# Class 8 NCERT Solutions – Chapter 14 Factorisation – Exercise 14.2

• Last Updated : 10 Mar, 2021

### (viii) a4+2a2b2+b4

Solution:

(i) a2+8a+16

Ans:

Given: a2+8a+16

Since, 8a and 16  can be substituted by 2×4×a and  42  respectively we get,

= a2+2×4×a+42

Therefore, by using the identity : (x+y)2 = x2+2xy+y2

a2+8a+16 = (a+4)2

(ii) p2–10p+25

Ans:

Given: p2–10p+25

Since, 10p and 25 can be substituted by  2×5×p and 52 respectively we get,

= p2-2×5×p+52

Therefore, by using the identity : (x-y)2 = x2-2xy+y2

p2–10p+25 = (p-5)2

(iii) 25m2+30m+9

Ans:

Given: 25m2+30m+9

Since, 25m2 , 30m and  9 can be substituted by (5m)2, 2×5m×3 and 32 respectively we get,

= (5m)2 + 2×5m×3 + 32

Therefore, by using the identity: (x+y)2 = x2+2xy+y2

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

(iv) 49y2+84yz+36z2

Ans:

Given: 49y2+84yz+36z

Since, 49y2, 84yz and 36z2 can be substituted by (7y)2, 2×7y×6z and (6z)2  respectively we get,

=(7y)2+2×7y×6z+(6z)2

Therefore, by using the identity: (x+y)2 = x2+2xy+y2

49y2+84yz+36z2 = (7y+6z)2

(v) 4x2–8x+4

Ans:

Given: 4x2–8x+4

Since, 4x2, 8x and 4 can be substituted by (2x)2, 2×4x and 22 respectively we get,

= (2x)2-2×4x+22

Therefore, by using the identity: (x-y)2 = x2-2xy+y2

4x2–8x+4 = (2x-2)2

(vi) 121b2-88bc+16c2

Ans:

Given: 121b2-88bc+16c2

Since, 121b2, 88bc and 16c2 can be substituted by (11b)2, 2×11b×4c and (4c)2 respectively we get,

= (11b)2-2×11b×4c+(4c)2

Therefore, by using the identity: (x-y)2 = x2-2xy+y2

121b2-88bc+16c2 = (11b-4c)2

(vii) (l+m)2-4lm (Hint: Expand (l+m)2 first)

Ans:

Given: (l+m)2-4lm

By expanding (l+m)2 using identity: (x+y)2 = x2+2xy+y2 , we get,

(l+m)2-4lm = l2+m2+2lm-4lm

(l+m)2-4lm = l2+m2-2lm  =  l2-2lm+m2

Therefore, by using the identity: (x-y)2 = x2-2xy+y2

(l+m)2-4lm = (l-m)2

(viii) a4+2a2b2+b4

Ans:

Given: a4+2a2b2+b4

Since, a4 and b4 can be substituted by (a2)2 and (b2)2 respectively we get,

= (a2)2+2×a2×b2+(b2)2

Therefore, by using the identity: (x+y)2 = x2+2xy+y2

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

### (viii) 25a2–4b2+28bc–49c2

Solution:

(i) 4p2–9q2

Ans:

Given: 4p2–9q2

Since, 4p2 and 9q2 can be substituted by (2p)2 and (3q)2 respectively we get,

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

Therefore, by using the identity: x2-y2 = (x+y)(x-y)

4p2–9q2 = (2p-3q)(2p+3q)

(ii) 63a2–112b2

Ans:

Given: 63a2–112b2

63a2–112b2 = 7(9a2 –16b2)

Since, 9a2 and 16b2 can be substituted by (3a)2 and (4b)2 respectively we get,

= 7((3a)2–(4b)2)

Therefore, by using the identity: x2-y2 = (x+y)(x-y)

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

(iii) 49x2–36

Ans:

Given: 49x2–36

Since, 49x2 and 36 can be substituted by (7x)2 and 62 respectively we get,

= (7x)2 – 62

Therefore, by using the identity: x2-y2 = (x+y)(x-y)

49x2–36 = (7x+6)(7x–6)

(iv) 16x5–144x3

Ans:

Given:  16x5 – 144x3

16x5 – 144x3 = 16x3(x2–9)

Since, 9 can be substituted by 32  respectively we get,

= 16x3(x2–32)

Therefore, by using the identity: x2-y2 = (x+y)(x-y)

16x5–144x3 = 16x3(x–3)(x+3)

(v) (l+m)2 – (l-m)2

Ans:

Given: (l+m)2 – (l-m)

By expanding (l+m)2 – (l-m)2 using identity: x2-y2 = (x+y)(x-y) , we get,

= {(l+m)-(l–m)}{(l +m)+(l–m)}

= (l+m–l+m)(l+m+l–m)

= (2m)(2l)

(l+m)2 – (l-m)2 = 4 ml

(vi) 9x2y2–16

Ans:

Given: 9x2y2–16

Since, 9x2y2 and 16 can be substituted by (3xy)2 and 42 respectively we get,

= (3xy)2-42

Therefore, by using the identity: x2-y2 = (x+y)(x-y)

9x2y2–16 = (3xy–4)(3xy+4)

(vii) (x2–2xy+y2)–z2

Ans:

Given: (x2–2xy+y2)–z2

By compressing x2–2xy+y2 using identity: (x-y)2 = x2-2xy+y2 , we get,

(x2–2xy+y2)–z2 = (x–y)2–z2

Therefore, by using the identity: x2-y2 = (x+y)(x-y)

= {(x–y)–z}{(x–y)+z}

(x2–2xy+y2)–z2 = (x–y–z)(x–y+z)

(viii) 25a2–4b2+28bc–49c2

Ans:

Given: 25a2–4b2+28bc–49c2

25a2–4b2+28bc–49c2 = 25a2–(4b2-28bc+49c2 )

Since, 25a2, 4b2, 28bc and 49c2 can be substituted by (5a)2, (2b)2, 2(2b)(7c) and (7c)2 respectively we get,

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

Therefore, by using the identity: (x-y)2 = x2-2xy+y2

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

and by using Identity: x2-y2 = (x+y)(x-y) , we get

25a2–4b2+28bc–49c2 = (5a+2b-7c)(5a-2b+7c)

### (ix)6xy–4y+6–9x

Solution:

(i) ax2+bx

Ans:

Given: ax2+bx

Taking x as common, we get

ax2+bx = x(ax+b)

(ii) 7p2+21q

Ans:

Given: 7p2+21q

Taking 7  as common, we get,

7p2+21q2 = 7(p2+3q2)

(iii) 2x3+2xy2+2xz2

Ans:

Given: 2x3+2xy2+2xz2

Taking 2x as common, we get,

2x3+2xy2+2xz2  = 2x(x2+y2+z2)

(iv) am2+bm2+bn2+an

Ans:

Given: am2+bm2+bn2+an2

Taking m2 and n2 as common, we get,

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

Taking (a+b) as common, we get,

am2+bm2+bn2+an2 = (a+b)(m2+n2)

(v) (lm+l)+m+1

Ans:

Given: (lm+l)+m+1

= lm+m+l+1

Taking m as common, we get,

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

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

(vi) y(y+z)+9(y+z)

Ans:

Given: y(y+z)+9(y+z)

Taking  (y+z) as common, we get,

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

(vii) 5y2–20y–8z+2yz

Ans:

Given: 5y2–20y–8z+2yz

Taking 5y and 2z as common, we get,

= 5y(y–4)+2z(y–4)

Taking (y-4) as common, we get,

5y2–20y–8z+2yz = (y–4)(5y+2z)

(viii) 10ab+4a+5b+2

Ans:

Given: 10ab+4a+5b+2

Taking 5b and 2 as common, we get,

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

Taking (2a+1) as common, we get,

10ab+4a+5b+2 = (2a+1)(5b+2)

(ix) 6xy–4y+6–9x

Ans:

Given: 6xy–4y+6–9x

= 6xy–9x–4y+6

Taking 3x and 2 as common, we get,

= 3x(2y–3)–2(2y–3)

Taking (2y-3) as common, we get,

6xy–4y+6–9x = (2y–3)(3x–2)

### (v) a4–2a2b2+b4

Solution:

(i) a4–b4

Ans:

Given: a4–b4

Since, a4 and b4 can be substituted by (a2)2 and (b2)2 respectively we get,

= (a2)2-(b2)2

Therefore, by using Identity: x2-y2 = (x+y)(x-y) , we get

= (a2-b2) (a2+b2)

a4–b4 = (a – b)(a + b)(a2+b2)

(ii) p4–81

Ans:

Given: p4–81

Since, p4 and 81 can be substituted by (p2)2 and (9)2 respectively we get,

= (p2)2-(9)2

Therefore, by using Identity: x2-y2 = (x+y)(x-y) , we get

= (p2-9)(p2+9)

= (p2-32)(p2+9)

p4–81 =(p-3)(p+3)(p2+9)

(iii) x4–(y+z)4

Ans:

Given: x4–(y+z)4

Since, x4 and (y+z)4 can be substituted by (x2)2 and [(y+z)2]2 respectively we get,

= (x2)2-[(y+z)2]2

Therefore, by using the identity: x2-y2 = (x+y)(x-y) , we get

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

= {(x –(y+z)(x+(y+z)}{x2+(y+z)2}

x4–(y+z)4  = (x–y–z)(x+y+z) {x2+(y+z)2}

(iv) x4–(x–z)4

Ans:

Given: x4–(x–z)4

Since, x4 and (x-z)4 can be substituted by (x2)2 and [(x-z)2]2 respectively we get,

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

By using Identity: x2-y2 = (x+y)(x-y) , we get

= {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)

x4–(x–z)4 = z(2x-z)( 2x2-2xz+z2)

(v) a4–2a2b2+b4

Ans:

Given: a4–2a2b2+b4

Since, a4 and b4 can be substituted by (a2)2 and (b2)2 respectively we get,

= (a2)2-2a2b2+(b2)2

Therefore, by using the identity: (x-y)2 = x2-2xy+y2

= (a2-b2)2

And by using Identity: x2-y2 = (x+y)(x-y) , we get

a4–2a2b2+b4  = ((a–b)(a+b))2

### (iii) p2+6p–16

Solution:

(i) p2+6p+8

Ans:

Given: p2+6p+8

Since, 6p and 8 can be substituted by 2p+4p and 4×2 respectively we get,

= p2+2p+4p+8

Taking p and 4 common terms, we get

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

Again taking (p+2) as common, we get

= (p+2)(p+4)

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

(ii) q2–10q+21

Ans:

Given: q2–10q+21

Since, 10q and 21 can be substituted by (-3q)+(-7q) and (-3)×(-7) respectively we get,

= q2–3q-7q+21

Taking q and 7 common terms, we get

= q(q–3)–7(q–3)

Again taking (q–3) as common, we get

= (q–7)(q–3)

q2–10q+21 = (q–7)(q–3)

(iii) p2+6p–16

Ans:

Given: p2+6p+16

Since, 6p and 16 can be substituted by 8p +(-2p) and (-2)×8 respectively we get,

= p2–2p+8p–16

Taking p and 8 common terms, we get

= p(p–2)+8(p–2)

Again taking (p-2) as common, we get

= (p+8)(p–2)

p2+6p–16 = (p+8)(p–2)

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