Phân tích đa thức sau thành nhân tử x^3-7x-6 (giải bằng nhiều cách)

Đáp án:

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

`(x-3)(x+2)(x+1)` 

Giải thích các bước giải:

`text{Cách 1:}`

\(x^3-7x-6=x^3+x^2-x^2-x-6x-6\)

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

\(=x^2.\left(x+1\right)-x.\left(x+1\right)-6.\left(x+1\right)\)

\(=\left(x+1\right).\left(x^2-x-6\right)\)

\(=\left(x+1\right).\left(x^2+2x-3x-6\right)\)

\(=\left(x+1\right).\left[\left(x^2+2x\right)-\left(3x+6\right)\right]\)

\(=\left(x+1\right).\left[x.\left(x+2\right)-3.\left(x+2\right)\right]\)

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

`text{Cách 2:}`

`x^3-7x-6` 

`=x^3+x^2-x^2-x-6x-6`

`=x^2(x+1)-x(x+1)-6(x+1)`

`=(x^2-x-6)(x+1)`

`=(x^2+2x-3x-6)(x+1)`

`=[x(x+2)-3(x+2)](x+1)`

`=(x-3)(x+2)(x+1)`