# Let’s factorise :i. ii. iii. iv. v. vi. vii. viii. ix. x. xi. xii. xiii. i. 225m2 – 100n2

= (15m)2 – (10n)2

[ a2 – b2 = (a – b)(a + b)]

Here, a = 15m, b = 10n

= (15m – 10n)(15m + 10n)

ii.  [ a2 – b2 = (a – b)(a + b)]

Here, a = 5x,  iii. 7ax2 + 14ax + 7a

= 7a(x2 + 2x + 1)

Since, a2 + 2ab + b2= (a + b)2

Here, a = x, b = 1

= 7a(x + 1)2

iv. 3x4 – 6x2a2 + 3a4

= 3(x4 – 2x2a2 + a4)

= 3[(x2)2 – 2(x2)(a2) + (a2)2]

Since, a2 + 2ab + b2= (a + b)2

Here, a = x2, b = a2

= 3(x2 – a2)2

[ a2 – b2 = (a – b)(a + b)]

Here, a = x, b = a

= 3[(x – a)(x + a)]2

= 3(x – a)2(x + a)2

v. 4b2c2 – (b2 + c2 – a2)2

= (2bc)2 – (b2 + c2 – a2)2

= (2bc + b2 + c2 – a2) (2bc – (b2 + c2 – a2)) [ a2 – b2 = (a – b)(a + b)]

= (b2 + c2 + 2bc – a2) (2bc – b2 – c2 + a2)

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

So, if a = b, b = c in b2 + c2 + 2bc then after applying the identity we get,

b2 + c2 + 2bc = (b + c)2

= [(b + c)2 – a2] [a2 – (b2 + c2 – 2bc)]

= (b + c – a) (b + c + a)[a2– (b – c)2]

[ (a – b)2 = a2 – 2ab + b2, Here a = b, b = c]

= (b + c – a) (b + c + a)(a – (b – c))(a + b – c)

= (b + c – a) (b + c + a)(a – b + c)(a + b – c)

vi. 64ax2 – 49a(x – 2y)2

= a(64x2 – 49(x – 2y)2)

= a[(8x)2 – (7(x – 2y))2]

[ a2 – b2 = (a – b)(a + b)]

Here, a = 8x, b = 7(x – 2y)

= a[8x + 7(x – 2y)][8x – 7(x – 2y)]

= a(8x + 7x – 14y)(8x - 7x + 14y)

= a(15x – 2y)(x + 14y)

vii. x2 – 9 – 4xy + 4y2

= x2 – 4xy + 4y2 – 9

= x2 – 2(x)(2y) + (2y)2 – 9

Since, a2 - 2ab + b2= (a - b)2

Here, a = x, b = 2y

= (x – 2y)2 - 32

[ a2 – b2 = (a – b)(a + b)]

Here, a = x – 2y, b = 3

= (x – 2y + 3)(x – 2y – 3) [ a2 – b2 = (a – b)(a + b)]

viii. x2 – 2x – y2 + 2y

= x2 – y2 – 2x + 2y

= (x + y)(x – y) – 2(x – y)

[ a2 – b2 = (a – b)(a + b), here a = x, b = y]

= (x – y)(x + y – 2) [Taking (x – y) as common]

ix. 3 + 2a – a2

= 2 + 1 + 2a – a2

= 2 + 2a + 12 – a2

= 2(1 + a) + (1 – a)(1 + a)

[ a2 – b2 = (a – b)(a + b), here a = 1, b = a]

= (1 + a)(2 + 1 – a) [Taking (1 + a) as common]

= (1 + a)(3 – a)

x. x4 – 1

= (x2)2 - 12

= (x2 – 1)(x2 + 1)

[ a2 – b2 = (a – b)(a + b), here a = x2, b = 1]

= (x – 1)(x + 1)(x2 + 1)

[ a2 – b2 = (a – b)(a + b), here a = x, b = 1]

xi. a2– b2 – c2 + 2bc

= a2 – (b2 + c2 – 2bc)

= a2 – (b - c)2

[ (a – b)2 = a2 – 2ab + b2, here a = b and b = c]

= (a + b – c)(a – (b – c))

[ a2 – b2 = (a – b)(a + b), here a = a, b = b - c]

= (a + b – c)(a – b +c)

xii. ac + bc + a + b

= ac + a + bc + b

= a(c + 1) + b(c + 1)

= (c + 1)(a + b) [Taking (c + 1) as common]

xiii. x4 + x2y2 + y4

= x4 + 2x2y2 – x2y2 + y4

= x4 + 2x2y2 + y4 – x2y2

= (x2)2 + 2(x2)(y2) + (y2)2 – (xy)2

= (x2 + y2)2 – (xy)2

[ (a + b)2 = a2 + 2ab + b2, here a = x2, b = y2]

= (x2 + y2 + xy)(x2 + y2 – xy)

[ a2 – b2 = (a – b)(a + b), here a = x2 + y2 and b = xy]

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