Question

8-50y square z square
$8 - 50y {}^{2} z {}^{2}$
Factorise the following

Answer 1

= > 8 - 50 y² z²


= > 2{ 4 - 25 y² z² }


= > 2{ ( 2 )² - ( 5 yz )² }



We know, a² - b² = ( a + b )( a - b )

Assume 2 as a and 5yz as b

So,



= > 2{ ( 2 + 5yz ) ( 2 - 5yz ) }


$.$

= > 2( 2 + 5yz )( 2 - 5yz )

Answer 2

Solution -  ​​​  

8 - 50y²z²      

= 2[4 - 25y²z²]  ​

= 2[(2)² - (5yz)²]  

= 2[(2 + 5yz) (2 - 5yz)]

Answer : 2[(2 + 5yz) (2 - 5yz)]  

Explaination -  

Identity used :    a² - b² = (a + b) (a - b)  

At first, we had taken the common factor of 8 and 50y²z², that is 2    

as 2 × 4 = 8  

2 × 25 × y² × z² = 50y²z²    

Then, we got    

2[4 - 25y²z²] = 2[(2)² - (5yz)²]  

Here, we had used the identity    

a² - b² = (a + b) (a - b) ​ ​

 where,    

a² = (2)²  

b² = (5yz)²  

According to identity,  

a² - b² = (2)² - (5yz)²  

= (a + b) (a - b) = (2 + 5yz) (2 - 5yz)  

And “2” was the common factor  

So, the answer is    

8 - 50y²z² = 2[(2 + 5yz) (2 - 5yz)    

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