Ok guys, I need some more help with a question for my girlfriend. Basically she was given a problem on a test/quiz and the only way I know how to do it is with a method that she hasnt learned in class yet. So pretty much I want you guys to look it over and let me know if there is another method to the problem that she might know. The problem is...

$$2x^3 + 3x^2 \lt 11x + 6$$

She has to solve it and give the answer in interval notation. So I would first move everything from the right side to the left to have a third order polynomial. My next step would be to use the rational zero test to start finding a zero using synthetic division. After I found one zero, I would factor it out of the function, and i would be left with $(x-a)$*2nd orderpolynomial. I could factor out the polynomial and find my 3 zero's.

The only problem is, she didnt learn synthetic divison, rational zero test, or long division of polynomials. is there another way she could do this problem with knowledge she might have? i tried to briefly teach her my way and it went over her head

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    $\begingroup$ @Greg: Mathematically, your questions are fine. I am sure you will receive an excellent answer soon enough. However, the title of a question should be informative enough that someone searching for a question like this one on the site, or on Google, would find it. Something like "How to solve a polynomial inequality?" would be better. $\endgroup$ Jul 19, 2011 at 0:59
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    $\begingroup$ @Greg She must have been taught some method to handle such problems. Without knowing what those methods are the problem cannot be answered. $\endgroup$ Jul 19, 2011 at 1:28
  • $\begingroup$ I've heard about people referring to themselves in the third person, but this is ridiculous!! $\endgroup$ Jul 19, 2011 at 5:36
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    $\begingroup$ @The Chaz, Nobody is referring to themself in the third person. Like I said, this isn't for me. I'm an engineering student at Georgia Tech, and I haven't done things like this since 8th grade. All I asked was if there is another method of going about it since clearly her teacher put it on the test before teaching the method that I know of. As you can see, I know the most common method but it wasn't taught to her yet $\endgroup$ Jul 19, 2011 at 18:30
  • $\begingroup$ related, but not quite directly answering what you're asking: math.stackexchange.com/questions/36266/… $\endgroup$
    – Isaac
    Jul 19, 2011 at 19:17

3 Answers 3


Factor like this $$2x^3 +3x^2 - 11x - 6 = (x-2)(2x+1)(x+3).$$ Then draw a sign chart.

  • $\begingroup$ I did this by checking the "obvious" possibilities for roots (rational roots theorem), then synthetic division for the rest. $\endgroup$ Jul 19, 2011 at 1:13
  • $\begingroup$ But, as stated, the student does not know the Rational Root Test, so such roots are not "obvious". $\endgroup$ Jul 19, 2011 at 1:31
  • $\begingroup$ The root I found was x = 2. This is very low-level and very unfancy. Perhaps calling it the "rational roots theorem" is overdoing it a bit. $\endgroup$ Jul 19, 2011 at 1:54
  • $\begingroup$ The OP explicitly stated that the student does not know any of these techniques (rational root test, synthetic division, etc). So this does not answer the question. $\endgroup$ Jul 19, 2011 at 1:58

As has been noted, the easiest way to solve a polynomial inequality $f(x) < 0$ (or $f(x) > 0$) is to find the roots of the polynomial $f$, and then form a sign chart.

I would like to mention the reason why sign charts work. In my limited experience as a teacher, knowing the reason why a certain methods works is usually just as important as knowing the method itself.

First, note that polynomials are very well behaved. In particular, polynomials are continuous. Intuitively, this means that in order to sketch the function $f(x)$, one does not need to lift their pencil from the paper. In other words, the function has no sudden jumps or holes. Consequently, the places where the function crosses the $x$-axis (i.e. the places where $f(x) = 0$ are exactly the places where the function can change sign. Therefore, solving polynomial inequalities amounts to finding the zeroes of the function and determining the sign of each interval. However, finding the sign of each interval is also straight-forward since every point in any given interval will have the same sign. Therefore, one just has to choose a test-point for each interval to determine the sign of the entire interval (which in turn solves inequality).

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    $\begingroup$ Indeed, the roots partition the real line into intervals where the polynomial has contant sign. See my answer here for more on this. $\endgroup$ Jul 19, 2011 at 1:56
  • $\begingroup$ Thanks for the info. You said it much better than I. $\endgroup$
    – JavaMan
    Jul 19, 2011 at 2:50

You could just move everything to one side and graph it to find approximate roots, then substitute them in. Or you could feed it to Wolfram Alpha

  • $\begingroup$ To whoever downvoted, this is a sensible answer given that "she didnt learn synthetic divison, rational zero test, or long division of polynomials" $\endgroup$
    – kuch nahi
    Jul 19, 2011 at 1:06
  • $\begingroup$ well the professor put it on a test ebfore they learned that material, so there must be another way $\endgroup$ Jul 19, 2011 at 1:18
  • $\begingroup$ A actually did polynomial long division. You learn quickly that if $a$ is a root of a polynomial $P$, that $x - a$ will divide $P(x)$ evenly. $\endgroup$ Jul 19, 2011 at 1:21
  • $\begingroup$ @kuch I didn't downvote, but keep in mind that graphing won't be of help generally if one needs to find exact rational roots of rational/integer coefficient polynomials. $\endgroup$ Jul 19, 2011 at 2:18
  • $\begingroup$ @Bill Dubuque: that is why I suggested substituting them in. I believe that if you find the roots any which way (whether by the rational root theorem, graphing, or having a little bird tell you) and verify them that works. For a class like this, I have my own rational root theorem: all roots of a polynomial are rational, have a denominator less than 5, and are $\leq$ 10 in absolute value. That gives only 121 possibilities and one can reasonably put them in a spreadsheet and check. $\endgroup$ Jul 19, 2011 at 2:34

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