# How do you solve (x-1)(x-2)(x-3)>=0?

Aug 22, 2016

$x \in \left[1 , 2\right] \cup \left[3 , \infty\right)$

#### Explanation:

$f \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x - 3\right) = {x}^{3} - 6 {x}^{2} + 11 x - 6$

is a cubic with positive leading coefficient and zeros of multiplicity $1$ at $x = 1$, $x = 2$ and $x = 3$.

As a result:

• $f \left(x\right)$ is continuous.

• $f \left(x\right)$ is positive for large positive values of $x$.

• $f \left(x\right)$ is negative for large negative values of $x$.

• $f \left(x\right)$ changes sign at each of its zeros.

Hence $f \left(x\right)$ is non-negative in the intervals $\left[1 , 2\right]$ and $\left[3 , \infty\right)$

graph{x^3-6x^2+11x-6 [-3.397, 6.603, -2.24, 2.76]}