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

• Last Updated : 10 Mar, 2021

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

Solution:

(i) a2+8a+16

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12.

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

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)

My Personal Notes arrow_drop_up