2 – 9)

We can factorise p2 – 9 further as

                       p2 – 9 = (p)2 – (3)2

                                 = (p + 3)(p – 3)

∴                  p4 – 81 = (p + 3)(p – 3)(p2 + 9)


(iii)  ∵ x4 – (y + z)4 = [x2]2 – [(y + z)2]2

                              = [(x)2 + (y + z)2][(x2) – (y + z)2]

                                                                 [Using a2 – b2 = (a + b)(a – b)]

We can factorise [x2 – (y + z)2] further as

       (x)2 – (y + z)2 = [(x) + (y + z)][(x) – (y + z)]

                             = (x + y + z)(x – y – z)

∴       x4 – (y + z)4 = (x + y + z)(x – y – z)[x2 + (y + z)2]


(iv) We have x4 – (x – z)4 = [x2]2 – [(x – z)2]2

                                        = [x2 + (x – z)2][x2 – (x – z)2]

Now, factorising x2 – (x – z)2 further, we have

                   x2 – (x – z)2 = [x + (x – z)][x – (x – z)]

                                       = (x + x – z)(x – x + z) = (2x – z)(z)

∴                x4 – (y – z)4 = z(2x – z)[x2 + (x – z)2]

                                      = z(2x – z)[x2 + (x2 – 2xz + z2)]

                                      = z(2x – z)[x2 + x2 – 2xz + z2]

                                      = z(2x – z)(2x2 – 2xz + z2)


(v) ∵   a4 – 2a2b2 + b4 = (a2)2 – 2(a2)(b2) + (b2)2

                                    = (a2 – b2)2

                                    = [(a2 – b2)(a2 + b2)]

                                    = [(a – b)(a + b)(a2 + b2)]

∴         a4 – 2a2b2 + b4 = (a – b)(a + b)(a2 + b2)



Q5: Factorise the following expressions.


(i) p2 + 6p + 8
(ii) q2 – 10q + 21
(iii) p2 + 6p – 16
Solution:
(i) We have     p2 + 6p + 8 = P2 + 6p + 9 – 1                [∵ 8 = 9 – 1]

                                          = [(p)2 + 2(p)(3) + 32] – 1

                                          = (p + 3)2 – 12      [Using a2 + 2ab + b2 = (a + b)2]

                                          = [(p + 3) + 1][(p + 3) – 1]

                                          = (p + 4)(p + 2)

∴                     p2 + 6p + 8 = (p + 4)(p + 2)


(ii) We have     q2 – 10q + 21 = q2 – 10q + 25 – 4       [∵ 21 = 25 – 4]

                                              = [(q)2 – 2(q)(5) + (5)2] – (2)2

                                              = [q – 5]2 – [2]2

                                              = [(q – 5) + 2][(q – 5) – 2]

                                                                         [Using a2 – b2 = (a + b) (a – b)]

                                              = (q – 3)(q – 7)


(iii) We have      p2 + 6p – 16 = p2 + 6p + 9 – 25          [∵ –16 = 9 – 25]

                                               = [(p)2 + 2(p)(3) + (3)2] – (5)2

                                               = [p + 3]2 – [5]2 [Using a2 + 2ab + b2 = (a + b)2]

                                               = [p + 3] + 5][(p + 3) – 5]

                                                                          [Using a2 – b2 = (a + b)(a – b)

                                               = (p + 8)(p – 2)



Division of Algebraic Expressions

For division of algebraic expression by another expression:

  1. We write them in the form of a fraction, such that the divisor is the denominator.
  2. Factorise the numerator as well as the denominator.
  3. Cancel the factors common to both the numerator and denominator.


Example 1. Divide 33(p4 + 5p3 – 24p2) by 11p(p + 8).
Solution: We have 33(p4 + 5p3 – 24p2) ÷ 11p(p + 8)
NCERT Solutions for Class 8 Maths - Factorisation- 1
[In numerator, p is taken out common from each term]
NCERT Solutions for Class 8 Maths - Factorisation- 1
[In numerator, 5p is splitted in to 8p – 3p such that (8p)(–3p) = –24p]
NCERT Solutions for Class 8 Maths - Factorisation- 1
[Cancelling the factors 11p and x + 8 common to both the numerator and denominator]
Thus, 33(p4 + 5p3 – 24p2) ÷ 11p(p + 8)
                                          = 3p(p – 3)

The document NCERT Solutions for Class 8 Maths - Factorisation- 1 is a part of the Class 8 Course NCERT Textbooks & Solutions for Class 8.
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FAQs on NCERT Solutions for Class 8 Maths - Factorisation- 1

1. What is factorisation and why is it important in mathematics?
Ans. Factorisation is the process of breaking down a number or an algebraic expression into its factors. It is important in mathematics as it helps in simplifying expressions, solving equations, finding common factors, and understanding the properties of numbers.
2. How do you factorise a quadratic expression?
Ans. To factorise a quadratic expression, we need to find two binomials whose product is equal to the given expression. We can use methods like the trial and error method, completing the square method, or the quadratic formula to factorise a quadratic expression.
3. Can we factorise any number or expression?
Ans. Not every number or expression can be factorised. Prime numbers, for example, cannot be factorised further as they only have two factors, 1 and the number itself. Similarly, some algebraic expressions may not have any common factors or may not be factorisable using simple methods.
4. What are the applications of factorisation in real life?
Ans. Factorisation has various applications in real life, such as simplifying fractions, finding the prime factors of a number, solving problems related to area and perimeter, and understanding patterns in mathematics and science.
5. How can factorisation be used to solve equations?
Ans. Factorisation can be used to solve equations by setting the equation equal to zero and factoring it into simpler expressions. By using the zero product property, we can then find the values of the variable that satisfy the equation. This method is particularly useful in solving quadratic equations.
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