Question

phân tích đa thức thành nhan tử x.(x+6).(x+10).(x+4)+128

 

Answer 1

ta có 

\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)

\(=\left[\left(x^2+10x+12\right)-12\right]\left[\left(x^2+10x+12\right)+12\right]+128\)

\(=\left(x^2+10x+12\right)^2-12^2+128=\left(x^2+10x+12\right)^2-16\)

\(=\left(x^2+10x+12-4\right)\left(x^2+10x+12+4\right)=\left(x^2+10x+8\right)\left(x^2+10x+16\right)\)

\(=\left(x+2\right)\left(x+8\right)\left(x^2+10x+8\right)\)