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I was sort of finding the roots by doing $x^3-4x>0=x(x^2-4)$

$x = 0, x = -2,x = 2$ for $x(x-2)(x+2)>0$

Then I stopped and thought; maybe I shouldn't be doing that? I am doubting myself! Can anyone confirm if I am doing the right thing to solve the equality?

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    Yes, I thi$n$k i will leave it. My latex is $n$ot good enough to cope with only two edits and no preview, so I won't attempt it again here!2012-08-12

4 Answers 4

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This is a good approach. You are almost there. A product is positive if it has an even number of negative terms. So it is positive if there are no negative terms, which is if $x \gt 2$, or if there are two negative terms ...

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    @Magpie: You are correct in the comment. You can see it in http://www.wolframalpha.com/input/?i=plot+x%5E3-4x The three roots, $-2, 0, 2$ are where the function crosses zero and divide the positive regions from the negative ones.2012-08-11
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I like to do tables to determine signs. It follows from Bolzano's Theorem the sign will be kept between the roots. You already factored this and obtained $x=2,x=-2,x=0$.

$\begin{array}{|c|c|c|c|} & (-\infty,-2) & (-2,0)& (0,2)&(2,+\infty)\\ \hline x-2 & - &- & -&+\\ \hline x+2 &- &+ &+ &+\\ \hline x & - & - &+ &+\\ \hline p(x)& -&+&-&+\\ \hline \end{array}$

Explanation: The top row is the real line divided into the intervals by the roots we have. The first three rows of $+/-$ account for the sign of the factors in each interval, which we obtain by inspection. Finally, the sign of $p$ is obtained by "multiplying" each of the values obtained, so $-\times-\times-=-$, $-\times+\times +=+$,$\&c$.

Try and do the same for $p(x)={x^3} + 4{x^2} + 3x - 2$.

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    Yikes! Not sure what's up, there. Might want to post to/ check the meta about such issues.2012-08-12
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A good way to do this is to draw a number line and mark the places where the polynomial is zero. The sign will be constant in the intervals so created. Check inside each one to see the sign (+ or -). Then it's easy to read the solution.

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    Correct. This procedure applies to polynomials, and more generally, to continuous functions. It does generalize for rational functions. Put a white dot at each zero of the denominator. That point is excluded from the domain. Along with the (filled) dots at the zeroes, the function's sign is constant between dots.2012-08-11
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The polynomial $x^3-4x$ is $0$ at $x=-2, 0,$ and $2.$ At $-3$ the value of the polynomial is $< 0$; at $3$ the value is $>0.$ At $-1,$ the value is $>0.$ At $1$ the value is $<0.$

From this we see the polynomial is $> 0$ for
$-22.$

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    @David - when a continuous function changes sign it passes through zero (intermediate value theorem) - so to find out where a continuous function is positive it is necessary to identify where (if anywhere) it takes the value zero - in some way or another this is what any analysis will do. A function need not change sign at a zero ($y=x^2$), so some attention is required at each zero. The method is perfectly systematic. [for no zero one can look to express the function in such a way that this becomes obvious eg $y=(x-4)^2+1$]2012-08-12