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Bank Exams Exam  >  Bank Exams Notes  >  NCERT Mathematics for Competitive Exams  >  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
∴Question: Factorise 
Question: Factorise 
Question: Factorise 
 

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: 
Q1: Find the common factors of the given terms.
(ii) ∵                         2y = 2 * y = (2 * y)
and                        22y = 2 * 11 X y = (2 X y) X 11
∴ the common factor   = 2 X y
                                   = 2y
Q1: Find the common factors of the given terms.
                                       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)
Division of Algebraic Expressions
[In numerator, p is taken out common from each term]
Division of Algebraic Expressions
[In numerator, 5p is splitted in to 8p - 3p such that (8p)(-3p) = -24p]
Division of Algebraic Expressions
[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)

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FAQs on NCERT Solutions: Factorisation- 1

1. What are the main methods of factorisation I need to know for bank exams?
Ans. Factorisation involves breaking expressions into simpler factors using several key methods: finding common factors, grouping terms, using algebraic identities like difference of squares, and trinomial factorisation. For competitive exams, mastering these techniques speeds up problem-solving significantly. Refer to flashcards and mind maps on EduRev to compare methods side-by-side and identify which approach suits each expression type.
2. How do I factor out the highest common factor from polynomial expressions?
Ans. The highest common factor (HCF) is the largest term dividing all components of an expression. Identify the GCD of coefficients and the lowest power of each variable present in every term, then divide the entire expression by this factor. For example, in 6x²y + 9xy², the HCF is 3xy. Practice with worksheets to strengthen recognition speed during timed bank exams.
3. When should I use factorisation by grouping instead of other methods?
Ans. Factorisation by grouping works best for four-term polynomials where no single common factor exists across all terms. Group pairs of terms sharing factors, then extract the common binomial. This technique proves invaluable in algebraic simplification questions appearing in bank competitive exams. Visual worksheets demonstrating grouped-term patterns help clarify when to apply this specific factorisation strategy.
4. Why does using algebraic identities like a²-b² = (a+b)(a-b) matter in exam questions?
Ans. Algebraic identities provide instant factorisation without lengthy calculations, saving critical exam time. Recognising patterns like difference of squares, perfect squares, and sum/difference of cubes allows direct factoring of complex expressions. These identity-based factorisation techniques frequently appear in bank exams because they test conceptual understanding rather than mechanical work.
5. What's the difference between factorising trinomials with coefficient 1 versus trinomials with larger leading coefficients?
Ans. Trinomials with leading coefficient 1 (x² + bx + c) require finding two numbers multiplying to c and adding to b. When the leading coefficient exceeds 1 (ax² + bx + c), use the AC method: multiply a×c, find factor pairs, split the middle term, then group. Bank exam questions test both forms; MCQ tests on EduRev clarify distinctions through varied examples.
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