Question

Phân tích đa thức sau thành nhân tử:

\(a,x^3+4x^2+4x+3\)

\(b,x^2-y^2+4y-4\)

\(c,x^4+x^3y-xy^3-y^4\)

Answer 1

Câu a : \(x^3+4x^2+4x+3\)

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

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

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

Câu b : \(x^2-y^2+4y-4\)

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

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

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

Câu c : \(x^4+x^3y-xy^3-y^4\)

\(=x^3\left(x+y\right)-y^3\left(x+y\right)\)

\(=\left(x+y\right)\left(x^3-y^3\right)\)

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