Find the number of connected components of the set $$\left\{x\in \mathbb R : x^3\left(x^2+5x-\frac{65}{3}\right)>70x^2-300x-297\right\}$$ under the usual topology on $\mathbb R$.
We solve the equation $\displaystyle x^3\left(x^2+5x-\frac{65}{3}\right)-70x^2+300x+297=0$ and the real roots we get give the partion for the real line into intervals. We then check the for the intervals satisfying the condition given in the set which will give the number of connected components.
I could not explicitly calculate the roots nor determine the precisely number of real roots. Is there a way to solve this as it is a question in a math examination ? Can we solve this by Descartes' rule of signs ?? I am aware this question has already been asked.
Using Descarte's rule of sign we see that the polynomial can have $2$ or $0$ positive real roots and $3$ or $1$ negative real roots.
Best Answer
Descartes's rule of sign will not give you the exact number as you have shown in your question. Here are some other methods that can be applied to a polynomial $p(x)$.
1) Check whether there are multiple (real or complex) roots. A root $a$ of $p(x)$ is a multiple root $a$ if and only if $a$ is a common root of $p(x), p'(x)$. To see this, write $p(x) = (x-a)^nq(x)$ with $n \ge 1$ and $q(a) \ne 0$. Then $p'(x) = n(x-a)^{n-1}q(x) + (x-a)^nq'(x)$. If $n > 1$, then $p'(a) = 0$ and if $n = 1$ then $p'(a) \ne 0$. We conclude that $p(x)$ has a multiple root if and only if the gcd of $p(x)$ and $p'(x)$ (which can easily be computed by the Euclidean algorithm) has degree $\ge 1$. In fact, if $p(x)$ has a root of $a$ of order $n > 1$, then $(x-a)^{n-1}$ divides $p(x), p'(x)$. Conversely, assume that the gcd $h(x)$ has degree $\ge 1$. Then $h(x)$ has a root $a$ and we see that $p(a) = p'(a) = 0$.
Applying this to your polynomial $p(x) = x^5+5x^4-\frac{65}{3}x^3-70x^2 +300x +297$ shows that no multiple roots exist. Hence it has $1$ or $3$ aor $5$ real roots of order $1$ (since non-real roots occur in complex conjugate pairs).
2) Consider $p'(x), p''(x),p'''(x)$ to find extrema and inflection points. This allows you to determine roots of $p(x)$ lying between these points.
Doing this with your polymomial yields $p'(x) = 5x^4+20x^3-65x^2-140x+300 = 5(x^4+4x^3-13x^2-28x+60)$, $p''(x) = 20x^3+60x^2-130x-140 = 10(2x^3+6x^2-13x-14)$, $p'''(x) = 60x^2+120x-130 = 10(6x^2-12x-13)$. It is easy to verify that $p'(x)$ has the roots $-5,-3,2$, where $2$ has order $2$ (try the divisors of $60$). We have $p''(-5) = -21 < 0$, $p''(-3) = 53 > 0$, $p''(2) = 0$, $p'''(2) = 250 \ne 0$. Hence $p(x)$ has a local maxinum at $-5$, a local minimum at $-3$ and an inflection point at $2$. Hence $p(x)$ is increasing on $(-\infty,-5]$, decreasing on $[-5,-3]$ and increasing on $[-3,\infty)$. We have $p(-5) = -\frac{734}{3}$, $p(-3) = -486$. Hence $p(x)$ does not have a root in $(-\infty,-3]$ and a single root of order $1$ n $[-3,\infty)$.