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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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Are you able to draw the graph for $y = x^3 - 4 x?$ – Will Jagy Aug 11 '12 at 23:14
I think so. I have it at $+\infty$ for $ x= 2$ $-\infty$ for $ x= -2$ and cutting through zero – Magpie Aug 11 '12 at 23:21
Please try the graph, with specific $x$ values $-3,-2,-1,0,1,2,3$ which give nice integer $y$ values. For example, $3^3 - 4 \cdot 3 = 27 - 12 = 15.$ Your manipulation of the $\infty$ symbol makes little sense. – Will Jagy Aug 11 '12 at 23:32
At $x=-2$ the value is not $-\infty$ but $0$ as your factorization shows. – Ross Millikan Aug 11 '12 at 23:40
Comments cannot be edited after some time (5 minutes?). You can delete and resubmit as a workaround, but sometimes it confuses things to take it out of order. – Ross Millikan Aug 12 '12 at 0:09
up vote 1 down vote accepted

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
$$-2<x<0 \text{ and } x>2.$$

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ah ok I didn't see this! I think this is right, so I am going to accept. – Magpie Aug 11 '12 at 23:48
Are we just supposed to find this out by trial and error? Isn't there a standard method? – David Aug 12 '12 at 1:40
@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 $y=(x-4)^2+1$] – Mark Bennet Aug 12 '12 at 7:15

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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Can someone fix the table? – Pedro Tamaroff Aug 12 '12 at 1:12
Yikes! Not sure what's up, there. Might want to post to/ check the meta about such issues. – Cameron Buie Aug 12 '12 at 1:42

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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You are assuming polynomials are continuous (which they are): this would not work for $\tan x = 0$. – Henry Aug 11 '12 at 23:51
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. – ncmathsadist Aug 11 '12 at 23:54

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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I wrote where I got with this in a comment above. I think I get it now, but I am not sure. – Magpie Aug 11 '12 at 23:44
@Magpie: You are correct in the comment. You can see it in The three roots, $-2, 0, 2$ are where the function crosses zero and divide the positive regions from the negative ones. – Ross Millikan Aug 11 '12 at 23:47

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