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Ncert Solutions for Class 8 Maths - Factorization Chapter 14 CBSE Part 2





In this page we have Ncert Solutions for Class 8 Maths - Factorization Chapter 14 for EXERCISE 2 . Free pdf download also available .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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