# Can anybody solve this equation using decompositionf, grouping, factoring etc.... x^3+3x^2-x-1=0.

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You should move the constant term to the right and then you may use the alternate form of equation, such that:

`x^3 + 3x^2 -x = 1`

You may use graphical method to solve the equation, hence, you need to sketch the graph of the function `y = x^3 + 3x^2 -x` and `y = 1` and you need to check if they intersect each other such that:

Notice that the line `y=1` intersect the curve `y=x^3 + 3x^2 -x` three times, for `x in (-3.5 ; -3), ` `x in (-1 ; 0)` and `x in (0.5 ; 1).`

**Hence, evaluating the solutions to the given equation yields `x_1 in (-3.5 ; -3), x_2 in (-1 ; 0)` and `x_3 in (0.5 ; 1).` **

x^3 + 3x^2 - x - 3 = 0

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

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

x^2 = 1 or x = -3

x= 1, -1, -3.

`x^3+3x^2-x-1=0`

`(x^3+3x^2)(-x-1)`

`x^2(x+3)-1(x+1)`

`(x^2-1)` `(x+3)` `(x+1)` set the parentheses equal to 0 and solve

`x= 1` `x=-1` `x=-3`

but that is not analytical solution !

the right equation is x^3 + 3x^2 -x - 1 = 0.

is it x^3 + 3x^2 -x - 1 = 0 or x^3 + 3x^2 - x - 3 = 0 ?

if the second one then it will have a close-form solution:

x^3 + 3x^2 - x - 3 = 0

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

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

either x^2 = 1 or x = -3

so solutions are:

x= 1, -1, -3.