Question

factorise x³-4x²+x+6 using factorisation theorem.(with all steps please)

Answer 1

x³ - 4 x² + x + 6

==> x³ - 2 x² - 2 x²  + x + 6 [ Split - 4 x² ]

==> x³ - 2 x² - ( 2 x² - x - 6 ) [ Take - 1 as common ]

==> x² ( x - 2 ) - ( 2 x² - 4 x + 3 x - 6 ) [ Take x² as common ]

==> x² ( x - 2 ) - [ 2 x ( x - 2 ) + 3 ( x - 2 ) ] [ Take commons ]

==> x² ( x - 2 ) - ( 2 x + 3 )( x - 2 )

==> ( x - 2 )( x² - 2 x - 3 ) [ x - 2 is common ]

==> ( x - 2 )( x² - 3 x + x - 3 ) [ Split and solve ]

==> ( x - 2 )( x ( x - 3 ) + 1 ( x - 3 ))

==> ( x - 2 )( x - 3 )( x +  1 )

The answer is ( x - 2 )( x - 3 )( x +  1 )

#JISHNU#

Hope it helps u :-)

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Answer 2

Answer like
x^3-4x^2+x+6

${x}^{3} - 3 {x}^{2} - {x}^{2} + 3x - 2x + 6$
${x}^{2} (x - 3) - x(x - 3) - 2(x - 3)$
$(x - 3)( {x}^{2} - x - 2)$
hope helpful