# how to solve x^3-3x^2-4=0

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### 1 Answer

You should use derivative method such that:

`(x^3 - 3x^2 - 4)' = 3x^2 - 6x`

You need to find the zeroes of derivative such that:

`3x^2 - 6x = 0 => x^2 - 2x = 0 => x(x - 2) = 0`

`x_1 = 0; x_2 = 2`

You need to evaluate the expression at x = 0 and x = 2 such that:

`x = 0 => 0^3 - 3*0^2 - 4= -4`

`x = 2 => 2^3 - 3*2^2 - 4 = -8`

`x->-oo => x^3 - 3x^2 - 4 -> -oo`

`x->oo => x^3 - 3x^2 - 4 -> oo`

Since the values of the expression are negative at x=2 and positive if x^3 - 3x^2 - 4 -> oo, then, the graph of the function intersects x axis once.

**Hence, you may find one real root in interval `(2,oo), ` the other two roots being complex.**