Question

x^3-5x^2+8x-4

Answer 1

Ta có \(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4.\)

 \(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-4x+4\right)\)

\(=\left(x-1\right)\left(x-2\right)^2\)

Vậy \(x^3-5x^2+8x-4=\left(x-1\right)\left(x-2\right)^2\)

Answer 2

x^3-5x^2+8x-4

=x^3-x^2-4x^2+4x+4x-4

=x^2(x-1)-4x(x-1)+4(x-1)

=(x-1)(x^2-4x+4)

=(x-1)(x-2)^2