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My questions are about a polynomial with the following form: $$q(x) = 1 + \alpha p(x)p(-x),~ \alpha \in \mathbb{R}^+$$ where $p(x)$ is a polynomial with real coefficients and degree $n$: $$ p(x) = \sum_{i=0}^{n} c_i x^i,~ c_i\in \mathbb{R}$$

I would like to factorize $q(x)$ in the following form: $$ q(x) = \prod_{i=1}^{2n}(x - r_{q,i})$$

If the roots of $p(x)$ are known:

  • Is there a simple relationship between $r_q$ and the roots of $p$, denoted $r_p$ ?

I observed $q(x)$ presents only even powers of $x$ and that its roots come in the form $\pm r_q$. (see Properties of roots of an even polynomial). In order to group the roots $r_q$ by type (complex conjugate pairs or reals), I'm wondering:

  • Is there a way to know how much roots of each type there will be, depending of the degree $2 n$ of $q(x)$ ?

Thank you for your help!

sekisushai
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  • Welcome to stackexchange. This potentially interesting question is vague and thus likely to be downvoted or closed. You're more likely to get an answer if you edit the question to clarify. Perhaps show us a few worked examples that lead you to believe the roots have some structure. And tell us what you mean by "straight". Perhaps tell us where the question comes from. – Ethan Bolker Feb 14 '18 at 23:50
  • @EthanBolker Thank you for your comments. I've edited the question and tried to be more specific – sekisushai Feb 15 '18 at 00:00
  • How many examples have you looked at? When $p(x) = x$ the number of roots of $q$ depends on the sign of $\alpha$. What about arbitrary linear polynomials $p$? I think you need to work more on this before asking. – Ethan Bolker Feb 15 '18 at 00:13
  • Please don't include a link to itself in a post. – fleablood Feb 15 '18 at 00:15
  • It doesn't contain roots of the form $\pm r$ because it is an even power. $x^2 - 5x + 6$ is an even power but has roots $2$ and $3$. – fleablood Feb 15 '18 at 00:17
  • @EthanBolker, sorry the coefficient $\alpha$ is positive in my case. – sekisushai Feb 15 '18 at 00:19
  • It has roots that are $\pm r$ because $q(-x) = q(x)$ always. Has nothing to do with the power of the polynomial. – fleablood Feb 15 '18 at 00:19
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    @fleablood Thank you for your comment. I've meant that $q(x)$ has no odd powers of $x$, it has an even degree but only with even powers of $x$. – sekisushai Feb 15 '18 at 00:23
  • Well, $\alpha$ is a red herring: you could include $\sqrt{\alpha}$ as a factor of $p$, and then you'd no longer need $\alpha$. So you really only need to look at $q(x) = 1 + p(x)p(-x)$; for this, if $r$ is a root, so is $-r$ (by direct substitution, and commutativity of multiplication). I'm not sure whether there's anything else useful to be said. – John Hughes Feb 15 '18 at 00:35
  • Also...just to be sure: do you care about complex roots, or only real roots? – John Hughes Feb 15 '18 at 00:37
  • @JohnHughes Thank you for your comments. In fact, I would like to factorize $q(x)$ and I'm wondering if I know the roots of $p(x)$ if there is a way to predict the roots of $q(x)$ from that. Concerning the roots, I would like to group the roots by complex conjugate pairs and reals ones. Is there a way to predict the nature of the roots (real or complex) from the degree of $q(x)$. I've edited the question including this details. – sekisushai Feb 17 '18 at 11:12

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