Question
Factorise : $5 - (3a^2- 2a) (6 - 3a^2+ 2a)$
✓
Answer
$5 - (3a^2- 2a) (6 - 3a^2+ 2a)$
$= 5 - ( 3a^2 - 2a )[ 6 - ( 3a^2 - 2a )]$
Assume that $3a^2 - 2a = x$
Therefore,
$5 - ( 3a^2 - 2a )( 6 - 3a^2 + 2a )$
$= 5 - x( 6 - x )$
$= 5 - 6x + x^2$
$= 5( 1 - x ) - x( 1 - x )$
$= ( 5 - x )( 1 - x )$
$= ( x - 5 )( x - 1 )$
$= ( 3a^2 - 2a - 5 )( 3a^2 - 2a - 1 )$
$= ( 3a^2 - 5a + 3a -5 )(3a^2 - 3a + a - 1 )$
$= [ a( 3a - 5 ) + 1( 3a - 5)][3a( a - 1) + 1( a - 1)]$
$= ( 3a - 5 )( a + 1 )( 3a + 1 )( a - 1 )$
$= 5 - ( 3a^2 - 2a )[ 6 - ( 3a^2 - 2a )]$
Assume that $3a^2 - 2a = x$
Therefore,
$5 - ( 3a^2 - 2a )( 6 - 3a^2 + 2a )$
$= 5 - x( 6 - x )$
$= 5 - 6x + x^2$
$= 5( 1 - x ) - x( 1 - x )$
$= ( 5 - x )( 1 - x )$
$= ( x - 5 )( x - 1 )$
$= ( 3a^2 - 2a - 5 )( 3a^2 - 2a - 1 )$
$= ( 3a^2 - 5a + 3a -5 )(3a^2 - 3a + a - 1 )$
$= [ a( 3a - 5 ) + 1( 3a - 5)][3a( a - 1) + 1( a - 1)]$
$= ( 3a - 5 )( a + 1 )( 3a + 1 )( a - 1 )$
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