Question

Phân tích đa thức thành nhân tử: 

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

Answer 1

x³-7x+6

=x³+3x²-3x²-9x+2x+6

=x²(x+3)-3x(x+3)+2(x+3)

=(x²-3x+2)(x+3)

=(x²-2x-x+2)(x+3)

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

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

Answer 2

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

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

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

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

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

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

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

Answer 3

\(x^3-7x+6\\ =x^3-x-6x+6\\ =\left(x^3-x\right)-\left(6x+6\right)\\ =x\left(x^2-1\right)=6\left(x+1\right)\\ =x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\\\)

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