By trial and error (or the way I did it was to plot the graph), you'll find that is a solution.
Indeed, by polynomial division, we see that:
and we know that cannot be factorised over the Real field as its determinant is less than 0.
Once again, by trial and error (or plot the graph), you'll find that is a solution.
Factorising this:
, for neater form:
We know we can't simplify this further as has a determinant which is less than 0, so this second question has and
In the event you don't know a way to start finding solutions to these (AlvinCY used, presumably logical, guesswork) you can use the "Rational Root Theorem" to come up with some possibilities.
Given , any rational roots of this polynomial equation will be of the form: . There may, of course, be other real solutions to the equation, but this will give you a list to start with, anyway.
-Dan
Can't EVER do too much Physics!
-Dan
I do apologise, I was doing it at late hours of night while chatting on MSN with others about their weekends.
Thanks for picking it up guys, and yes, I do mean discriminants... and no it wasn't negative, therefore you CAN factorise it.
I certainly look like a little bit of a fool at the moment
Thanks guys.