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

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

Solution:

(i) a2+8a+16

Ans:

Given: a2+8a+16

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

= a2+2Ă—4Ă—a+42

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

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

(ii) p2â€“10p+25

Ans:

Given: p2â€“10p+25

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

= p2-2Ă—5Ă—p+52

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

p2â€“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, 2Ă—5mĂ—3 and 32 respectively we get,

= (5m)2 + 2Ă—5mĂ—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, 2Ă—7yĂ—6z and (6z)2  respectively we get,

=(7y)2+2Ă—7yĂ—6z+(6z)2

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

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

(v) 4x2â€“8x+4

Ans:

Given: 4x2â€“8x+4

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

= (2x)2-2Ă—4x+22

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

4x2â€“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, 2Ă—11bĂ—4c and (4c)2 respectively we get,

= (11b)2-2Ă—11bĂ—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+2Ă—a2Ă—b2+(b2)2

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

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

### (viii) 25a2â€“4b2+28bcâ€“49c2

Solution:

(i) 4p2â€“9q2

Ans:

Given: 4p2â€“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)

4p2â€“9q2 = (2p-3q)(2p+3q)

(ii) 63a2â€“112b2

Ans:

Given: 63a2â€“112b2

63a2â€“112b2 = 7(9a2 â€“16b2)

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

= 7((3a)2â€“(4b)2)

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

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

(iii) 49x2â€“36

Ans:

Given: 49x2â€“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)

49x2â€“36 = (7x+6)(7xâ€“6)

(iv) 16x5â€“144x3

Ans:

Given:  16x5 â€“ 144x3

16x5 â€“ 144x3 = 16x3(x2â€“9)

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

= 16x3(x2â€“32)

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

16x5â€“144x3 = 16x3(xâ€“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)-(lâ€“m)}{(l +m)+(lâ€“m)}

= (l+mâ€“l+m)(l+m+lâ€“m)

= (2m)(2l)

(l+m)2 – (l-m)2 = 4 ml

(vi) 9x2y2â€“16

Ans:

Given: 9x2y2â€“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)

9x2y2â€“16 = (3xyâ€“4)(3xy+4)

(vii) (x2â€“2xy+y2)â€“z2

Ans:

Given: (x2â€“2xy+y2)â€“z2

By compressing x2â€“2xy+y2 using identity: (x-y)2 = x2-2xy+y2 , we get,

(x2â€“2xy+y2)â€“z2 = (xâ€“y)2â€“z2

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

= {(xâ€“y)â€“z}{(xâ€“y)+z}

(x2â€“2xy+y2)â€“z2 = (xâ€“yâ€“z)(xâ€“y+z)

(viii) 25a2â€“4b2+28bcâ€“49c2

Ans:

Given: 25a2â€“4b2+28bcâ€“49c2

25a2â€“4b2+28bcâ€“49c2 = 25a2â€“(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

25a2â€“4b2+28bcâ€“49c2 = (5a+2b-7c)(5a-2b+7c)

### (ix)6xyâ€“4y+6â€“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) 5y2â€“20yâ€“8z+2yz

Ans:

Given: 5y2â€“20yâ€“8z+2yz

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

= 5y(yâ€“4)+2z(yâ€“4)

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

5y2â€“20yâ€“8z+2yz = (yâ€“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) 6xyâ€“4y+6â€“9x

Ans:

Given: 6xyâ€“4y+6â€“9x

= 6xyâ€“9xâ€“4y+6

Taking 3x and 2 as common, we get,

= 3x(2yâ€“3)â€“2(2yâ€“3)

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

6xyâ€“4y+6â€“9x = (2yâ€“3)(3xâ€“2)

### (v) a4â€“2a2b2+b4

Solution:

(i) a4â€“b4

Ans:

Given: a4â€“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)

a4â€“b4 = (a â€“ b)(a + b)(a2+b2)

(ii) p4â€“81

Ans:

Given: p4â€“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)

p4â€“81 =(p-3)(p+3)(p2+9)

(iii) x4â€“(y+z)4

Ans:

Given: x4â€“(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 â€“(y+z)(x+(y+z)}{x2+(y+z)2}

x4â€“(y+z)4  = (xâ€“yâ€“z)(x+y+z) {x2+(y+z)2}

(iv) x4â€“(xâ€“z)4

Ans:

Given: x4â€“(xâ€“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)

x4â€“(xâ€“z)4 = z(2x-z)( 2x2-2xz+z2)

(v) a4â€“2a2b2+b4

Ans:

Given: a4â€“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

a4â€“2a2b2+b4  = ((aâ€“b)(a+b))2

### (iii) p2+6pâ€“16

Solution:

(i) p2+6p+8

Ans:

Given: p2+6p+8

Since, 6p and 8 can be substituted by 2p+4p and 4Ă—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) q2â€“10q+21

Ans:

Given: q2â€“10q+21

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

= q2â€“3q-7q+21

Taking q and 7 common terms, we get

= q(qâ€“3)â€“7(qâ€“3)

Again taking (qâ€“3) as common, we get

= (qâ€“7)(qâ€“3)

q2â€“10q+21 = (qâ€“7)(qâ€“3)

(iii) p2+6pâ€“16

Ans:

Given: p2+6p+16

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

= p2â€“2p+8pâ€“16

Taking p and 8 common terms, we get

= p(pâ€“2)+8(pâ€“2)

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

= (p+8)(pâ€“2)

p2+6pâ€“16 = (p+8)(pâ€“2)

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