Re: factoring quadratics
- From: Bart Goddard <goddardbe@xxxxxxxxxxxx>
- Date: 4 Jan 2007 03:11:15 GMT
zxcv_890@xxxxxxxxxxx wrote:
That is to say, given the
quadratic ax^2 +bx+c, to show that such a quadratic is not factorable,
The first little hurdle here is that, to be precise, we don't say
"factorable", but things like "factorable over the integers" or
"factorable over the rationals" or "factorable over the reals".
ALL quadratics are factorable over the complex numbers.
I'm guessing that you mean here "factorable over the rationals",
so in:
does it suffice to show that there are no such numbers a, b, c, such
that np+mq=b and (np)(mq)=ac? If so, how would one show this?
you want n,m,and q to be rational numbers. To show that a
quadratic is not factorable over the rationals, note that
if you let x=(-p/m) and plug it into your factored expression,
you get 0. So your quadratic has a rational root. The converse
of this idea is also true.
So to prove that a quadratic is not factorable over the rationals,
you just have to show that it's roots are NOT rational.
E.g., the roots of x^2+3x+1 are x=(-3 +/- sqrt(3^2-4*1*1))/2
or x=-3/2 +/-sqrt(5)/2. Since these aren't rational, the
quadratic doesn't factor.
B.
--
The man without a .sig
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