Question

solve the equation by factorisation method

Answer 1

= abx² - (b²-ac)x - bc

= abx² - b²x + acx - bc

= bx(ax-b) + c(ax-b)

= (bx+c)(ax-b)

Hence the roots are (bx+c)(ax-b)
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Solution → x = -c/b and b/a
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Answer 2

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( i ) abx² + ( b² - ac )x - bc = 0

General Solutions Include : 
     abx² + b²x - acx - bc = 0
=> ( bx - c )( ax + b ) = 0
=>  x = ( c / b ) ; ( -b / a )

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Proper method to do this is :
♦ abx² + ( b² - ac ) x - bc = 0 

♦ A = ab ; B = ( -bc ) 
=> AB = ( -ab²c ) = ( b² )( -ac ) 

♦ The above step assures that the Factorization is altered so that the middle term split turns out to be CONVENIENT .

Once it is done, by middle term split :
♦ abx² + b²x - acx - bc = 0

=> bx ( ax + b ) - c ( ax + b ) = 0
=> ( bx - c )( ax + b ) = 0

Further, it cannot be possible for the equation to hold if both of the factors turn out to be non-zero 
=> Either ( bx - c ) = 0    or    ( ax + b ) = 0
=> Either [ x = ( c / b )   or  x = ( -b / a ) ]

Thus, the roots are given by :
$x_1 = \frac{c}{b} \ \ \ \ \ x_2 = \frac{-b}{a}$

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Hope that helps