1. ## Polynomial Question

Hey there,

I was reading an article, where the author claims that the following polynomial:
$x^{2}-(1+\frac{1}{\theta }+\frac{1}{\theta\alpha })x+\frac{1}{\theta }=0
$

has two roots, first one root is $\lambda$, which is between 0 and 1, and the other can be expressed as $\frac{1}{\theta\lambda }$

note that this polynomial is obtained after factoring a second order difference equation with forward and lag operators. Now I have no idea how he arrives at that identity, could anyone explain me the trick behind it?

And while you are at it, could you tell me a similar identity between the roots of the following polynomial:

$x^{2}-(1+\frac{1}{\beta }+\frac{\gamma\kappa }{\beta })x+\frac{1}{\beta }=0$

So for example let $\lambda$, be the root between 0 and 1 of this polynomial, what is the second root in terms of $\lambda$?

Thanks a lot

Burak

2. Originally Posted by ichoosetonotchoosetochoos
Hey there,

I was reading an article, where the author claims that the following polynomial:
$x^{2}-(1+\frac{1}{\theta }+\frac{1}{\theta\alpha })x+\frac{1}{\theta }=0
$

has two roots, first one root is $\lambda$, which is between 0 and 1, and the other can be expressed as $\frac{1}{\theta\lambda }$

note that this polynomial is obtained after factoring a second order difference equation with forward and lag operators. Now I have no idea how he arrives at that identity, could anyone explain me the trick behind it?

And while you are at it, could you tell me a similar identity between the roots of the following polynomial:

$x^{2}-(1+\frac{1}{\beta }+\frac{\gamma\kappa }{\beta })x+\frac{1}{\beta }=0$

So for example let $\lambda$, be the root between 0 and 1 of this polynomial, what is the second root in terms of $\lambda$?

Thanks a lot

Burak
The constant term in a monic quadratic (one that has coefficien of $x^2$ equal to $1$) is the product of the two roots. So for a monic quadratic if the constant term is $c$ and one root is $\lambda$ then the other root is $c/\lambda$

CB