6
$\begingroup$

The actual problem is to find the product of all the real roots of this equation,I am stuck with his factorization:

$x^{101} – 3x^{100} + 2x^{99} + x^{98} – 3x^{97} + 2x^{96} + \cdots + x^2 – 3x + 2 = 0$

By just guessing I noticed that $(x^2 – 3x + 2)$ is one factor and then dividing that whole thing we get $(x^{99}+x^{96}+x^{93} + \cdots + 1)$ as the other factor , but I really don't know how to solve in those where wild guessing won't work! Do we have any trick for factorizing this kind of big polynomial?

Also I am not sure how to find the roots of $(x^{99}+x^{96}+x^{93} + \cdots + 1)=0$,so any help in this regard will be appreciated.

  • 0
    A good idea when factoring a polynomial (especially one that comes from a homework problem) is to look for obvious roots. For example, it's very easy to evaluate the polynomial at $0$ (constant term), $1$ (sum of coefficients) and $-1$ (alternating sum of coefficients), trying other value can be done with Horner's scheme. Applying this, you could find that $-1$, $1$ and $2$ are roots (finding $2$ is admittedly the hardest, but you actually found it). Once you clear them out, you end with a polynomial with no real root (as is explained in Zev's answer below).2011-09-07

2 Answers 2

10

Note that $t^{34}-1=(t^{33}+t^{31}+\cdots+t+1)(t-1)$ and so, substituting $t=x^3$, we get $x^{102}-1=(x^{99}+x^{96}+\cdots+x^3+1)(x^3-1)$ So any real root of $x^{99}+x^{96}+\cdots+x^3+1$ will be a real root of $x^{102}-1$ (and those should be easy to find). But note that, for example, $1$ is a real root of $x^{102}-1$, but is not a root of $x^{99}+x^{96}+\cdots+x^3+1$, since $34=1+1+\cdots+1\neq0$. So, once you find the real roots of $x^{102}-1$ and determine which of them is in fact a root of $x^{99}+x^{96}+\cdots+x^3+1$, you can combine with the real roots of $x^2-3x+2=(x-1)(x-2)$ to get the answer.

To factorize $x^{99}+x^{96}+\cdots+x^3+1$ into irreducibles over $\mathbb{Z}$ (which, it turns out, is equivalent to factoring into irreducibles over $\mathbb{Q}$ in this case), we use the fact that $x^{99}+x^{96}+\cdots+x^3+1=\frac{x^{102}-1}{x^3-1}$ combined with the fact that $x^{102}-1=\prod_{d\mid 102}\Phi_d(x)=\Phi_{102}(x)\Phi_{51}(x)\Phi_{34}(x)\Phi_{17}(x)\Phi_6(x)\Phi_3(x)\Phi_2(x)\Phi_1(x)$ where $\Phi_d(x)$ is the $d$th cyclotomic polynomial. The cyclotomic polynomials are all irreducible over $\mathbb{Q}$. Any irreducible polynomial in $\mathbb{R}[x]$, though, is either a linear $x-a$ for $a\in\mathbb{R}$, or a quadratic $x^2+ax+b$ for which $a^2-4b<0$. The factorization into irreducibles over $\mathbb{R}$ is just $(x-1)$, $(x+1)$, and then a bunch of quadratics $x^2-(\zeta_{102}^k+\overline{\zeta_{102}}^k)x+1=(x-\zeta_{102}^k)(x-\overline{\zeta_{102}}^k)$ where $\zeta_{102}$ is a primitive $102$th root of unity and $0.

Of course, the factorization into irreducibles over $\mathbb{C}$ is just $(x-1)(x-\zeta_{102})(x-\zeta_{102}^2)\cdots(x-\zeta_{102}^{50})(x+1)(x-\zeta_{102}^{52})\cdots(x-\zeta_{102}^{101})$ Wolfram Alpha has a nice printout with the conjugate pairs, it may be helpful.

  • 0
    @Joel Sorry, the "it" in my comment is really $x^{102} - 1$. I think the factorization in, for example, the last display of leonbloy's answer is very useful!2011-09-07
6

In regard to the first part of your question ("wild guessing"), the point was to note that the polynomial can be expressed as the sum of three polynomials, grouping same coefficients:

$ P(x)= x^{101} – 3x^{100} + 2x^{99} + x^{98} – 3x^{97} + 2x^{96} + \cdots + x^2 – 3x + 2 = A(x)+B(x)+C(x)$

with

$\begin{eqnarray} A(x) &= x^{101} + x^{98} + \cdots + x^2 &= x^2 (x^{99} + x^{96} + \cdots + 1) \\ B(x) &= - 3 x^{100} -3 x^{97} - \cdots -3 x &= - 3 x (x^{99} + x^{96} + \cdots + 1)\\ C(x) &= 2 x^{99} + 2 x^{96} + \cdots + 2 &= 2 (x^{99} + x^{96} + \cdots + 1) \\ \end{eqnarray} $

so $P(x) = (x^2 - 3x +2) (x^{99} + x^{96} + \cdots + 1) $

and applying the geometric finite sum formula:

$P(x)=(x^2 - 3x +2) ({(x^{3})}^{33} + {(x^{3})}^{32} + \cdots + 1) = (x^2 - 3x +2) \frac{x^{102}-1}{x^3-1} $

As Andre notes in the comments, your "guessing" was dictated by the very particular structure of the polynomial, you can't hope for some general guessing recipe...