factorise quest.no 15 i m very heartly thanful to you
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hope it will help you
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Hello Mate !!
आपका समाधान यहां है:
= ( 3a - 1 )^2 - 6a + 2
By taking 2 as common ,
= ( 3a - 1 )^2 - 2 ( 3a - 1 )
By taking ( 3a - 1 ) as common ,
= ( 3a - 1 ) ( 3a - 1 - 2 )
= ( 3a - 1 ) ( 3a - 3 )
By taking 3 as common ,
= ( 3a - 1 ) 3 ( a - 1 )
Rearranging the terms ,
= 3 ( a - 1 ) ( 3a - 1 )
===============================
आशा करता हूँ की ये काम करेगा !
आपका समाधान यहां है:
= ( 3a - 1 )^2 - 6a + 2
By taking 2 as common ,
= ( 3a - 1 )^2 - 2 ( 3a - 1 )
By taking ( 3a - 1 ) as common ,
= ( 3a - 1 ) ( 3a - 1 - 2 )
= ( 3a - 1 ) ( 3a - 3 )
By taking 3 as common ,
= ( 3a - 1 ) 3 ( a - 1 )
Rearranging the terms ,
= 3 ( a - 1 ) ( 3a - 1 )
===============================
आशा करता हूँ की ये काम करेगा !
Anonymous:
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