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Ncert Solutions for Factorization Class 8 CBSE Part 2



In this page we have NCERT book Solutions for Class 8th Maths:Factorization for EXERCISE 2 . Hope you like them and do not forget to like , social share and comment at the end of the page.

Question 1

Factorize the following expressions.

(i) a² + 8a + 16

(ii) p² – 10 p + 25

(iii) 25m² + 30m + 9

(iv) 49y² + 84yz + 36z²

(v) 4x² – 8x + 4

(vi) 121b² – 88bc + 16c²

(vii) (l + m) ² – 4lm

(viii) a4 + 2a²b² + b4

Answer

We have to make use of following identities to factorize them

(a+b)2= a2 +b2 +2 ab

(a-b)2= a2 +b2 -2 ab

a2 –b2 = (a-b)(a+b)

 

1) a² + 8a + 16

= a2 + 2×a× 4 +  42

So from first identity, it can be written as

    =(a+4)2

 

2)  p² – 10 p + 25

= p2 -  2×p× 5 +  52

So from second identity, it can be written as

    =(p-5)2

3) 25m² + 30m + 9

= (5m)2 + 2×5m× 3 +  32

So from first identity, it can be written as

    =(5m+3)2

 

4) 49y² + 84yz + 36z²

= (7y)2 + 2×7y× 6z +  (6z)2

So from first identity, it can be written as

    =(7y+6z)2

 

5) 4x² – 8x + 4

= (2x)2 -  2×2x× 2 +  22

So from second identity, it can be written as

   =(2x-2)2

= 4(x-1)2  taking common factor 2 out of square

 

6) 121b² – 88bc + 16c²

= (11b)2 - 2×11b× 4c +  (4c)2

 

So from second identity, it can be written as

    =(11b-4c)2

 

7) (l + m) ² – 4lm

=l2 + m2 +2lm -4lm

= l2 + m2 -2lm

So from second identity, it can be written as

    =(l-m)2

 

8) a4 + 2a²b² + b4

= (a2)2 + 2a²b²+(b2)2

So from first identity, it can be written as

=(a²+b²)²

 

Question 2

 Factorize.

(i) 4p² – 9q²

(ii) 63a² – 112b²

(iii) 49x² – 36

(iv) 16x5 – 144x³

 (v) (l + m) ² – (l – m) ²

(vi) 9x² y² – 16

(vii) (x² – 2xy + y²) – z²

 (viii) 25a² – 4b² + 28bc – 49c²

Answer

We have to make use of following identities to factorize them

(a+b)2= a2 +b2 +2 ab

(a-b)2= a2 +b2 -2 ab

a2 –b2 = (a-b)(a+b)

 

1) 4p² – 9q²

=(2p)2 –(3q)2

So from third identity, it can be written as

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

 

2) 63a² – 112b²

= 7( 9a2 -16b2)

=7[ (3a)2 –(4b)2]

So from third identity, it can be written as

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

3) 49x² – 36

=(7x)2 –(6)2

So from third identity, it can be written as

=(7x-6)( 7x+6)

4) 16x5-144x3

= x³(16x²-144)

= x³(4x+12)(4x-12)

5) (l + m) ² – (l – m) ²

= [ l+m +l-m][l+m-l+m]

=2l×2m

=4lm

 

6) 9x² y² – 16

= (3xy-4)(3xy+4)

7) (x² – 2xy + y²) – z²

=  (x-y)2 –z2   as (a-b)2= a2 +b2 -2 ab

Now as a2 –b2 = (a-b)(a+b)

= (x-y+z)(x-y-z)

8) 25a² – 4b² + 28bc – 49c²

Factorizing each tem

= (5a)2 –(2b)2 + 2×2b×7c –(7c)2

Rearranging the terms

=(5a)2 –[(2b)2 - 2×2b×7c +(7c)2]

Now as (a-b)2= a2 +b2 -2 ab

=(5a)2 –(2b-7c)2

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

 

Question 3

 Factorize 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

 

Answer

1) ax² + bx

=x(ax+b)

 

2) 7p² + 21q²

=7(p2+3q2)

 

3) 2x³ + 2xy² + 2xz²

=2x(x²+y²+z²)

4) am² + bm² + bn² + an²

Rearranging the terms

= am²+ an²+ bm² + bn²

=a(m2 +n2) +b(m2 +n2)

=(a+b) (m2 +n2)

 

5) (lm + l) + m + 1

=l(m+1) +1(m+1)

=(l+1)(m+1)

 

 

6) y (y + z) + 9 (y + z)

= (y+z)(y+9)

 

7) 5y² – 20y – 8z + 2yz

=5y(y-4) +2z(y-4)

=(5y+2z)(y-4)

 

8) 10ab + 4a + 5b + 2

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

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

 

9) 6xy – 4y + 6 – 9x

=2y(3x-2) +3(2-3x)

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

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

 

Question 4

 Factorize.

(i) a4 – b4

 (ii) p4 – 81

 (iii) x4 – (y + z)4

 (iv) x4 – (x – z)4

 (v) a4 – 2a²b² + b4

Answer:

i) a4-b4 = (a²+b²)(a²-b²)

ii) p4 – 81

=(p²+9)(p²-9)

iii) x4 – (y + z)4

= (x²+(y+z) ²)(x²-(y+z) ²)

= (x²+(y+z) ²)[(x+y+z)(x-y-z)]

iv) x4 – (x – z)4

=(x²-(x-z) ²)(x²+(x-z) ²)

=[(x+x-z)(x-x+z)][(x²+(x-z) ²]

=z(2x-z) [(x²+(x-z) ²]

v) a4 – 2a²b² + b4

=(a2 –b2)2

 

Question 5

 Factorize the following expressions.

(i) p² + 6p + 8

 (ii) q² – 10q + 21

 (iii) p² + 6p – 16

Answer

1)  p²+6p+8

=p(p+6)+8

2) q²-10q+21

=q(q-10)+21

3)  p²+6p-16

=p(p+6)-16

 

 


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