This problem is on my Calculus Readiness Test and I was having a lot of trouble with it. The problem is $$(x-3)(x-2)(x-1)\gt0$$ I know how to solve $(x-3)\gt0$ but I have never seen these type of problem before. I've tried to distribute everything but it gets messy and doesn't really simplify neatly. How should I go about solving this?


  • $\begingroup$ how about considering the possibilities for x and how that affects the sign of each term? $\endgroup$ – Ralth Sep 8 '12 at 2:12

As the others have mentioned, you must find the roots or zeroes of the function before solving the inequality. Therefore, the roots for $f(x) =(x-3)(x-2)(x-1)$ are $1,2,3$. There are numerous ways to solve this but I would like to show you how you can do it graphically. Here is the graph of the function:

graph of (x-3)(x-2)(x-1)

Note, that the question asks you to find when $f(x) \gt 0$ Therefore, look for all the spots on the graph where the graph is above zero and not equal to zero. Firstly, we can disregard any value that is less than or equal to $1, 2, 3$ because at those points the function is not greater than zero.

Next, we can see from the graph that between the $1$ and $2$, the $f(x)$ is greater than zero. So we can say that one of the solutions is: $$ 1\lt x\lt2 $$ because at points between that interval, the value of $f(x) \gt 0$.

Next, we disregard the interval $ 2 \le x \le 3$ because at that interval $x \le 0$ or in other words, no matter what $f(x)$ you find between that interval, you will always have value that is $\le 0$.

So, all we are left with is $x\gt3$ because according to the graph, all values above $3$ are greater than $0$.

Therefore, the solution through graph analysis is $$1 \lt x \lt 2 \\ x \gt 3$$ or in interval notation $$ (1,2) \cup (3,\infty) $$

  • $\begingroup$ Wow! You really explained this well. Thank you for taking the time to respond to my older questions :). $\endgroup$ – Kot Feb 28 '13 at 3:11
  • $\begingroup$ @StevenN Anytime! :) $\endgroup$ – Jeel Shah Feb 28 '13 at 9:45

In fact you don’t want to multiply it out: the factored form is much more useful here. You have a product of three numbers; for the moment just call them $a,b$, and $c$. When is such a product positive: it’s certainly positive if all three of $a,b$, and $c$ are positive. But it’s also positive if exactly two of them are negative and the remaining one is positive. There are the only ways to get a positive product from three factors.

Now, here’s a chart of the signs of the factors $x-3,x-2$, and $x-1$ and their product:

                         1            2            3
     x-3:       -        -      -     -      -     0          +
     x-2:       -        -      -     0      +     +          +
     x-1:       -        0      +     +      +     +          +
 product:       -        0      +     0      -     0          +

When $x<1$, all three factors are negative, and so is their product. When $x=1$, one factor is $0$, and so is the product. When 1or $x>3$. In interval notation, the solution set of the inequality is $(1,2)\cup(x,\infty)$.

This technique works whenever you’re comparing a product with $0$.

  • $\begingroup$ @gekkostate: I want the semi-graphical chart, not an array, and I want ‘for the moment’, not ‘for a moment’. $\endgroup$ – Brian M. Scott Feb 27 '13 at 14:04
  • $\begingroup$ @BrianMScott, sorry about that. I thought that having the array would look nicer and still get the message across. $\endgroup$ – Jeel Shah Feb 27 '13 at 14:06
  • 1
    $\begingroup$ @gekkostate: No harm done. I realize that the array is prettier, but it doesn’t have the pictorial aspect with the number line. I just wanted to let you know that I wasn’t rolling it back out of pique: I really did have a preference. $\endgroup$ – Brian M. Scott Feb 27 '13 at 14:12

One way of doing this is to first find the zeros of $f(x) = (x - 3)(x - 2)(x - 1)$. They are 1, 2, and 3. If you want, you can then draw a line and label the points 1, 2, and 3. Then deternine the sign of points less than 1, between 1 and 2, between 2 and 3, and greater than 3. Evaluate these points under the function $f(x) = (x - 3)(x - 2)(x - 1)$. The interval for which the points you choose evaluate to sometime greater than 0 are the solutions to the inequality.

For your particular example, if $x < 1$, then $f(x) < 0$; if $1 < x < 2$, then $f(x) > 0$; if $2 < x < 3$, then $f(x) < 0$; and if $3 < x$, then $f(x) > 0$. Therefore, the solution to the inequality $f(x) > 0$ is $(1,2) \cup (3, \infty)$.

Note that above, a nice simple way of determining, for example, if the sign of $f(x)$ for $x < 1$ if simply take a point $x < 1$ and evaluate the function at that point. For example $0 < 1$, $f(0) = (-3)(-2)(-1) = -6$, which is negative. So $f(x) < 0$ for all $x < 1$.

By the way the justification of this sort trick is by using the continuity of the function and the intermediate value theorem.

  • $\begingroup$ Thank you, I am currently taking Pre-Calculus and we just reviewed Set unions and intersections. I will try to make sense of your answer. $\endgroup$ – Kot Sep 8 '12 at 2:18

Consider graphing $f(x) = (x-3)(x-2)(x-1)$. (It doesn't even need to be terribly accurate. The factored form of the cubic polynomial tells you where the zeros are, and the only other information you need is where the curve is above the $x$-axis, and where it is below the $x$-axis.)

Once you've (roughly) graphed the this function, you can see where $(x-3)(x-2)(x-1)$ is greater than 0 (that is, the intervals on the $x$-axis where $(x-3)(x-2)(x-1)$ is above the $x$-axis).


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