Question

1.Phân tích thành nhân tử

`a) x(x+4)(x+6)(x+10)+128`

`b)x^4+6x^3+7x^2-6x+1`

Answer 1

a: \(=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)

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

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

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

b: \(=\left(x^2+3x-1\right)^2\)