wiz-icon
MyQuestionIcon
MyQuestionIcon
7
You visited us 7 times! Enjoying our articles? Unlock Full Access!
Question

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

Open in App
Solution

(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)


flag
Suggest Corrections
thumbs-up
26
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Factorisation by Regrouping Terms
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon