Question

Phân tích đa thức thành nhân tử:

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

Answer 1

\(x^3-7x-6=x^3+3x^2+2x-3x^2-9x-6\)

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

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

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

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

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

x-3=0 x=3

x+1=0 x=-1

x+2=0 x=-2

          

Answer 2

    x³ -7x +6

= x³ -x²+x²-x-6x+6 
= x²(x-1)+x(x-1)-6(x-1) 
= (x-1)(x² +x-6) 
= (x-1)(x²-2x+3x-6) 
=(x-1)(x-2)(x+3)