Phân tích đa thức thành nhân tử:
a) \({x^3} - 2{x^2} + x - x{y^2}\) ;
b) \({x^2} - 7x + 12\) ;
c) \({x^2} - x - 6\) ;
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d) \(2{x^2} + x - 6\) ;
e) \({x^3} - 2x - 4\) .
\(\eqalign{ & a)\,\,{x^3} - 2{x^2} + x - x{y^2} = x\left( {{x^2} - 2x + 1 - {y^2}} \right) \cr & \,\,\,\,\, = x\left[ {{{\left( {x - 1} \right)}^2} - {y^2}} \right] = x\left( {x - 1 - y} \right)\left( {x - 1 + y} \right) \cr & b)\,\,{x^2} - 7x + 12 = {x^2} - 4x - 3x + 12 \cr & \,\,\,\,\, = x\left( {x - 4} \right) - 3\left( {x - 4} \right) = \left( {x - 4} \right)\left( {x - 3} \right) \cr & c)\,\,\,{x^2} - x - 6 = {x^2} - 3x + 2x - 6 \cr & \,\,\,\,\, = x\left( {x - 3} \right) + 2\left( {x - 3} \right) = \left( {x - 3} \right)\left( {x + 2} \right) \cr & d)\,\,2{x^2} + x - 6 = 2{x^2} + 4x - 3x - 6 \cr & \,\,\,\,\, = 2x\left( {x + 2} \right) - 3\left( {x + 2} \right) = \left( {x + 2} \right)\left( {2x - 3} \right) \cr & e)\,\,{x^3} - 2x - 4 = {x^3} - 2x - 8 + 4 \cr & \,\,\,\,\, = \left( {{x^3} - 8} \right) - \left( {2x - 4} \right) = \left( {{x^3} - {2^3}} \right) - 2\left( {x - 2} \right) \cr & \,\,\,\,\, = \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) - 2\left( {x - 2} \right) \cr & \,\,\,\,\, = \left( {x - 2} \right)\left( {{x^2} + 2x + 4 - 2} \right) \cr & \,\,\,\,\, = \left( {x - 2} \right)\left( {{x^2} + 2x + 2} \right) \cr} \)