Question

Phân tích đa thức thành nhân tử:

\(a,3y^3+6xy^2+3x^2y\)

\(b,x^3-3x^2-4x+12\)

\(c,x^3+3x^2-3x-1\)

Answer 1

\(3y^3+6xy^2+3x^2y=3y\left(y^2+2xy+x^2\right)=3y\left(x+y\right)^2\)

\(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

\(x^3+3x^2-3x-1=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)=\left(x-1\right)\left(x^2+x+1+3x\right)\)

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

Tham khảo nhé~