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Factorise : (A2 + B2 - 4c2)2 - 4a2b2 - Mathematics

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प्रश्न

Factorise : (a2 + b2 - 4c2)2 - 4a2b2

योग

उत्तर

(a2 + b2 - 4c2)2 - 4a2b2
= ( a2 + b2 - 4c2 )2 - ( 2ab )2
= ( a2 + b2 - 4c2 - 2ab )( a2 + b2 - 4c2 + 2ab )         [ ∵ a2 - b2 = ( a + b )( a - b )]
= ( a2 + b2 - 2ab - 4c2 )( a2 + b2 + 2ab - 4c2 )
= [ ( a - b )2 - ( 2c )2 ][ ( a + b )2 - ( 2c )2]
= ( a - b + 2c )( a - b - 2c )( a + b + 2c )( a + b - 2c )

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Method of Factorisation : Difference of Two Squares
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Factorisation - Exercise 5 (C) [पृष्ठ ७३]

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सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 5 Factorisation
Exercise 5 (C) | Q 22 | पृष्ठ ७३
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