x square+4y square+4yx-64 factorise
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x² + 4y² + 4yx - 64
= ( x )² + ( 2y )² + 4xy - 64
= [ ( x )² + ( 2y )² + 2 ( x ) ( 2y ) ] - 64
= ( x + 2y )² - 64
= ( x + 2y )² - ( 8 )²
= ( x + 2y + 8 ) ( x + 2y - 8 )
= ( x )² + ( 2y )² + 4xy - 64
= [ ( x )² + ( 2y )² + 2 ( x ) ( 2y ) ] - 64
= ( x + 2y )² - 64
= ( x + 2y )² - ( 8 )²
= ( x + 2y + 8 ) ( x + 2y - 8 )
Answered by
1
Answer:
Step-by-step explanation:
x^2 + 4y^2 + 4xy - 64 =
= (x + 2y)^2 - 8^2 =
= (x + 2y -8)(x + 2y + 8)
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