How to solve x^3-3x-2=0x33x2=0 ?

2 Answers

The roots are -1,-1,21,1,2

Explanation:

It is easy to see by inspection that x = -1x=1 satisfies the equation :

(-1)^3-3times(-1)-2 = -1+3-2=0(1)33×(1)2=1+32=0

To find the other roots let us rewrite x^3-3x-2x33x2 keeping in mind that x+1x+1 is a factor:

x^3-3x-2 = x^3+x^2-x^2-x-2x-2x33x2=x3+x2x2x2x2
qquadqquad = x^2(x+1)-x(x+1)-2(x+1)
qquadqquad = (x+1)(x^2-x-2)
qquadqquad = (x+1)(x^2+x-2x-2)
qquadqquad = (x+1){x(x+1)-2(x+1)}
qquadqquad = (x+1)^2(x-2)

Thus, our equation becomes

(x+1)^2(x-2)=0

which obviously has roots -1,-1,2

We can also see it in the graph:

graph{x^3-3x-2}

Jun 4, 2018

x_1=x_2=-1 and x_3=2

Explanation:

x^3-3x-2=0

x^3+1-(3x+3)=0

(x+1)(x^2-x+1)-3(x+1)=0

(x+1)(x^2-x+1-3)=0

(x+1)(x^2-x-2)=0

(x+1)(x+1)(x-2)=0

(x+1)^2*(x-2)=0

Thus x_1=x_2=-1 and x_3=2