Phân tích đa thức thành nhân tử:
a) \({x^2} + 2x \)
b) \({x^3} - 5{x^2} + 5x - 1 \)
A.\(\begin{array}{l}a)\,\,x\left( {x + 2} \right)\\b)\,\,\left( {x - 1} \right)\left( {{x^2} - 4x + 1} \right)\end{array}\)
B.\(\begin{array}{l}a)\,\,x\left( {x + 2} \right)\\b)\,\,\left( {x - 1} \right){\left( {x - 2} \right)^2}\end{array}\)
C.\(\begin{array}{l}a)\,\,x\left( {x - 2} \right)\\b)\,\,\left( {x + 1} \right){\left( {x - 2} \right)^2}\end{array}\)
D.\(\begin{array}{l}a)\,\,x\left( {x - 2} \right)\\b)\,\,\left( {x + 1} \right)\left( {{x^2} - 4x + 1} \right)\end{array}\)
a) \({x^2} + 2x \)
b) \({x^3} - 5{x^2} + 5x - 1 \)
A.\(\begin{array}{l}a)\,\,x\left( {x + 2} \right)\\b)\,\,\left( {x - 1} \right)\left( {{x^2} - 4x + 1} \right)\end{array}\)
B.\(\begin{array}{l}a)\,\,x\left( {x + 2} \right)\\b)\,\,\left( {x - 1} \right){\left( {x - 2} \right)^2}\end{array}\)
C.\(\begin{array}{l}a)\,\,x\left( {x - 2} \right)\\b)\,\,\left( {x + 1} \right){\left( {x - 2} \right)^2}\end{array}\)
D.\(\begin{array}{l}a)\,\,x\left( {x - 2} \right)\\b)\,\,\left( {x + 1} \right)\left( {{x^2} - 4x + 1} \right)\end{array}\)
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