Question

Phân tích các đa thức thành nhân tử:

x^5+x^3-x^2-1

x^4-3x^3-x+3

x^3-x^2y-xy^2+y^3

3x+3y-x^2-2xy-y^2

Answer 1

\(A=x^5+x^3-x^2-1=x^3\left(x^2+1\right)-\left(x^2+1\right)=\left(x^2+1\right)\left(x^3-1\right)=\left(x^2+1\right)\left(x-1\right)\left(x^2+x+1\right)\)\(B=x^4-3x^3-x+3=x^3\left(x-3\right)-\left(x-3\right)=\left(x-3\right)\left(x^3-1\right)=\left(x-3\right)\left(x-1\right)\left(x^2+x+1\right)\)\(C=x^3-x^2y-xy^2+y^3=x^2\left(x-y\right)-y^2\left(x-y\right)=\left(x-y\right)\left(x^2-y^2\right)=\left(x-y\right)^2\left(x+y\right)\)\(D=3x+3y-x^2-2xy-y^2=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)

Answer 2