Question

Phân tích đa thức thành nhân tử

\(x\left(x+2\right)\left(x+3\right)\left(x+5\right)+9\)

\(x^2+2xy+y^2+2x+2y-15\)

\(4x^4y^4+1\)

 

 

Answer 1

\(x\left(x+2\right)\left(x+3\right)\left(x+5\right)+9\)

\(=\left(x^2+5x+6\right)\left(x^2+5x\right)+9\)

Đặt \(t=x^2+5x\)ta được;

\(t\left(t+6\right)+9=t^2+6t+9\)

\(=\left(t+3\right)^2=\left(x^2+5x+3\right)^2\)

b)\(x^2+2xy+y^2+2x+2y-15\)

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

\(=\left(x+y+1+4\right)\left(x+y+1-4\right)\)

\(=\left(x+y-3\right)\left(x+y+5\right)\)

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