Solution for the inequality: $(x-1)(x-2)(x-3)>0$ Recently I came across an inequality like this:
$$(x-1)(x-2)(x-3)>0$$
The question was: Which solution for this in inequality is correct?


*

*$x>3$ or $x<1$ 

*$x>3$ or $1<x<2$ 

*$x<1$ or $2<x<3$ 

*There are no solutions


There was a timelimit of 3 minutes for solving this. I freaked out a  bit as I haven't worked that much with these kind of inequalities yet. I thought about multiplying out the terms on the left side, and then checking the solutions, but even after I tried that on paper, there is no way I could have done that in 3 minutes under pressure. 
What is the expected, fast way to solve this inequality? Or was it just expected to "see" it, based on the given solutions?
 A: Hint. It is easy to see that the product is positive for $x>3$. Moreover the sign of the product $(x-1)(x-2)(x-3)$ changes along the real line as the variable $x$ crosses any of its distinct roots $1$, $2$, and $3$.
A: HINT: it is only $$x>3$$ or $$1<x<2$$ one factor must be positive and the other two negative or all three positive
A: There are solutions since the limit of the expression at $+\infty$ is $+\infty$. Note there are  $3$ pivot-values, $1,2,3$ and that there is a change in sign at each of these values, and only at these values by the I.V.T. 
We readily deduce a table of signs per interval:
$$\begin{matrix}&1&&2&&3&\\
\hline -&0&+&0&-&0&+\end{matrix}$$
A: See the table below 
just take the product from each column. since the of $(x-1)(x-2)(x-3)$ is given by the product of sign of $(x-1)$, $(x-2)$, and $(x-3)$. In each column you have the sign of each  $(x-1)$, $(x-2)$, and $(x-3)$.
A: Draw an axle from left to right, and mark 1, 2, 3 on it. Put a point for x at 0. then move x from left to right. 
When x < 1, x is at the left of 1, 2, and 3, therefore (x-1), (x-2), and (x-3) are all negative. We know answer (1) and (3) is not right.
When x > 3, x is at the right of 1, 2, and 3, therefore (x-1), (x-2), and (x-3) are all positive. This is a solution, at least. Then We know answer (4) is not right.
A: For a multiple-choice problem, some simple reasoning can narrow down the options quickly.
If $x$ is a large positive number, all three terms in the product are positive.  This eliminates options 4 and 3.
If $x$ is a large negative number. all three terms in the product are negative.  This eliminates option 1.
That leaves option 2 as the only remaining possibility.  
