Question

Factorise the following.

iam waiting for answer plz answer me fast​

Answer 1

Question:-

Factorize the following:-

a) $\sf{(a-b)^2 - (b-c)^2}$

b) $\sf{64 - 25x^2 + 60xy - 36y^2}$

c) $\sf{x^4 - 1}$

d) $\sf{(x-2y)^2-(2x+y)^2}$

e) $\sf{36x^2 - 49(y-x)^2}$

f) $\sf{(p-q)^2-81}$

g) $\sf{4p^2 - 12pq + 9q^2 - 64}$

h) $\sf{(4b-3c)^2 - (3b+2c)^2}$

i) $\sf{c^2 - a^2 - b^2 - 2ab}$

Solutions:-

a) $\sf{(a-b)^2 - (b-c)^2}$

Here we can see that the equation is in the form of (a² - b²) where we will take $\sf{a=(a-b)^2-(b-c)^2}$. So we will apply:-

$\sf{a^2-b^2 = (a+b)(a-b)}$

= $\sf{(a-b)^2 - (b-c)^2}$

= $\sf{[(a-b)+(b-c)][(a-b)-(b-c)]}$

= $\sf{(a-b+b-c)(a-b-b+c)}$

(-b and +b) cancel out and (-b and -b) add up

= $\sf{(a-c)(a-2b-c)}$

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b) $\sf{64 - 25x^2 + 60xy - 36y^2}$

Taking (-) minus as common,

= $\sf{64 - [25x^2 - 60xy + 36y^2]}$

= $\sf{64-[(5x)^2-2\times5x\times6y+(6y)^2]}$

$\sf{(5x)^2-2\times5x\times6y+(6y)^2}$ is the expanded form of (5x - 6y)²

= $\sf{(8)^2 - (5x-6y)^2}$

= $\sf{[(8)-(5x-6y)][(8)+(5x-6y)]\:\:\:\:[\because a^2 - b^2 = (a+b)(a-b)]}$

= $\sf{(8-5x+6y)(8+5x-6y)}$

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c) $\sf{x^4-1}$

= $\sf{(x^2)^2 - (1)^2}$

By applying a² - b² = (a + b)(a -b)

= $\sf{(x^2+1)(x^2-1)}$

= $\sf{(x^2+1)(x^2-1^2)}$

Breaking x² - 1² again,

$\sf{(x^2+1)(x+1)(x-1)}$

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d) $\sf{(x-2y)^2-(2x+y)^2}$

By applying a² - b² = (a + b)(a -b)

= $\sf{[(x-2y)-(2x+y)][(x-2y)+(2x+y)]}$

= $\sf{(x-2y-2x-y)(x-2y+2x+y)}$

= $\sf{(-3y-x)(3x-y)}$

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e) $\sf{36x^2-49(y-x)^2}$

= $\sf{(6x)^2 - [7(y-x)]^2}$

By applying a² - b² = (a + b)(a -b)

= $\sf{[6x+7(y-x)][6x-7(y-x)]}$

= $\sf{(6x+7y-7x)(6x-7y+7x)}$

= $\sf{(7y-x)(13x-7y)}$

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f) $\sf{(p+q)^2-81}$

= $\sf{(p+q)^2-(9)^2}$

= $\sf{[(p+q)-(9)][(p+q)+9]}$

= $\sf{(p+q-9)(p+q+9)}$

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g) $\sf{4p^2 - 12pq + 9q^2 - 64}$

= $\sf{[(2p)^2 - 2\times2p\times3q + (3q)^2]-(8)^2}$

(2p)² - 2×2p×3q + (3q)² is the expanded form of (2p - 3q)²

= $\sf{(2p-3q)^2 - (8)^2}$

By applying a² - b² = (a + b)(a -b)

= $\sf{(2p-3q+8)(2p-3q-8)}$

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h) $\sf{(4b-3c)^2 - (3b+2c)^2}$

By applying a² - b² = (a + b)(a -b)

= $\sf{[(4b-3c)+(3b+2c)][(4b-3c)-(3b+2c)]}$

= $\sf{(4b-3c+3b+2c)(4b-3c-3b-2c)}$

= $\sf{(7b-c)(b-5c)}$

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i) $\sf{c^2 - a^2 - b^2 - 2ab}$

Taking (-) minus common.

= $\sf{c^2 - (a^2+b^2+2ab)}$

a² + b² + 2ab is the expanded form of (a+b)²

= $\sf{c^2-(a+b)^2}$

By applying a² - b² = (a + b)(a -b)

= $\sf{[c-(a+b)][c+(a+b)]}$

= $\sf{(c-a-b)(c+a+b)}$

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