Question

Phân tích đa thức thành nhân tử :

\(P=x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)

Answer 1

\(P=x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)

\(=x^2\left(y-x+x-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)

\(=x^2\left(y-x\right)-z^2\left(y-x\right)+x^2\left(x-z\right)-y^2\left(x-z\right)\)

\(=\left(x^2-z^2\right)\left(y-x\right)+\left(x^2-y^2\right)\left(x-z\right)\)

\(=\left(x-z\right)\left(x+z\right)\left(y-x\right)-\left(y-x\right)\left(x-z\right)\left(x+y\right)\)

\(=\left(x-z\right)\left(y-x\right)\left(x+z-x-y\right)\)

\(=\left(x-z\right)\left(y-x\right)\left(z-y\right)\)

Answer 2

x^2(y-z)+y^2(z-x)+z^2(x-y)=x2(y−z)+y2(z−y+y−x)+z2(x−y)=x2(y−z)−y2(y−z)−y2(x−y)+z2(x−y)=(x−y)(x+y)(y−z)+(z−y)(y+z)(x−y)=(x−y)(y−z)(x−z)
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