NCERT solution class 8 chapter 14 Factorisation

Exercise 14.1   Exercise 14.2   Exercise 14.3   Exercise 14.4

Exercise 14.2

Question 1 

Factorise the following expressions.

(i) a2 + 8a + 16

(ii) p2 − 10p + 25

(iii) 25m2 + 30m + 9

(iv) 49y2 + 84yz + 36z2

(v) 4x2 − 8x + 4

(vi) 121b2 − 88bc + 16c2

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

(viii) a4 + 2a2b2 + b4

Sol :

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

= (a + 4)2 [(x + y)2 = x2 + 2xy + y2]

(ii) p2 − 10p + 25 = (p)2 − 2 × p × 5 + (5)2

= (p − 5)2 [(ab)2 = a2 − 2ab + b2]

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

= (5m + 3)2 [(a + b)2 = a2 + 2ab + b2]

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

= (7y + 6z)2 [(a + b)2 = a2 + 2ab + b2]

(v) 4x2 − 8x + 4 = (2x)2 − 2 (2x) (2) + (2)2

= (2x − 2)2 [(ab)2 = a2 − 2ab + b2]

= [(2) (x − 1)]2 = 4(x − 1)2

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

= (11b − 4c)2 [(ab)2 = a2 − 2ab + b2]

(vii) (l + m)2 − 4lm = l2 + 2lm + m2 − 4lm

= l2 − 2lm + m2

= (l m)2 [(ab)2 = a2 − 2ab + b2]

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

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

 

 

Question 2

Factorise

(i) 4p2 − 9q2

(ii) 63a2 − 112b2

(iii) 49x2 − 36

(iv) 16x5 − 144x3

(v) (l + m)2 − (lm)2

(vi) 9x2y2 − 16

(vii) (x2 − 2xy + y2) − z2

(viii) 25a2 − 4b2 + 28bc − 49c2

Sol :

(i) 4p2 − 9q2 = (2p)2 − (3q)2

= (2p + 3q) (2p − 3q) [a2b2 = (ab) (a + b)]

(ii) 63a2 − 112b2 = 7(9a2 − 16b2)

= 7[(3a)2 − (4b)2]

= 7(3a + 4b) (3a − 4b) [a2b2 = (ab) (a + b)]

(iii) 49x2 − 36 = (7x)2 − (6)2

= (7x − 6) (7x + 6) [a2b2 = (ab) (a + b)]

(iv) 16x5 − 144x3 = 16x3(x2 − 9)

= 16 x3 [(x)2 − (3)2]

= 16 x3(x − 3) (x + 3) [a2b2 = (ab) (a + b)]

(v) (l + m)2 − (lm)2 = [(l + m) − (lm)] [(l + m) + (lm)]

[Using identity a2b2 = (ab) (a + b)]

= (l + ml + m) (l + m + lm)

= 2m × 2l

= 4ml

= 4lm

(vi) 9x2y2 − 16 = (3xy)2 − (4)2

= (3xy − 4) (3xy + 4) [a2b2 = (ab) (a + b)]

(vii) (x2 − 2xy + y2) − z2 = (xy)2 − (z)2 [(ab)2 = a2 − 2ab + b2]

= (xyz) (xy + z) [a2b2 = (ab) (a + b)]

(viii) 25a2 − 4b2 + 28bc − 49c2 = 25a2 − (4b2 − 28bc + 49c2)

= (5a)2 − [(2b)2 − 2 × 2b × 7c + (7c)2]

= (5a)2 − [(2b − 7c)2]

[Using identity (ab)2 = a2 − 2ab + b2]

= [5a + (2b − 7c)] [5a − (2b − 7c)]

[Using identity a2b2 = (ab) (a + b)]

= (5a + 2b − 7c) (5a − 2b + 7c)

 

 

Question 3

Factorise 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

Sol :

(i) ax2 + bx = a × x × x + b × x = x(ax + b)

(ii) 7p2 + 21q2 = 7 × p × p + 3 × 7 × q × q = 7(p2 + 3q2)

(iii) 2x3 + 2xy2 + 2xz2 = 2x(x2 + y2 + z2)

(iv) am2 + bm2 + bn2 + an2 = am2 + bm2 + an2 + bn2

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

= (a + b) (m2 + n2)

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

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

= (l + l) (m + 1)

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

(vii) 5y2 − 20y − 8z + 2yz = 5y2 − 20y + 2yz − 8z

= 5y(y − 4) + 2z(y − 4)

= (y − 4) (5y + 2z)

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

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

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

(ix) 6xy − 4y + 6 − 9x = 6xy − 9x − 4y + 6

= 3x(2y − 3) − 2(2y − 3)

= (2y − 3) (3x − 2)

 

 

Question 4

Factorise

(i) a4b4

(ii) p4 − 81

(iii) x4 − (y + z)4

(iv) x4 − (xz)4

(v) a4 − 2a2b2 + b4

Sol :

(i) a4b4 = (a2)2 − (b2)2

= (a2b2) (a2 + b2)

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

(ii) p4 − 81 = (p2)2 − (9)2

= (p2 − 9) (p2 + 9)

= [(p)2 − (3)2] (p2 + 9)

= (p − 3) (p + 3) (p2 + 9)

(iii) x4 − (y + z)4 = (x2)2 − [(y +z)2]2

= [x2 − (y + z)2] [x2 + (y + z)2]

= [x − (y + z)][ x + (y + z)] [x2 + (y + z)2]

= (xyz) (x + y + z) [x2 + (y + z)2]

(iv) x4 − (xz)4 = (x2)2 − [(xz)2]2

= [x2 − (xz)2] [x2 + (xz)2]

= [x − (xz)] [x + (xz)] [x2 + (xz)2]

= z(2xz) [x2 + x2 − 2xz + z2]

= z(2xz) (2x2 − 2xz + z2)

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

= (a2 b2)2

= [(ab) (a + b)]2

= (ab)2 (a + b)2

 

 

Question 5

Factorise the following expressions

(i) p2 + 6p + 8

(ii) q2 − 10q + 21

(iii) p2 + 6p − 16

Sol :

(i) p2 + 6p + 8

It can be observed that, 8 = 4 × 2 and 4 + 2 = 6

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

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

= (p + 2) (p + 4)

(ii) q2 − 10q + 21

It can be observed that, 21 = (−7) × (−3) and (−7) + (−3) = − 10

q2 − 10q + 21 = q2 − 7q − 3q + 21

= q(q − 7) − 3(q − 7)

= (q − 7) (q − 3)

(iii) p2 + 6p − 16

It can be observed that, 16 = (−2) × 8 and 8 + (−2) = 6

p2 + 6p − 16 = p2 + 8p − 2p − 16

= p(p + 8) − 2(p + 8)

= (p + 8) (p − 2)

 

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