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Factorise : (A2 - A) (4a2 - 4a - 5) - 6

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Question

Factorise : (a2 - a) (4a2 - 4a - 5) - 6

Sum
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Solution

Let us assume, a2 - a = x
Then the given expression is,
(a2 - a) (4a2 - 4a - 5) - 6
= x( 4x - 5 ) - 6
= 4x2 - 5x - 6
= 4x2 - 8x + 3x - 6
= 4x( x - 2 ) + 3( x - 2 )
= ( 4x + 3 )( x - 2 )
= [ 4( a2 - a ) + 3 ]( a2 - a - 2 )          [ resubstitute the value of x ]
= [ 4a2 - 4a + 3 ]( a2 - a - 2 )
= [ 4a2 - 4a + 3 ]( a2 - 2a + a - 2 )
= [ 4a2 - 4a + 3 ][ a( a - 2 ) + 1( a - 2 )]
= [ 4a2 - 4a + 3 ]( a - 2 )( a + 1 )

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Chapter 5: Factorisation - Exercise 5 (E) [Page 76]

APPEARS IN

Selina Concise Mathematics [English] Class 9 ICSE
Chapter 5 Factorisation
Exercise 5 (E) | Q 10 | Page 76

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