Given: x³-6x+5
Notice that 6x=x+5x, so the equation can be restated as follows:
x³-6x+5=x³-x-5x+5=0 ⇒ x(x²-1)-5(x-1)=0 ⇒ x(x+1)(x-1)-5(x-1)=0 ⇒
(x-1){x(x+1)-5}=0 ⇒ (x-1)(x²+x-5)=0
Thus, we have: x-1=0 or x²+x-5=0
From x-1=0, we have: x=1 To evaluate x²+x-5=0, use the quadratic formula.
We have: x=(-1-sqrt(21))/2 or (-1+sqrt(21))/2
The answer : x=(-1-sqrt(21))/2, 1, or (-1+sqrt(21))/2