ta có:B(x)=x^3-7x-6=0
=>x^3 + x^2 - x^2 - 6x - x - 6
=> (x^3 + x^2) - (x^2 + x) - (6x + 6)
=> x^2(x + 1) - x(x + 1) - 6(x + 1)
=> (x + 1)(x^2 - x - 6)
=> (x + 1)(x^2 - 3x + 2x - 6)
=> (x + 1){(x^2 - 3x) + (2x - 6)}
=> (x + 1){(x(x - 3) + 2(x - 3)}
=> (x + 1)(x - 3)(x + 2)=0
=>x=-1;-2 và 3
ta có:B(x)=x^3-7x-6=0
=>x^3 + x^2 - x^2 - 6x - x - 6=0
=> (x^3 + x^2) - (x^2 + x) - (6x + 6)=0
=> x^2(x + 1) - x(x + 1) - 6(x + 1)=0
=> (x + 1)(x^2 - x - 6)=0
=> (x + 1)(x^2 - 3x + 2x - 6)=0
=> (x + 1){(x^2 - 3x) + (2x - 6)}=0
=> (x + 1){(x(x - 3) + 2(x - 3)}=0
=> (x + 1)(x - 3)(x + 2)=0
=>x=-1;-2 và 3