Q. 114.3( 9 Votes )

# Let’s factorise :

i.

ii.

iii.

iv.

v.

vi.

vii.

viii.

ix.

x.

xi.

xii.

xiii.

Answer :

_{i. 225m}^{2} – 100n^{2}

= (15m)^{2} – (10n)^{2}

[∵ a^{2} – b^{2} = (a – b)(a + b)]

Here, a = 15m, b = 10n

= (15m – 10n)(15m + 10n)

ii.

[∵ a^{2} – b^{2} = (a – b)(a + b)]

Here, a = 5x,

iii. 7ax^{2} + 14ax + 7a

= 7a(x^{2} + 2x + 1)

Since, a^{2} + 2ab + b^{2}= (a + b)^{2}

Here, a = x, b = 1

= 7a(x + 1)^{2}

iv. 3x^{4} – 6x^{2}a^{2} + 3a^{4}

= 3(x^{4} – 2x^{2}a^{2} + a^{4})

= 3[(x^{2})^{2} – 2(x^{2})(a^{2}) + (a^{2})^{2}]

Since, a^{2} + 2ab + b^{2}= (a + b)^{2}

Here, a = x^{2}, b = a^{2}

= 3(x^{2} – a^{2})^{2}

[∵ a^{2} – b^{2} = (a – b)(a + b)]

Here, a = x, b = a

= 3[(x – a)(x + a)]^{2}

= 3(x – a)^{2}(x + a)^{2}

v. 4b^{2}c^{2} – (b^{2} + c^{2} – a^{2})^{2}

= (2bc)^{2} – (b^{2} + c^{2} – a^{2})^{2}

= (2bc + b^{2} + c^{2} – a^{2}) (2bc – (b^{2} + c^{2} – a^{2})) [∵ a^{2} – b^{2} = (a – b)(a + b)]

= (b^{2} + c^{2} + 2bc – a^{2}) (2bc – b^{2} – c^{2} + a^{2})

[∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

So, if a = b, b = c in b^{2} + c^{2} + 2bc then after applying the identity we get,

b^{2} + c^{2} + 2bc = (b + c)^{2}

= [(b + c)^{2} – a^{2}] [a^{2} – (b^{2} + c^{2} – 2bc)]

= (b + c – a) (b + c + a)[a^{2}– (b – c)^{2}]

[∵ (a – b)^{2} = a^{2} – 2ab + b^{2}_{,} 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. 64ax^{2} – 49a(x – 2y)^{2}

= a(64x^{2} – 49(x – 2y)^{2})

= a[(8x)^{2} – (7(x – 2y))^{2}]

[∵ a^{2} – b^{2} = (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. x^{2} – 9 – 4xy + 4y^{2}

= x^{2} – 4xy + 4y^{2} – 9

= x^{2} – 2(x)(2y) + (2y)^{2} – 9

Since, a^{2} - 2ab + b^{2}= (a - b)^{2}

Here, a = x, b = 2y

= (x – 2y)^{2} - 3^{2}

[∵ a^{2} – b^{2} = (a – b)(a + b)]

Here, a = x – 2y, b = 3

= (x – 2y + 3)(x – 2y – 3) [∵ a^{2} – b^{2} = (a – b)(a + b)]

viii. x^{2} – 2x – y^{2} + 2y

= x^{2} – y^{2} – 2x + 2y

= (x + y)(x – y) – 2(x – y)

[∵ a^{2} – b^{2} = (a – b)(a + b), here a = x, b = y]

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

ix. 3 + 2a – a^{2}

= 2 + 1 + 2a – a^{2}

= 2 + 2a + 1^{2} – a^{2}

= 2(1 + a) + (1 – a)(1 + a)

[∵ a^{2} – b^{2} = (a – b)(a + b), here a = 1, b = a]

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

= (1 + a)(3 – a)

x. x^{4} – 1

= (x^{2})^{2} - 1^{2}

= (x^{2} – 1)(x^{2} + 1)

[∵ a^{2} – b^{2} = (a – b)(a + b), here a = x^{2}, b = 1]

= (x – 1)(x + 1)(x^{2} + 1)

[∵ a^{2} – b^{2} = (a – b)(a + b), here a = x, b = 1]

xi. a^{2}– b^{2} – c^{2} + 2bc

= a^{2} – (b^{2} + c^{2} – 2bc)

= a^{2} – (b - c)^{2}

[∵ (a – b)^{2} = a^{2} – 2ab + b^{2}, here a = b and b = c]

= (a + b – c)(a – (b – c))

[∵ a^{2} – b^{2} = (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. x^{4} + x^{2}y^{2} + y^{4}

= x^{4} + 2x^{2}y^{2} – x^{2}y^{2} + y^{4}

= x^{4} + 2x^{2}y^{2} + y^{4} – x^{2}y^{2}

= (x^{2})^{2} + 2(x^{2})(y^{2}) + (y^{2})^{2} – (xy)^{2}

= (x^{2} + y^{2})^{2} – (xy)^{2}

[∵ (a + b)^{2} = a^{2} + 2ab + b^{2}, here a = x^{2}, b = y^{2}]

= (x^{2} + y^{2} + xy)(x^{2} + y^{2} – xy)

[∵ a^{2} – b^{2} = (a – b)(a + b), here a = x^{2} + y^{2} and b = xy]

Rate this question :

Let’s factorise :

i.

ii.

iii.

iv.

v.

vi.

vii.

viii.

ix.

x.

xi.

xii.

xiii.

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andLet’s form a whole square of the expressions given below using the above identities-

i.

ii.

iii.

iv.

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andLet’s find out the square of the expressions given below using the above identities –

i.

ii.

iii.

iv.

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Let’s find the product by formulae;

i.

ii.

iii.

iv.

v.

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Let’s express the following expressions as a difference of two squares:

i.

ii.

iii. x

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Let’s fill up the table below :

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Let’s fill up the table below:

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Let’s write the area of the rectangular region whose the length is 8x+3y cm. and breadth is 8x – 3y cm.

West Bengal - MathematicsLet’s solve:

i.

ii.

iii.

iv.

v.

vi.

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