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
∴


 

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: 

(ii) ∵                         2y = 2 * y = (2 * y)
and                        22y = 2 * 11 X y = (2 X y) X 11
∴ the common factor   = 2 X y
                                   = 2y

                                       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 = [x^2]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)

[In numerator, p is taken out common from each term]

[In numerator, 5p is splitted in to 8p – 3p such that (8p)(–3p) = –24p]

[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)