Question

factorize x^2-(x-y)^2​

Answer 1

Given :

x² - ( x-y )²

To find :

Factor

Solution :

Method 1)

• x² - ( x-y )²

[ Note - Using identity ( a² - b²) = (a+b)(a-b) ]

Here ,

• a = x
• b = ( x-y )

➝ x² - ( x-y )² = [ (x) + (x-y) ] × [ (x) - (x-y ) ]

➝ x² - ( x-y )² = [ x + x - y ] × [ x - x + y ]

➝ x² - ( x-y )² = ( 2x - y ) ( y)

➝ x² - ( x-y )² = y(2x-y)

• Therefore factor are , y and (2x-y)

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Method 2)

• x² - ( x-y )²

[ Note - Using identity ( p-q )² = p² + q² - 2pq

Here

• p = x
• q = y

➝ x² - ( x-y )² = x² - [ (x)² + (y)² - (2)(x)(y) ]

➝ x² - ( x-y )² = x² - x² - y² + 2xy

➝ x² - ( x-y )² = 2xy - y²

➝ x² - ( x-y )² = y(2x-y)

• Therefore factor are , y and (2x-y)

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ANSWER : y(2x-y)

Answer 2

✾$\pink{\textsf{Required Solution}}\!\!$✾

Method 1:

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

$\dashrightarrow \underbrace{ \sf \{x + (x - y) \} \{x - (x - y) \}}_{ \tiny\pink{\boxed {\sf{}identity \: used \to{a}^{2} - {b}^{2} = (a + b)(a - b) }}}$

$\dashrightarrow \sf \{x + x - y \} \{x - (x - y) \}$

$\dashrightarrow \sf \{x + x - y \} \{x - x + y \}$

$\dashrightarrow \sf \{x + x - y \} \{ \cancel{x } - \cancel{x} + y \}$

$\dashrightarrow \sf \{x + x - y \} \{ 0 + y \}$

$\dashrightarrow \sf \{x + x - y \} \{ y \}$

$\dashrightarrow \sf \{2x - y \} \{ y \}$

$\dashrightarrow \sf y\{2x - 1\}$

Method 2:

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

$\dashrightarrow \sf {x}^{2} - \underbrace{(x {}^{2} + y {}^{2} - 2 \: \times \: x \: \times \: y ) }_{\tiny\pink{\boxed {\sf identity \: used \to {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab}}}$

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

$\dashrightarrow \sf {x}^{2} - x {}^{2} - y {}^{2} + 2 x y$

$\dashrightarrow \sf\cancel{ {x}^{2}} -\cancel{ x {}^{2} } - y {}^{2} + 2 x y$

$\dashrightarrow \sf- y {}^{2} + 2 x y$

$\dashrightarrow \sf y(2 x - 1)$

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Some more formals related to this question:

→(a+b)²=a²+b²+2ab

→(a-b)(a+b)=a²-b²

→(a +b+c)²= a² +b² +c²+2ab + 2bc + 2ca

→(a +b-c)²= a² +b² + c²+2ab - 2bc- 2ca

→(a - b + c)²= a²+ b² + c² - 2ab - 2bc + 2ca

→(a - b - c)² = a² + b²+ c²- 2ab+2bc-2ca

→(a+b)³=a³+b³+3ab(a+b)

→(a-b)³=a³+b³-3ab(a-b)

→a³ - b³ = (a - b)(a²+ ab + b²)