Ta có: \(x^3-5x^2+8x-4\)
\(=x^3-4x^2+4x-x^2+4x-4\)
\(=x\left(x^2-4x+4\right)-\left(x^2-4x+4\right)\)
\(=\left(x-2\right)^2\left(x-1\right)\)
Vậy \(A=\left(x-1\right)\left(x-2\right)^2\)
Ta có : \(x^3-5x^2+8x-4\)\(\Leftrightarrow x^3-x^2-4x^2+4x+4x-4\)\(\Leftrightarrow x^2.\left(x-1\right)-4x.\left(x-1\right)+4.\left(x-1\right)\)\(\Leftrightarrow\left(x-1\right).\left(x^2-4x+4\right)\)\(\Leftrightarrow\left(x-1\right).\left(x-2\right)^2\)
<=>x3-x2-4x2+4x+4x-4
<=>x2(x-1)-4x(x-1)+4(x-1)
<=>(x2-4x+4)(x-1)
<=>(x-2)2(x-1)
\(x^3-5x^2+8x-4\)
\(=\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)\)
\(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
