How to show that the polynomial $$x^3-3x^2-3x+7$$ has a positive real root? I can graph it and see that it is indeed true, but can we prove it rigorously?
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Along with @Stefano's hint, keep in mind that every odd polynomial has at least one real root because of the intermediate value theorem. – nullgeppetto Oct 24 '14 at 20:48
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@nullgeppetto, it has at least one real root, but it is not necessarily positive. – Galc127 Oct 24 '14 at 20:53
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@Galc127, yes you are right of course. Just thought it would help Stanley in general. Sorry if does not help. – nullgeppetto Oct 24 '14 at 21:02
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It helps. Thanks. – Stanley Oct 24 '14 at 21:04
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I have a beautiful solution for this by solving the equation, although you can do other things to prove the existence of roots.
$$x^3-3x^2-3x+7=0 \iff (x-1)^3-6(x-1)+2=0 \stackrel{y:=x-1}{\iff} y^3-6y+2=0\; \stackrel{y:=\frac{2}{z}+z}{\iff}\; 2-6\left (\frac{2}{z}+z \right ) +\left (\frac{2}{z}+z \right )^3 =0 \; \stackrel{\cdot z^3}{\iff}\; z^6+2z^3+8=0 \; \stackrel{\theta:=z^3}{\iff}\; \theta^2+2\theta+8=0$$
You have a long way back, but fear not if you meet complex numbers, they'll get scared and go away in the final solution.
UserX
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Let's denote $p(x)=x^3-3x^2-3x+7.$
In this case the Descartes' rule of signs http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs is not of use, because there are two changes of sign. This means that the number of positive roots is $0$ or $2,$ but we can discard the possibility of having $0$ positive roots.
So we need to use a different king of argument. The idea is to use Bolzano's theorem: If $p$ is continuous on $[a,b]$ and $p(a)<0$ and $p(b)>0$ then there exists $c\in (a,b)$ such that $p(c)=0.$
$p(x)$ is a continuous function and computing some values we have that $p(1)=2,$ $p(2)=-3,$ $p(3)=-2$ and $p(4)=13.$ So, it follows from Bolzano's theorem the existence of a root in $(2,3)$ and another root in $(3,4).$
mfl
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