# How to solve this equation in different ways ? x^3+3x^2-x-1=0

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### 1 Answer

You should use derivative of the given function to find its zeroes such that:

`f(x) = x^3+3x^2-x-1 => f'(x) =3x^2 + 6x - 1`

You need to evaluate the roots of derivative such that:

`3x^2 + 6x - 1 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (-6+-sqrt(36 + 12))/6`

`x_(1,2) = (-6+-sqrt48)/6 => x_(1,2) = (-6+-4sqrt3)/6 => x_(1,2) = (-3+-2sqrt3)/3`

You need to evaluate the function at `x_(1,2) = (-3+-2sqrt3)/3` such that:

`f(-1 + 2sqrt3/3) = 0.154^3+3(0.154)^2-0.154-1`

`f(-1 + 2sqrt3/3) = 0.003 + 0.071 - 0.154 - 1 = -1.08`

`f(-1- 2sqrt3/3) = -0.003 + 0.071+ 0.154 - 1 = -0.7`

`f(-oo) = -oo ; f(oo) = oo`

Notice that f(-1+2sqrt3/3) is negative and f(oo) is positive, hence, the graph of the function intercepts x axis one time at least.

**Hence, notice that there is a root of the function for `x in (-1 + 2sqrt3/3 ; oo).` **

You may find the next two roots sketching the graph of the function such that:

**Notice that the second root is in (-1,0) and the third root is in (-4,-3).**