1
$\begingroup$

I am interested in finding the number of real roots of the polynomial equation

$$ x^9 + \frac{9}{8}x^6 + \frac{27}{64}x^3 - x + \frac{219}{512} = 0. $$

I know that graphing it would tell me how many real roots it has: the graph cuts the x-axis three times. But the coefficients are telling me some factorization is possible. I tried to write it like

$$ \left(x^3 + \frac{3}{8}\right)^3 = x - \frac{3}{8}, $$

but what next? Or is a graphical solution is the only possibility?

$\endgroup$
  • 2
    $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – apnorton Aug 12 '15 at 4:36
2
$\begingroup$

We have

$$ \begin{align} &x^9 + \frac{9}{8}x^6 + \frac{27}{64}x^3 - x + \frac{219}{512} \\ &\qquad = \frac{1}{512} (2 x-1) \left(4 x^2+2 x-3\right) \left(64 x^6+64 x^4+48 x^3+64 x^2+24 x+73\right). \end{align} $$

Since

$$ \begin{align} &64 x^6+64 x^4+48 x^3+64 x^2+24 x+73 \\ &\qquad \geq 64 x^6+48 x^3+24 x+73 \\ &\qquad > 28 x^6+48 x^3+24 x+44 \\ &\qquad = 4 (x+1)^2 \left(7 x^4-14 x^3+21 x^2-16 x+11\right) \\ &\qquad \geq 4 (x+1)^2 \left(7 x^4-14 x^3+15 x^2-16 x+8\right) \\ &\qquad = 4 (x+1)^2 (x-1)^2 \left(7 x^2+8\right) \\ &\qquad \geq 0, \end{align} $$

the only real roots are $x=1/2$ and

$$ x = \frac{-1 \pm \sqrt{13}}{4}. $$

$\endgroup$
  • $\begingroup$ So, just like a quadratic , the sixth degree equation being always positive has no real roots. Does this hold for all even degree polynomials? And how did you come about with such a factorisation? $\endgroup$ – Irrational 3.14 Aug 12 '15 at 16:50
  • $\begingroup$ Any function which is always positive has no roots. Or do I misunderstand your question? The factorization was obtained by Mathematica's Factor command. $\endgroup$ – Antonio Vargas Aug 12 '15 at 18:01
  • $\begingroup$ Oh! My question was point less. Can I somehow know the number of real roots (without actually finding them)? Drawing the graph is a cumbersome affair. $\endgroup$ – Irrational 3.14 Aug 12 '15 at 18:51
  • $\begingroup$ Descartes' rule of signs (and its descendants) are pretty much as good as you can get in terms of general methods. $\endgroup$ – Antonio Vargas Aug 12 '15 at 18:57
1
$\begingroup$

The next level beyond Descartes' rule of signs is Sturm's theorem. Using it, you can find the number of real roots, or the number of roots in any interval, for any polynomial with real coefficients.

You start by finding the canonical Sturm sequence of the polynomial. In this case the Sturm sequence (normalized so the leading coefficients are $\pm 1$) is $$\pmatrix{{x}^{9}+{ {9\,{x}^{6}}/{8}}+{ {27\,{x}^{3}}/{64}}-x+{ { 219}/{512}} \cr{x}^{8}+3\,{x}^{5}/4+{ {9\,{x}^{2}}/{64}}-1/9\cr -{x}^{6}- 3\,{x}^{3}/4+{ {64\,x}/{27}}-{ {73}/{64}}\cr -x^3 + { {27\,{x}^{2} }/{64}}+{ {3}/{64}}\cr x^2-{ {1069962688\,x}/{175592043}}+{ {19822249}/{6503409}}\cr x -{ {44951085}{81718064}}\cr1 }$$
If $\sigma(a)$ and $\sigma(b)$ are the number of sign changes in the sequence at $a$ and at $b$ (where $a < b$, and $a$ and $b$ are not multiple roots of the polynomial), then the number of distinct roots in the interval $(a,b]$ is $\sigma(a) - \sigma(b)$.

For the total number of real roots, you just have to look at the leading terms of each polynomial in the sequence. Thus when $b \to +\infty$, the pattern of signs in the Sturm sequence is $+,+,-,-,+,+,+$: $\sigma(b) = 2$. When $a \to -\infty$, the pattern of signs is $-,+,-,+,+,-,+$: $\sigma(a)= 5$. $\sigma(a) - \sigma(b) = 3$, so there are three real roots.

$\endgroup$
  • $\begingroup$ You missed the last element in the sequence. The sign changes are $++--+++$ and $-+-++-+,$ so $\sigma(a) = 5$. $\endgroup$ – Antonio Vargas Oct 21 '15 at 1:14
  • $\begingroup$ Oops, yes. I'll edit. $\endgroup$ – Robert Israel Oct 21 '15 at 7:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.