Mathematics Chapter 14 Factorisation

Here, we have a common factor 3 in all the terms.

∴    9x + 18y + 6xy + 27 = 3[3x + 6y + 2xy +9]
  We find that      3x + 6y = 3(x + 2y) and 2xy + 9 = 1(2xy+9)
i.e. a common factor in both the groups does not exist,
 Thus, 3x + 6y + 2xy + 9 cannot be factorised.
On regrouping the terms, we have
   3x + 6y + 2xy + 9 = 3x + 9 + 2xy + 6y
                            = 3(x + 3) + 2y(x + 3)
                            = (x + 3) (3 + 2y)
Now, 3[3x + 6y + 2xy + 9] = 3[(x + 3)(3 + 2y)]
Thus, 9x + 18y + 6xy + 27 = 3(x+3) (2y+3)

We have    12x =
                   36 =

∴        12x + 36 = [(223)x] + (223) 3
                         =

We have                   22y = 2 × 11 × y
                                33z = 3 × 11 × z

∴                     22y - 33z = [2 × (11) × y] + [3 × (11) × z]
                                      
                                      = (11)[2 × y - 3 × z]
                                      = 11[2y - 3z]

We have            14pq = 2 × 7 × p × q = 2 × (7 × p × q)
                         35pqr = 5 × 7 × p × q × r = 5 × (7 × p × q) × r

∴          14pq + 35 pqr = 2 × (7 × p × q) + 5 × (7 × p × q) × r
                                   = (7 × p × q)[2 + 5 × r]
                                   = 7pq(2+5r)







 

∵                                       12x = 2 × 2 × 3 × x = (2 × 2 × 3) × x
and                                    36 = 2 × 2 × 3 × 3 = (2 × 2 × 3) × 3

∴          the common factor = 2 × 2 × 3
                                          = 12

∵                  2y = 2 × y = (2 × y)
and             22xy = 2 × 11x × y = (2 × y) × 11x

∴   the common factor = 2 × y
                                   = 2y

∵                      14pq = 2 × 7 × p × q = (2 × 7 × p × q)
and                 = 2 × 2 × 7 × p × p × q × q
                                      = (2 × 7 × p × q) × 2 × p × q

∴  the common factor = (2 × 7 × p × q)
                                  = 14pq

∵                                   2x = 1 × 2 × x = 1 × 2 × x
                                   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.

 ∵                      6abc = 2 × 3 × a × b × c = (2× 3 × a × b)× c
                     24 = 2 × 2 × 2 × 3 × a × b × b = (2 × 3 × a × b) × 2 × 2 × b
                 12 = 2 × 2 × 3 × a × a × b = (2 × 3 × a × b) × 2 × a

∴    the common factor = 2 × 3 × a × b
                               = 6ab

∵             16 = 2 × 2 × 2 × 2 × x × x × x = (2 × 2 × x) × 2 × 2 × x × x
               -4 = -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

∵                 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

∵            3x² y³ = 3 × x × x × y × y × y
                       = (x × x × y × y) × 3 × y
             10x³ y² = 2 × 5 × x × x × x × y × y
                         = (x × x × y × y) × 2 × 5 × x
              6x² y²z = 2 × 3 × x × x × y × y × z
                         = (x × x × y × y) × 2 × 3 × z

∴   the common factor = (x × x × y × y)
                             
                              = x² y²

 

∵                     7x = 7 × x = (7) × x
                      42 = 2 × 7 × 3 = (7) × 2 × 3

∴             7x - 42 = 7[(x) - (2 × 3)]
                         = 7[x - 6]

∵                    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]

∵          -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]

  

∵                 20 l²m = 2 × 2 × 5 × l × l × m = (2 × 5 × l × m) × 2 × l
                  30 alm = 2 × 3 × 5 × a × l × m = (2 × 5 × l × m) × 3 × a

∴        20 l²m + 30 alm = (2 × 5 × l × m)[2 × l + 3 × a]
                                 = 10lm[21 + 3a]

∵         5x²y = 5 × x × 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)

∵                   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]
 

∵        -4a² = (-1) × 2 × 2 × a × a = (2 × 2 × a)[(-1) × a]
          4ab  = 2 × 2 × a × b = (2 × 2 × a)[b]
          4ca = 2 × 2 × c × a = (2 × 2 × a)[c]

∴   -4a² + 4ab - 4ca = (2 × 2 × a)[(-1) × a + b - c]
                            = 4a[-a + b - c]

∵          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)

∵                   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]

∴   a x² y + b x y² + c x y z = (x × y) [a × x + b × y + c × z]
                                      = xy[ax + by +cz)

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

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

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

Regrouping the terms, we have
15pq + 15 + 9q + 25p = 15 pq + 9q + 25 p + 15
                              = 3q[5p + 3] + 5[5p +3]
                              = (5p + 3)[3q + 5]

Regrouping the terms, we have
z - 7 + 7xy - xyz = z - 7 - xyz + 7xy
                        = 1[z - 7] - xy[z - 7]
                        = (z - 7) (1 - xy)

x(5y + 3)

3x(x+2)

2(x + 2)

5x(y + 2)

3ab(4a + 5b)

2x²(7x² - 9x + 5)

(3x - 2) (2y - 3)

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)
        

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)

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

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

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

∴      

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

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)

We have
                (l + m)² - 4lm = (l² + 2lm + m²) - 4lm
                                        [ Collecting the like terms 2lm and -4lm]
                                     = l² + (2lm - 4lm) + m²
                                     = l² + 2lm + m²
                                     =
                                     =

∴ 

We have          
                                            

∴          

∵         
                           = (2p - 3q) (2p + 3q) [Using ]

∴      

We have
       
                               =

∵       
                             = (3a + 4b) (3a - 4b)  [Using

∴   

∵ 
                     = (7x - 6) (7x + 6)   [Using

∴      = (7x - 6)(7x + 6)

We have        and

∴    
                             =
                             =
                             =   [Using a²-b² = (a-b) (a+b)]

∴ 

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

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)

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)

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)

∵                  ax² = a × x × x = (x)[ax]
                    bx = b × x = (x)[b]

∴              ax² + bx = x[ax + b]

We have   7p² + 21q²= 7 × p × p + 7 × 3 × q × q
                              = 7[ p × p + 3 × q × q]
                              = 7[p² + 3q²]

Taking out 2x as common from each term, we have
          2x³ + 2xy² + 2xz² = 2x[x² + y² - z²]

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]

We have                     lm + l = l(m + 1)

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

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

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

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]

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]

Regrouping the terms, we have
          6xy - 4y + 6 - 9x = 6xy - 4y - 9x + 6
                                  = 2y[3x - 2] - 3[3x - 2]
                                  = (3x - 2)[2y - 3]

Using              a² - b² = (a - b) (a + b), we have
                     a⁴ - b⁴= (a²)² - (b²)²
                               =
                             
                              = (a + b)(a - b)
                              
                             

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)

∵   x⁴ - (y + z)⁴ = [x²]² - [(y + z)²]²
                       = [(x²) + (y + z)²] [(x²) - (y + z)²]
                                                
                                                     [Using a² - b² = (a+b)(a-b)]
We can factorise [x² - (y + z)²] further as
       x² - (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]

∵   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)

We have    p²+ 6p + 8  = p² + 6p + 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)

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)

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)

(x + 4)(x + 4)

(2y - 3)(2y - 3)

(7y - 6) (7y + 6)

(a - b - c) (a - b + c)

(m-4)(m+4)(m²+16)

(x + 2) (x + 3)

(y - 3) (y - 4)

(x - 6) (x + 2)

3(y +1) (y + 2)

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)

24xy²z³  6yz² =
                      =
                      =

∴         24xy²x³ 6yz² = 4xyz

63a²b⁴c⁶ 7a²b²c³ =
                            =
                           =
                           = 
                           = 9 × a + b² × c³
                          = 9 × 1 × b² × c³ = 9b²c³

∴        63a²b⁴c⁶ by 7a²b²c³ = 9b³c³

We have             28x⁴ 56x =
                                        =
                                        =

∴          28x⁴   56x =

We have                     -36y³ = (-1) x 2 x 2 x 3 x 3 x y x y x y
and                             9y² = 3 x 3 x y x y

∴          -36y³9y² =
                             =

We have            66pq²r³ = 2 × 3 × 11 × p × q × q × r × r × r
and                       11qr² = 11 × q × r × r

∴          66pq²r3 11qr² =
                                   =
                                   =

We have 
                                            =
                                            =
                                            =
                                            =

We have        
                                       =
                                        =
                                        =
                                        =
                                                     

∵           (5x² - 6x) 3x =
                                   =

∴      (5x² - 6x) 3x =

∵ 
                                       =

∴     

∵ 
                                    =
                                    =
                                    =

∴     = 2(x + y + z)

∵  
                                        =
                                        =

∴     

∵   
                                       

∴ 

∵                 10x - 25 = 5(2x - 5)

∴                   
Thus,          (10x - 25) 5 = 2x - 5

∵           10x - 25 = 5(2x - 5)

∴          
Thus, = 5

∵                  6y + 21 = 3(2y + 7)

∴          
                                    = 2 × y × 3 = 6y

∴  10y(6y + 21) 5(2y + 7) = 6y

∵             3z - 24 = 3(z-8)

∴      
                                   =
                                   = xy

∴             

∵           3a - 12 = 3(a - 4)
             5b - 30 = 5(b - 6)
                96 =
and            144 = 2 × 2 × 2 × 2 × 3 × 3

∴          
           
             = 2 × 5 × a × b × c = 10abc
Thus, 96abc (3a - 12)(5b - 36) 144(a - b)(b - 6) = 10abc
 

We have 5(2x + 1)(3x + 5) (2x + 1) =
                                                    =

∴       5(2x + 1)(3x + 5) (2x + 1) = 5(3x +5)

We have 26xy(x + 5) (y - 4) 13x(y - 4) =
                                                        =

∴       26xy(x + 5) (y - 4) 13x(y - 4) = 2y(x + 5)

We have    
             =

∴         52pqr(p + q)(q + r)(r + p) 104pq(q + r)(r + p) =

∵ 20 = 2 × 2 × 5

∴     
                                            = 2 × 2 × (y² + 5y + 3) = 4(y² + 5y + 3)

x(x + 1)(x + 2)(x + 3) x(x + 1)
We have    

∴     x(x+1)(x+2)(x+3) x(x + 1) = (x + 2)(x + 3)

∵          

                              [Splitting 7y in 2y + 5y such that 2y × 5y = 10y²]
                                = y(y + 2) + 5(y + 2)
                                = (y + 2)(y + 5)

∴        

∴     = y + 2

∵      m² -14m - 32 = (m² - 16m + 2m - 32]   [∵  -14 = -16 + 2]

                           = m(m - 16) + 2(m - 16)       [16 × 2 = 32]
                           = (m - 16)(m + 2)

∴ 
Thus,

∵      5p² - 25p + 20 = 5(p² - 5p + 4)               [∵ 1 × 4 = 4]
                            = 5[p(p - 1) - 4(p - 1)]   [-1 + (-4) = -5]
                            = 5[(p - 1)(p - 4)]

∴    
Thus,   (5p² - 25p + 20) (p - 1) = 5(p-4)

∵             z² + 6z -16 = z² + 8z - 2z - 16               ∵ 8 - 2 = 6
                               = z(z + 8) - 2(z + 8)        ∵ 8 × 2 = 16
                               = (z + 8)(z - 2)

∴      
                                     =
Thus        

∵                p² - q² = (p - q) (p + q)
                                           [Using a² - b² = (a + b)(a - b)]

∴      
 Thus,        

∵     9x² - 16y² = (3x)² - (4y)²
                      = (3x - 4y)(3x + 4y)

∴
                              =

∴  

∵             50y² - 98 = 2(25y² - 49)
                            = 2[(5y)² - (7)²]
                            =2[(5y - 7)(5y + 7)]

∴     
                                   =
Thus, 39y³(50y² - 98) / 26y²(5y + 7) = 3y(5y - 7)

3x²

-2x²

8x

3(x + y + z)

7x

4x(x + 3)

(z - 4)

4(x - 5) = 4x - 5
The  given statement is incorrect.
The correct statement is:
4(x - 5) = 4x - 20                                (∴  4 × 5 = 20)

x(3x + 2) = 3x² + 2
It is an incorrect statement.
 
The correct statement is:
x(3x + 2) = 3x² + 2x

2x + 3y = 5xy
It is an incorrect statement
The correct statement is:
2x + 3y = 2x + 3y

x + 2x + 3x = 5x

∵    1 + 2 + 3 = 5 is an incorrect statement.

∴       The correct statement is:
x + 2x + 3x = 6x

5y + 2y + y - 7y = 0
It is an incorrect statement.

∵  5x + 2y + y = 8y       and   8y - 7y = y

∴  The correct statement is
           5y + 2y + y - 7y = y

3x + 2x = 5x²
It is an incorrect statement.
3x + 2x = 5x