24 Oct 2007 12:52:09 -0700 from Dougsd1r <dougsdir@[EMAIL PROTECTED]
>:
> How do I fully solve this equation?
>
> x^3 + x^2 -5x + 3 = 0
>
> According to MS Math there are three solutions x=-3, x=1 and x=1
>
> here is what i have so far:
>
> x^3 + x^2 -5x = -3
>
> (x)(x^2 + x - 5) = -3
This is a common student mistake. But factoring works *only* when you
have 0 on the right-hand side. So subtracting 3 from both sides was a
mistake.
There are methods to solve a cubic equation, but they're horrendous.
Instead, you want to try the Rational Root method[1]. When the
coefficients are all integers, then *if* there are any rational roots
they will have the form p/q where p is a factor of the constant term
and q is a factor of the leading coefficient. (There may or may not
be any rational roots, but if they exist they can have only the
indicated values.)
Here, the leading coefficient is 1 so the factors are 1 and -1. The
constant term is 3; its factors are 3, -3, -1, and 1. So the only
possible rational roots are
3/1, 3/-1, -3/1, -3/-1, 1/1, 1/-1, -1/1, -1/-1
As you see, there are some duplications. The non-duplicates are
3, -3, 1, -1
Try plugging each into the equation, starting with 1 because it's
easiest. 1^2 + 1^2 - 5*1 + 3 = 0 so 1 is a solution and x-1 is a
factor of the left-hand side. Divide, using long division or
synthetic division[2], and you get
x^3 + x^2 - 5x + 3 = (x-1)(x^2 + 2x - 3) = 0
Now you just have to solve x^2 + 2x + 3 = 0 ==> x = 1 or -3
So your solutions are 1 (a double root) and -3.
[1] http://oakroadsystems.com/math/polysol.htm#RationalRoots
[2] http://oakroadsystems.com/math/polysol.htm#DivideBy
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com/
"If there's one thing I know, it's men. I ought to: it's
been my life work." -- Marie Dressler, in /Dinner at Eight/
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