Phân tích đa thức thành nhân tử a)x^4-2x^3+2x-1 b)x^4+2x^3+2x^2+2x+1

\[\begin{array}{l} a){x^4} - 2{x^3} + 2x - 1\\ = \left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right) - 2x\left( {{x^2} - 1} \right)\\ = \left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right) - 2x\left( {x - 1} \right)\left( {x + 1} \right)\\ = \left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1 - 2x} \right)\\ = \left( {x - 1} \right)\left( {x + 1} \right){\left( {x - 1} \right)^2}\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ b){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = \left( {{x^4} + 2{x^3} + {x^2}} \right) + \left( {{x^2} + 2x + 1} \right)\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = {x^2}{\left( {x + 1} \right)^2} + {\left( {x + 1} \right)^2}\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array}\]

Đáp án:

Giải thích các bước giải: \[\begin{array}{l} a)\,\,\,{x^4} - 2{x^3} + 2x - 1\\ = \left( {{x^4} - 1} \right) - 2x\left( {{x^2} - 1} \right)\\ = \left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right) - 2x\left( {{x^2} - 1} \right)\\ = \left( {{x^2} - 1} \right)\left( {{x^2} - 2x + 1} \right)\\ = \left( {x - 1} \right)\left( {x + 1} \right){\left( {x - 1} \right)^2}\\ = \left( {x + 1} \right){\left( {x - 1} \right)^3}\\ b)\,\,\,{x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = \left( {{x^4} + 2{x^2} + 1} \right) + \left( {2{x^3} + 2x} \right)\\ = {\left( {{x^2} + 1} \right)^2} + 2x\left( {{x^2} + 1} \right)\\ = \left( {{x^2} + 1} \right)\left( {{x^2} + 1 + 2x} \right)\\ = \left( {{x^2} + 1} \right){\left( {x + 1} \right)^2} \end{array}\]