Ncert Solutions For Class 8 Maths Ex 14.2

Ncert Solutions For Class 8 Maths Chapter 14 Ex 14.2

Question1. Factorize the given terms:

( i ) x2 + 8y  + 16                 ( ii ) a2– 10a  +  25

( iii )25a2 + 30a  + 9          ( iv ) 49a2 +  84bc  +  36c2

( v )121x2 -88xy + 16y2       ( vi )(a + b)2 -4ab

( vii ) x4 + 2a2b2+b4              

 

sol.

( i )  x2 + 8y +16 = x2+ (4 + 4) + 4 * 4

Making use of identity;  a2 + (p + q)a + ab = (x + p)(x + q)

Here, x =a, p = 4 and q= 4

x2 + 8y +16 = (x + 4)(x + 4) = (x + 4)2

 

( ii )  a – 10a + 25 = a2 +(-5 – 5)a + (-5)(-5)

Making use of identity ;

x2+(p + q)x +pq = (x + p)(x + q)

Here, x = p, a=-5 and b= -5

a – 10a + 25 = (a – 5)(a – 5) = (a – 5)2

 

( iii ) 25a2 + 30a + 9 = (5a)2 + (2*5a*3) + 32

Making use of identity;

x2 +2xy + y2 = (x + y)2, here x=5a and y =3

25a2 + 30a + 9 = (5a + 3)2

 

( iv )  49a2 +  84ac  +  36c2 = (7a)2 + 2 * 7a * 6c + (6c)2

Making use of identity;

x2 + 2xy + b2 =(x + y)2, here x = 7a and  y=6c

49a2 +  84bc  +  36c2 = (7a + 6c)2

 

( v ) 121x2 -88xy + 16y2 = (11x)2 – 2 * 11x * 4y + (4y)2

Making use of identity;

a2 – 2ab + b2 = (a – b)2, here a =11x and b =4y

121x2 -88xy + 16y2 = (11x – 4)

 

( vi )  (a + b)2 – 4ab = a2 + 2ab + b2 -4ab          . . . . . .   [Since, ( a + b )2 = a2 + 2ab + b2]

= a2 – 2ab +b2

= (a -b)2                                   . . . . . . [Since,  (a-b)2 =a2 – 2ab +b2]

 

( vii ) x4 + 2a2b2+b4 = (x2)2 + 2 * a2*b2 + (b2)2

=(x2 + b2)2                                                . . . . . . . [Since, ( a + b )2 = a2 + 2ab + b2]

 

Question 2. Factorize the following expressions:

( i ) (25a)– (36b)2            ( ii ) 72a2 – 200b2

 ( iii ) 4a2– 36                       ( iv ) 16a5 – 144a3

( v ) (a2-2ab + b2)-c          ( vi ) 25a 2 – ( 49a2 +  84bc  +  36c2)

( vii ) 9a2b2 -25

 

sol.

( i ) (25a)– (36b)2 = (5a)2 – (6b)2

= (5a – 6b)(5a + 6b)              ..  . . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( ii ) 72a2 – 200b2 = 8(9a2 – 25b2)

                                        = 8((3a)2 – (5b)2)

= 8(3a + 5b)(3a – 5b)               . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( iii ) 4a2 – 36 = (2a)2 – 62

= (2a – 6)(2a + 6)                               . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( iv ) 16a5 – 144a3= 16a3(a2 – 9)

= 16a3(a2 – 32)

=  16a3(a + 3)(a – 3)               . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( v ) (a2-2ab + b2)-c = (a -b)2 – c                       . . . . . . [Since,(a – b2)=a2-2ab + b2]

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

 

( vi ) 25a 2 -( 49a2 +  84bc  +  36c2) = 25a2 –[ (7a)2 + 2 * 7b * 6c + (6c)2]

= (5a)2 – (7a + 6c)2

= (5a -7a -6c)(5a + 7a +6c)

= (12a + 6c)(-2a – 6c)

 

(vii) 9a2b2 -25 = (3ab)2 – 52

=  (3ab -5)(3ab + 4)                    [Since, a2-b2 =(a + b)(a-b) ]

 

Question 3. Factorize the following expressions:

                ( i ) ax2 + bx                                        ( ii ) 7a2 +21b2

                ( iii ) ap2+ bp2 + bq2 +aq2              ( iv ) a(a + b) +4 (a + b)

                ( v ) 5a2 – 20a -8b + 2ab                 ( vi )6ab – 4b + 6 – 9a

                ( vii ) (ap+a) + 1+p

sol.

( i ) ax2 + bx = x(a+b)

 

( ii ) 7a2+ 21b2 = 7(a2 + 3b2)

 

( iii ) ap2+ bp2 + bq2 +aq2

=p2(a + b) + q2(a + b)

=(p2 + q2)(a + b)

 

( iv ) a(a + b) +4 (a + b) = (a + 4)(a + b)

 

( v ) 5a2 – 20a -8b + 2ab = 5a(a – 4)+ 2b(a-4)

=    (a – 4)(5a + 2b)

 

( vi )  6ab – 4b + 6 – 9a = 6ab – 9a – 4b + 6

= 3a(2b – 3) – 2(2b – 3)

= (3a – 2)(2b -3)

 

( vii ) (ap+a) + 1+p = a(p +1) + 1(p + 1)

=  (a + 1)(p + 1)

 

Question 4. Factorize the following terms:

                ( i ) x4 – y4                            ( ii )a4 – 81

                ( iii ) a4 –(a + b)4                      ( iv )a 4– (a – b)4

                ( v ) a4-2a2b2 + b4

sol.

( i ) x4 – y4 =(x2)2  – (y2)2

= (x2 – y2)(x2 + y2)

= (x – y)(x + y)(x2 +y2)        . . . . . . . . .. . . . . . .        [Since, a2-b2 =(a + b)(a-b) ]

 

( ii ) a4 – 81 = (a2)2 -92

= (a2+ 9)(a2 -9)

= (a2 – 32)(a2 + 9)                 . . . .   . . . . .  . . . .   [Since, a2-b2 =(a + b)(a-b)]

= (a-3)(a + 3)(a2 +9)

 

( iii ) a4 – ( c + b)4 = (a2)2 – [(c +b)2]2

= [a2 – (c + b)2][a2 + (c + b)2]        . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

= [a –(c + b )][a + c +b][a2 + (c+b)2]

 

( iv ) a 4– (a – b)4= (a2)2 – [(a – b)2]2

= [a2 –(a-b)2][a2 + (a – b)2]            . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

= [a + (a-b)][a – (a-b)][a2+(a – b)2]

= [a + (a-b)][a – (a-b)][2a2 + b2 – 2ab]

 

( v )a4-2a2b2 + b4 = (a2)2 -2a2b2 + (b2)2

= (a2 –b2)2                           . . . . . . . .  . . . . .  [Since, (a-b)2 = a2 -2ab +b2]

=[(a +b)(a-b)]2                  . . . . . . . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

= [(a + b)2(a – b)2]             . . . .. . .. . .. . . . . [Since, (ab)q = aqbq]

 

Question 5.Factorize the given expressions:

( i ) a2 + 6a + 8                   ( ii )a2 +6a -16

( iii )a2 -10a +21

 

sol.

 ( i ) a2 + 6a + 8 = a2 + (2 +4)a + (4*2)

= a2+ 2a +4a +(4 * 2)

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

= (a +2)(a+4)

 

( ii )a2 + 6a -16    = a2 +(8 -2)a – (8*2)

= a2+8a -2a – (8 *2)

= a(a + 8) – 2(a + 8)

=(a-2)(a+8)

 

( iii ) a2 -10a + 21 = a2 –(7+3)a +(7*3)

= a2 -7a – 3a  + (7 * 3)

= a(a-7)-3(a -7)

= (a – 3)(a – 7)