Q. 114.3( 9 Votes )

Let’s factorise :

i.

ii.

iii.

iv.

v.

vi.

vii.

viii.

ix.

x.

xi.

xii.

xiii.

Answer :

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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