# Solve the inequality `x^3-6x^2+11x-6 >= 0`

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### 1 Answer

The inequality `x^3-6*x^2+11*x-6>= 0` has to be solved.

`x^3-6*x^2+11*x-6>= 0 `

=> `x^3 - x^2 - 5x^2 + 5x + 6x - 6 >= 0`

=> `x^2(x - 1) - 5x(x - 1) + 6(x - 1) >= 0`

=> `(x - 1)(x^2 - 5x + 6) >= 0`

=> `(x - 1)(x^2 - 2x - 3x + 6) >= 0`

=> `(x - 1)(x(x - 2) - 3(x - 2)) >= 0`

=> `(x - 1)(x - 2)(x - 3) >= 0`

This is true if neither of the three terms is negative or 2 of them are negative.

- `x - 1 >= 0` , `x - 2 >= 0` and `x - 3 >= 0`

=> `x >= 1` , `x >= 2` and `x >= 3`

This is satisfied if `x >= 3`

- `x - 1 >= 0` , `x - 2 <0` , `x - 3< 0`

=> `x >= 1` , `x < 2` and `x < 3`

This is satisfied for x in the set [1, 2)

- `x - 1 < 0` , `x - 2 >= 0` ,` x - 3< 0`

=> `x < 1` , `x >= 2` and `x < 3`

This is not possible for any value of x

- `x - 1 < 0` , `x - 2 < 0` , `x - 3 >= 0`

=> `x < 1` , `x < 2` and `x >= 3`

This is not possible for any value of x

**The solution of the inequality `x^3-6*x^2+11*x-6>= 0` is `[3, oo)U [1, 2)` **