Math, asked by mohitrack12300, 1 year ago

factorise the following using appropriate identifies (i) 9x² + 6xy + y² , (ii) 4y² - 4y + 1 , (iii) x² - y²/100

Answers

Answered by originaljustice
51
may this help u.......
Attachments:

RMS0307: Superb...
himanshu7068: \mathbb{ ANSWER }ANSWER :

(i) 9x² + 9xy + y²

[ using : (a+b)² = a² + 2ab + b² ]
Here, 
a = 3x , b = y

So, 
=> (3x)² + 2*3x*y + (y)²
=> (3x + y)²
=> (3x + y) (3x + y) 

➖➖➖➖➖➖➖➖➖➖

(ii) 4y² - 4y + 1

[using : (a - b)² = a² - 2ab + b² ]
Here, 
a = 2y , b = 1

So, 
=> (2y)² / 2*2y*1 + (1)²
=> (2y - 1)²
=> (2y - 1) (2y - 1)

➖➖➖➖➖➖➖➖➖➖

(iii) x² - y²/100

[ using : a² - b² = (a+b) (a-b)
sprao534: the solution of first is wrong
sprao534: 9xv2+6x+1 but not 9xv2+9x+1
Answered by BloomingBud
64
\mathbb{ ANSWER } :

(i) 9x² + 9xy + y²

[ using : (a+b)² = a² + 2ab + b² ]
Here,
a = 3x , b = y

So,
=> (3x)² + 2*3x*y + (y)²
=> (3x + y)²
=> (3x + y) (3x + y)

➖➖➖➖➖➖➖➖➖➖

(ii) 4y² - 4y + 1

[using : (a - b)² = a² - 2ab + b² ]
Here,
a = 2y , b = 1

So,
=> (2y)² / 2*2y*1 + (1)²
=> (2y - 1)²
=> (2y - 1) (2y - 1)

➖➖➖➖➖➖➖➖➖➖

(iii) x² - y²/100

[ using : a² - b² = (a+b) (a-b) ]
Here,
a = x
b = \frac{y}{100}

So,
 = > {(x)}^{2} - {( \frac{y}{10}) }^{2} \\ \\ = >( x + \frac{y}{10} )(x - \frac{y}{10} )

sprao534: 9xv2+6x+1 but not 9xv2+9x+1
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