chapter -14 Factorisation

EXERCISE- 14.1

Find the common factors of the given terms from question 1 to 8

1.  12x, 36

Explanation:

Factors of 12x= 2 × 2 × 3 × x

Factors of 36= 2 × 2 × 3 × 3

Common factors of 12x and 36 are 2, 2 and 3 

And 2 x 2 × 3= 12


2. 2y, 22xy

Explanation:

 Factors of 2y= 2 × y

 Factors of 22×y= 2 × 11 × x×  y

 The common factors are 2 and y

 And 2 × y= 2y


3. 14 pq, 28p2q2

Explanation:

Factors of 14pq= 2 × 7 × p × q

Factors of 28p2q2= 2 × 2 × 7 × p × p × q × q

Common factors of 14pq and 28p2q2= 2, 7, p, q

 And 2 × 7 × p × q= 14pq


4. 2x, 3x2, 4

Explanation:

Factors of 2x = 2×x

Factors of 3x2= 3×x×x

Factors of 4 = 2×2

Common factors of 2x, 3x2 and 4 is 1.


5. 6 abc, 24ab2, 12a2b

Explanation:

Factors of 6abc = 2 × 3 × a × b × c

Factors of 24ab2 = 2 × 2 × 2 × 3 × a × b × b

Factors of 12 a2 b = 2 × 2 × 3 × a × a × b

So, the common factors are 2, 3, a, b

and, 2×3×a×b = 6ab


6. 16 x3, – 4x2 , 32 x

Explanation:

Factors of 16 x3 = 2×2×2×2×x×x×x

Factors of – 4x2 = -1×2×2×x×x

Factors of  32x = 2×2×2×2×2×x

Common factors are 2,2 and x

and, 2×2×x = 4x


7.  10 pq, 20qr, 30 rp

Explanation:

 Factors of 10 pq = 2×5×p×q

Factors of  20qr = 2×2×5×q×r

Factors of 30rp= 2×3×5×r×p

Common factors of 10 pq, 20qr and 30rp are 2, 5

and, 2×5 = 10


8. 3x2y3 , 10x3y2 , 6x2y2z

Explanation:

Factors of  3x2y3 = 3×x×x×y×y×y

Factors of  10x3 y2 = 2×5×x×x×x×y×y

Factors of  6x2y2z = 3×2×x×x×y×y×z

Common factors of 3x2y3, 10x3y2 and 6x2y2z are x2, y2

and, x2×y2 = x2y2


Factorise the following expressions from question 9 to 18

9.7x–42

Explanation:

Factors of 7x= 7 × x

Factors of 42= 2 × 3 × 7

Common factor is 7

Therefore 7x-42= (7 × x)- ( 2 × 3 × 7)

                              =7 ( x - 6 )


10. 6p–12q

Explanation:

Factors of 6p= 2 × 3 × p

Factors of 12q= 2 × 2 × 3 × q

Common factor is 2, 3 and 2 × 3= 6

Therefore, 6p – 12q= (2 × 3 × p) – ( 2 ×  2 ×  3 ×  q)

                                      = 6( p – 2q)


11.  7a2+ 14a

Explanation:

Factors of 7a2= 7 × a × a

Factors of 14a= 2 × 7 × a

Common factors are 7 and a

 And 7 × a= 7a

 Therefore, 7a2 + 14a=  (7 × a × a) + (2 × 7 × a )

                                        = 7a ( a + 2)


12.  -16z+20 z3

Explanation:

Factors of -16z= -2 × 2 × 2 × 2 × z

Factors of 20z3= 2 × 2 × 5 × z × z × z

Common factors are 2, 2 and z

Therefore, -16z + 20z3= (-2 × 2 × 2 × 2 × z ) + ( 2 × 2 × 5 × z × z × z)

                                         = 4z ( -4 + 5z2)


13.  20l2m+30alm

Explanation:

Factors of 20l2m= 2 × 2 × 5 × l × l × m

Factors of 30alm= 2 × 3 × 5 × a × l × m

Common factors are 2, 5, l, m

Therefore, 20l2m+30alm= ( 2 × 2 × 5 × l × l × m) + ( 2 × 3 × 5 × a × l × m)

                                        = 10lm ( 2l + 3a)


14. 5x2y-15xy2

Explanation:

 5x2y-15xy2

 Factors of 5x2y= 5 × x × x × y

 Factors of 15xy2= 3 × 5 × x × y × y

 Common factors are 5, x, y

 Therefore, 5x2y-15xy2= ( 5 × x × x × y) – ( 3 × 5 × x × y × y)

                                         = 5xy (x – 3y)


15. 10a2-15b2+20c2

Explanation:

10a2-15b2+20c2

Factors of 10a2 = 2×5×a×a

Factors of – 15b2 = -1×3×5×b×b

Factors of 20c2 = 2×2×5×c×c

Common factor is 5

Therefore, 10a2-15b2+20c2= ( 2×5×a×a ) – ( 2×5×a×a ) + ( 2×2×5×c×c)

                                           =10a2-15b2+20c2

                                                               = 5(2a2-3b2+4c2 )


16.   -4a2+4ab–4 ca

Explanation:

Factors of – 4a2 = -1×2×2×a×a

Factors of 4ab = 2×2×a×b

Factors of – 4ca = -1×2×2×c×a

Common factor are 2, 2, a i.e. 4a

Therefore, – 4a2+4ab-4ca= ( -1×2×2×a×a ) + ( 2×2×a×b ) – ( -1×2×2×c×a )

                                         =– 4a2+4 ab-4 ca

                                          = 4a(-a+b-c)


17.  x2yz+xy2z +xyz2

Explanation:

Factors of x2yz = x×x×y×z

Factors of xy2z = x×y×y×z

Factors of xyz2 = x×y×z×z

Common factor are x, y, z i.e. xyz

Therefore, x2yz+xy2z+xyz2 = xyz(x+y+z)


18. ax2y+bxy2+cxyz

Explanation:

Factors of ax2y = a×x×x×y

Factors of bxy2 = b×x×y×y

Factors of cxyz = c×x×y×z

Common factors are xy

Now, ax2y+bxy2+cxyz = xy(ax+by+cz)


Factorise from question 19 to 23

19. x2+xy+8x+8y


Explanation:

 x2+xy+8x+8y 

 x × x + x × y + 8 × x + 8 × y

 x ( x + y) + 8 ( x + y)

  (x + 8) ( x + y)


20. 15xy–6x+5y–2

Explanation:

 15xy–6x+5y–2

3 × 5 × x × y – 3 × 2 × x + 5 × y – 2

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

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


21.  ax+bx–ay–by

Explanation:

ax+bx–ay–by

a × x + b × x – a × y – b × y

x ( a + b ) -y ( a + b)

( x – y) ( a + b)


22. 15pq+15+9q+25p

Explanation:

15pq+15+9q+25p

3 × 5 × p × q + 3 × 5 + 3 × 3 × q + 5 × 5 × q

3q( 5p + 3) + 5( 5p + 3) 

 ( 3q+5) ( 5p + 3)


23.  z–7+7xy–xyz

Explanation:

z–7+7xy–xyz

z – 7 + 7 × x × y – x × y × z

z( 1 + xy) +7 ( 1 – xy)

(z + 7) ( 1 + xy)


Exercise 14.2 

Factorise the following expressions from queation 24 to 31

24.  a2+8a+16

Explanation:

a2+8a+16

= a2+2×4×a+42

= (a+4)2                                 ( Since, (x + y)2= x2+ 2ab + b2)


25. p2–10p+25

Explanation:

 p2–10p+25

= p2-2×5×p+52

= (p-5)2                                 ( Since, (x + y)2= x2+ 2ab + b2)


26. 25m2+30m+9

Explanation:

25m2+30m+9

= (5m)2+2×5m×3+32

= (5m+3)2                    ( Since, (x + y)2= x2+ 2ab + b2)


27.  49y2+84yz+36z2

Explanation:

49y2+84yz+36z2

=(7y)2+2×7y×6z+(6z)2

= (7y+6z)2                    ( Since, (x + y)2= x2+ 2ab + b2)


28. 4x2–8x+4

Explanation:

 4x2–8x+4

= (2x)2-2×4x+22

= (2x-2)2                        ( Since, (x - y)2= x2- 2ab + b2)


29.  121b2–88bc+16c2

Explanation:

121b2-88bc+16c2

= (11b)2-2×11b×4c+(4c)2

= (11b-4c)2                    ( Since, (x - y)2= x2- 2ab + b2)


30.  (l+m)2–4lm (Hint: Expand (l+m)2 first)

Explanation:

(l+m)2- 4lm 

Expand (l+m)2 using the identity (x+y)2 = x2+2xy+y2

(l+m)2-4lm 

= l2+m2+2lm-4lm

= l2+m2-2lm

= (l-m)2                            ( Since, (x - y)2= x2- 2ab + b2)


31. a4+2a2b2+b4

Explanation:

a4+2a2b2+b4

= (a2)2+2×a2×b2+(b2)2

= (a2+b2)2                  ( Since, (x + y)2= x2+ 2ab + b2)


 Factorise from question 32 to 39

32.4p2–9q2


Explanation:

 4p2–9q2

= (2p)2- (3q)2

= (2p-3q)(2p+3q)

Since, x2-y2 = (x+y)(x-y)


33. 63a2–112b2

Explanation:

 63a2–112b2

= 7(9a2 –16b2)

= 7((3a)2–(4b)2)

= 7(3a+4b)(3a-4b)

Since, x2-y2 = (x+y)(x-y)


34. 49x2–36

Explanation:

 49x2–36

= (7x)2 -62

= (7x+6)(7x–6)

Since, x2-y2 = (x+y)(x-y)


35.  16x5–144x3 differ

Explanation:

 16x5–144x3

= 16x3(x2–9)

= 16x3(x2–9)

= 16x3(x–3)(x+3)

We know, x2-y2 = (x+y)(x-y)


36.  (l+m)2-(l-m) 2

Explanation:

(l+m) 2-(l-m) 2

[(l+m)+(l-m)] [(l+m) – ( l – m )]

Since, x2-y2 = (x+y)(x-y)

=( l+m+l–m) (l+m–l+m)

= (2m)(2l)

= 4 ml


37.  9x2y2–16

Explanation:

 9x2y2–16

= (3xy)2-42

= (3xy–4)(3xy+4)

Since, x2-y2 = (x+y)(x-y)


38. (x2–2xy+y2)–z2

Explanation:

(x2–2xy+y2)–z2

= (x–y)2–z2

Since, (x-y)2 = x2-2xy+y2

= {(x–y)–z}{(x–y)+z}

= (x–y–z)(x–y+z)

Since, x2-y2 = (x+y)(x-y)


39.  25a2–4b2+28bc–49c2

Explanation:

 25a2–4b2+28bc–49c2

= 25a2–(4b2-28bc+49c2 )

= (5a)2-{(2b)2-2(2b)(7c)+(7c)2}

= (5a)2-(2b-7c)2

= (5a+2b-7c)(5a-2b+7c)                         Since, x2-y2 = (x+y)(x-y)


Factorise the expressions from question 40 to 48

40.ax2+bx

Explanation:

 ax2+bx

Taking x as common, we get

 = x(ax+b)


41. 7p2+21q2

Explanation:

7p2+21q2 

Taking 7 as common, we get

= 7(p2+3q2)


42. 2x3+2xy2+2xz2

Explanation:

 2x3+2xy2+2xz2 

Taking 2x as common, we get

= 2x(x2+y2+z2)


43. am2+bm2+bn2+an2

Explanation:

am2+bm2+bn2+an2

Taking m2 and n2 as common, we get

 = m2(a+b)+n2(a+b)

 = (a+b)(m2+n2)


44. (lm+l)+m+1

Explanation:

 (lm+l)+m+1 

= lm+m+l+1 

= m(l+1)+(l+1) 

= (m+1)(l+1)

(vi) y(y+z)+9(y+z) 

= (y+9)(y+z)


45. y(y+z)+9(y+z)

Explanation:

 y(y+z)+9(y+z) 

= (y+9)(y+z)


46. 5y2–20y–8z+2yz

Explanation:

5y2–20y–8z+2yz 

Taking 5y and 2z as common, we get

= 5y(y–4)+2z(y–4)

= (y–4)(5y+2z)


47.  10ab+4a+5b+2

Explanation:

 10ab+4a+5b+2 

= 5b(2a+1)+2(2a+1) 

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


48. 6xy–4y+6–9x

Explanation:

 6xy–4y+6–9x 

= 6xy–9x–4y+6 

= 3x(2y–3)–2(2y–3) 

= (2y–3)(3x–2)


Factorise from 49 to 53

49.a4–b4


Explanation:

a4–b4

= (a2)2-(b2)2

= (a2-b2) (a2+b2)                             [Since, a2-b2= (a+b) (a-b)]

= (a – b)(a + b)(a2+b2)


50.  p4–81

Explanation:

 p4–81

= (p2)2-(9)2

= (p2-9)(p2+9)                                  [Since, a2-b2= (a+b) (a-b)]     

= (p2-32)(p2+9)

=(p-3)(p+3)(p2+9)


51.  x4–(y+z) 4

Explanation:

x4–(y+z) 4 = (x2)2-[(y+z)2]2

= {x2-(y+z)2}{ x2+(y+z)2}

= {(x –(y+z)(x+(y+z)}{x2+(y+z)2}

= (x–y–z)(x+y+z) {x2+(y+z)2}


52.  x4–(x–z) 4

Explanation:

 x4–(x–z) 4 = (x2)2-{(x-z)2}2

= {x2-(x-z)2}{x2+(x-z)2}

= { x-(x-z)}{x+(x-z)} {x2+(x-z)2}

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

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


53. a4–2a2b2+b4

Explanation:

 a4–2a2b2+b4 = (a2)2-2a2b2+(b2)2

= (a2-b2)2

= ((a–b)(a+b))2

= (a – b)2 (a + b)2


Factorise the following expressions from 54 to 56

54. p2+6p+8

Explanation:

p2+6p+8

We observe, 8 = 4×2 and 4+2 = 6

We can write as, p2+2p+4p+8

Taking Common terms, we get

= p2+2p+4p+8 

= p(p+2)+4(p+2)

= (p+2)(p+4)

Therefore, p2+6p+8 = (p+2)(p+4)


55. q2–10q+21

Explanation:

 q2–10q+21

We observe, 21 = -7×-3 and -7+(-3) = -10

q2–10q+21 

= q2–3q-7q+21

Taking common we get

= q(q–3)–7(q–3)

= (q–7)(q–3)

Therefore, q2–10q+21 = (q–7)(q–3)


56. p2+6p–16

Explanation:

p2+6p–16

We observe, -16 = -2×8 and 8+(-2) = 6

p2+6p–16 

= p2–2p+8p–16

Taking common we get

= p(p–2)+8(p–2)

= (p+8)(p–2)

Therefore, p2+6p–16 = (p+8)(p–2)


EXERCISE- 14.3  

Carry out the following divisions from question 57 to 61

57. 28x4 ÷ 56x

Explanation:

 Factors of 28x4 = 2×2×7×x×x×x×x

   Factors of 56x = 2×2×2×7×x

  28x4/56x= 2×2×7×x×x×x×x / 2×2×2×7×x

          = x3/2

              = 1 / 2x3


58.  –36y3 ÷ 9y2

Explanation:

Factors of -36y3= -2 × 2 × 3 × 3 × y × y × y

Factors of 9y2= 3 × 3 × y × y

    -36y3/ 9y2 = -2*2*3*3*y*y*y / 3*3*y*y 

               = -4y

59. 66pq2r3 ÷ 11qr2

Explanation:

 Factors of 66pq2r3= 2 × 3 × 11 × p × q × q × r × r × r

    Factors of 11qr2= 11 × q × r × r

    66pq2r3 ÷ 11qr2= 2 × 3 × 11 × p × q × q × r × r × r

                                         11 × q × r × r

             = 6pqr


60. 34x3y3z3 ÷ 51xy2z3

Explanation:

Factors of 34x3y3z3= 2 × 17 × x × x × x × y × y × y × z × z × z

Factors of 51xy2z3= 3 × 17 × x × y × y × z × z × z

     34x3y3z3 ÷ 51xy2z3= 2 × 17 × x × x × x × y × y × y × z × z × z

                                                   3 × 17 × x × y × y × z × z × z

= 23 x2y


61.  12a8b8 ÷ (– 6a6b4)

Explanation:

Factors of 12a8b8= 2 × 2 × 3 × a8

    Factors of -6a6b4= -2 × 3 × a6 × b4

    12a8b8 ÷ (– 6a6b4) = 2 × 2 × 3 × a8

                                    -2 × 3 × a6 × b4

                                  = -2a2b4


 Divide the given polynomial by the given monomial from question 62 to 66

62.(5x2–6x) ÷ 3x

Explanation:

 (5x2–6x) ÷ 3x

   x(5x - 6) ÷ 3x = x(5x-6) / 3x

                     = (5x-6)/3      


63. (3y8–4y6+5y4) ÷ y4

Explanation:

(3y8–4y6+5y4) ÷ y4

    y4( 3y4- 4y2+ 5) ÷ y4

The numerator y2 and denominator y2 is cancelled and we are left with the answer

    3y4- 4y2+ 5


64.  8(x3y2z2+x2y3z2+x2y2z3)÷ 4x2 y2 z2

Explanation:

 8(x3y2z2+x2y3z2+x2y2z3)÷ 4x2 y2 z2

      8x2y2z2( x + y + z)÷ 4x2 y2 z2

      2 (x + y + z)


65. (x3+2x2+3x) ÷2x

Explanation:

(x3+2x2+3x) ÷2x 

    x (x2+2x+3) ÷2x

    1/2 (x2+2x+3)


66. (p3q6–p6q3) ÷ p3q3

Explanation:

 (p3q6–p6q3) ÷ p3q3

    p3q3( q3- p3) ÷ p3q3

    p3 - q3


Work out the following divisions from question 67 to 71

67. (10x–25) ÷ 5

Explanation:

 (10x–25) ÷ 5 

= 5(2x-5) / 5

= 2x – 5


68.  (10x–25) ÷ (2x–5)

Explanation:

(10x–25) ÷ (2x–5)

Here, we take 5 common so that we can cancle the numerator and denominator.

 = 5(2x-5)/( 2x-5)

 = 5


69. 10y(6y+21) ÷ 5(2y+7)

Explanation:

10y(6y+21) ÷ 5(2y+7) 

Here, we take 3 common so that we can cancle the numerator and denominator.

= 10y×3(2y+7)/5(2y+7) 

= 6y


70. 9x2y2(3z–24) ÷ 27xy(z–8)

Explanation:

9x2y2(3z–24) ÷ 27xy(z–8) 

Here, we take 3 common so that we can cancle the numerator and denominator.

= 9x2y2×3(z-8)/27xy(z-8) 

= xy


71. 96abc(3a–12)(5b–30) ÷ 144(a–4)(b–6)

Explanation:

 96abc(3a–12)(5b–30) ÷ 144(a–4)(b–6)

= 9635abca–4b-6144(a–4)(b–6)

= 10abc


 Divide as directed from 72 to 76

72.5(2x+1)(3x+5)÷ (2x+1)

Explanation:

5(2x+1)(3x+5) / 2x+1

= 5(3x+5)


73. 26xy(x+5)(y–4)÷13x(y–4)

Explanation:

 26xy(x+5)(y–4) / 13x(y-4)

=  2y (x+5)


74.  52pqr(p+q)(q+r)(r+p) ÷ 104pq(q+r)(r+p)

Explanation:

52pqr(p+q)(q+r)(r+p) / 104pq(q+r)(r+p)

= r2 (p+q)


75.  20(y+4) (y2+5y+3) ÷ 5(y+4)

Explanation:

 20(y+4) (y2+5y+3) 

            5(y+4)

= 4 (y2+5y+3)


76.  x(x+1) (x+2)(x+3) ÷ x(x+1)

Explanation:

 x(x+1) (x+2)(x+3) 

             x(x+1)

= (x+2)(x+3)


 Factorise the expressions and divide them as directed from 77 to 83

77.(y2+7y+10)÷(y+5)

Explanation:

 (y2+7y+10)÷(y+5)

On simplifying (y2+7y+10) we get,

(y2+7y+10)

= y2+2y+5y+10 

= y(y+2)+5(y+2) 

= (y+2)(y+5)

Thus, (y2+7y+10)÷(y+5)

= y+2y+5y+5

 = y+2


78. (m2–14m–32)÷(m+2)

Explanation:

(m2–14m–32)÷ (m+2)

On simplifying, m2–14m–32, we get

m2–14m–32 

= m2+2m-16m–32 

= m(m+2)–16(m+2) 

= (m–16)(m+2)

Thus, (m2–14m–32)÷(m+2) 

= (m–16)(m+2) / m+2

= m-16


79.  (5p2–25p+20)÷(p–1)

Explanation:

(5p2–25p+20)÷(p–1)

On simplifying, 5p2–25p+20 we get,

 = 5(p2–5p+4)                                 [taking 5 common]

5(p2–5p+4 )

= 5(p2–p-4p+4)

= 5(p–1)(p–4)

Now, (5p2–25p+20)÷(p–1) 

= 5(p–1)(p–4) / p-1

= 5(p–4)


80.  4yz(z2+6z–16)÷2y(z+8)

Explanation:

 4yz(z2 + 6z–16)÷ 2y(z+8)

Factorising z2+6z–16, we get

z2+6z–16 

= z2-2z+8z–16 

= z(z-2)+8(z-2)

= (z–2)(z+8)

Now, 4yz(z2+6z–16) ÷ 2y(z+8) 

= 4yz(z–2)(z+8) / 2y(z+8)

 = 2z(z-2)


81.  5pq(p2–q2)÷2p(p+q)

Explanation:

5pq(p2–q2) ÷ 2p(p+q)

5pq(p+q)(p-q) ÷ 2p(p+q)                         [Since, a2–b2 = (a–b)(a+b)]

= 5pq(p–q)(p+q) / 2p(p+q)

= 5q(p–q)/2


82.  12xy(9x2–16y2)÷4xy(3x+4y)

Explanation:

12xy(9x2–16y2) ÷ 4xy(3x+4y)

Factorising 9x2–16y2 , we have

9x2–16y2 

= (3x)2–(4y)2 

= (3x+4y)(3x-4y)                                             [Since, a2–b2 = (a–b)(a+b)]

Now, 12xy(9x2–16y2) ÷ 4xy(3x+4y) 

= 12xy(3x+4y)(3x-4y) / 4xy(3x+4y)


= 3(3x-4y)


83. 39y3(50y2–98) ÷ 26y2(5y+7)

Explanation:

39y3(50y2–98) ÷ 26y2(5y+7)
       On simplifying, 39y3(50y2–98) we get

       39y3 x 2(25y2-49)

       78y3[(5y)2- 72]

        78y3(5y+7)(5y-7) [Using the identity a2-b2= (a+b)(a-b)]

Now, 78y3(5y+7)(5y-7)÷ 26y2(5y+7)

          78y3(5y+7)(5y-7)

                26y2(5y+7)

       = 3y(5y-7)


EXERCISE- 14.4

84. 4(x–5) = 4x–5

Explanation:

4(x- 5)

= 4x – 20 ≠ 4x – 5 ( RHS )

The correct statement is 4(x-5) = 4x–20


85.  x(3x+2) = 3x2+2

Explanation:

x(3x+2) 

= 3x2+2x ≠ 3x2+2 ( RHS )

The correct statement is x(3x+2) = 3x2+2x


86.  2x+3y = 5xy

Explanation:

LHS= 2x+3y ≠ R. H. S

The correct statement is 2x+3y = 2x+3 y


87. x+2x+3x = 5x

Explanation:

LHS = x+2x+3x 

         = 6x ≠ RHS

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


88. 5y+2y+y–7y = 0

Explanation:

LHS = 5y+2y+y–7y = y ≠ RHS

The correct statement is 5y+2y+y–7y = y


89. 3x+2x = 5x2

Explanation:

LHS = 3x+2x = 5x ≠ RHS

The correct statement is 3x+2x = 5x


90.  (2x) 2+4(2x)+7 = 2x2+8x+7

Explanation:

LHS = (2x) 2+4(2x)+7 

         = 4x2+8x+7 ≠ RHS

The correct statement is (2x) 2+4(2x)+7 = 4x2+8x+7


91.  (2x) 2+5x = 4x+5x = 9x

Explanation:

LHS = (2x) 2+5x 

         = 4x2+5x ≠ 9x = RHS

The correct statement is(2x) 2+5x = 4x2+5x


92. (3x + 2) 2 = 3x2+6x+4

Explanation:

LHS = (3x+2) 2 

         = (3x)2+22+2x2x3x

         = 9x2+4+12x ≠ RHS

The correct statement is (3x + 2) 2 = 9x2+4+12x


 Substituting x = – 3 in from question 93 to 95

93. x2 + 5x + 4 gives (– 3) 2+5(– 3)+4 = 9+2+4 = 15

Explanation:

Substituting x = – 3 in x2+5x+4, we get

x2+5x+4 = (– 3) 2+5(– 3)+4 

               = 9–15+4 = – 2


94.  x2 – 5x + 4 gives (– 3) 2– 5( – 3)+4 = 9–15+4 = – 2

Explanation:

Substituting x = – 3 in x2–5x+4

x2–5x+4

 = (–3) 2–5(– 3)+4 

= 9+15+4 = 28


95.  x2 + 5x gives (– 3) 2+5(–3) = – 9–15 = – 24

Explanation:

Substituting x = – 3 in x2+5x

x2+5x = (– 3) 2+5(–3) 

= 9–15 = -6


96. (y–3)2 = y2–9

Explanation:

LHS = (y–3)2  

(y – 3)2 = y2+(3) 2–2y×3                       [ Since, (a–b) 2 = a2+b2-2ab ]

= y2+9 –6y ≠ y2 – 9 = RHS

The correct statement is (y–3)2 = y2 + 9 – 6y


97. (z+5) 2 = z2+25

Explanation:

LHS = (z+5)2  

(z+5) 2 = z2+52+2×5×z                       [Since, (a+b) 2 = a2+b2+2ab ]

= z2+25+10z ≠ z2+25 = RHS

Thus, the correct statement is (z+5) 2 = z2+25+10z


98.  (2a+3b)(a–b) = 2a2–3b2

Explanation:

LHS = (2a+3b)(a–b) 

= 2a(a–b)+3b(a–b)

= 2a2–2ab+3ab–3b2

= 2a2+ab–3b2 ≠ 2a2–3b2 = RHS

The correct statement is (2a +3b)(a –b) = 2a2+ab–3b2


99. (a+4)(a+2) = a2+8

Explanation:

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

 = a(a+2)+4(a+2)

= a2+2a+4a+8

= a2+6a+8 ≠ a2+8 = RHS

Thus, the correct statement is (a+4)(a+2) = a2+6a+8


100. (a–4)(a–2) = a2–8

Explanation:

LHS = (a–4)(a–2) 

= a(a–2)–4(a–2)

= a2–2a–4a+8

= a2–6a+8 ≠ a2-8 ( RHS )

The correct statement is (a–4)(a–2) = a2–6a+8


101. 3x2/3x2 = 0

Explanation:

LHS = 3x2/3x2 = 1 ≠ 0 =(RHS)

Thus, the correct statement is 3x2/3x2 = 1


102.  (3x2+1)/3x2 = 1 + 1 = 2

Explanation:

LHS = (3x2+1)/3x2 

= (3x2/3x2)+(1/3x2) 

= 1+(1/3x2) ≠ 2 ( RHS )

The correct statement is (3x2+1)/3x2 = 1+(1/3x2)


103.  3x/(3x+2) = 1/2

Explanation:

LHS = 3x/(3x+2) ≠ 1/2 = RHS

The correct statement is 3x/(3x+2) = 3x/(3x+2)


104.  3/(4x+3) = 1/4x

Explanation:

LHS= 3 / 4x+3

       = 3 / 4x + 1 ≠ RHS

So, the correct statement is  3 / 4x + 1


105. (4x+5)/4x = 5

Explanation:

LHS = (4x+5)/4x 

         = 4x4x+ 54x

         = 1 + 54x ≠ 5 ( RHS )

Therefore, the correct statement is (4x+5)/4x = 1 + (5/4x)


106. 7x + 5 /5  = 7x

Explanation:

LHS = 7x+5 / 5

          = 7x / 5+ 5 /5

          = (7x/5)+1 ≠ 7x (RHS)

The correct statement is (7x+5)/5 = (7x/5) +1