| Try as potential roots the values of the form +/- n/d where n
divides the (absolute value of the) constant term and d
divides the (absolute value of the) coefficient of the highest
degree term.
Here, that means d is 1 and n is in {1,2,4,8,16} so the trial
roots are {-16,-8,-4,-2,-1,1,2,4,8,16}.
Both -2 and 2 will turn out to be roots, so divide by (x - -2)(x - 2)
= (x+2)(x-2) and try to factor what is left, which is x^2 - 4x + 4,
which factors easily to (x-2)^2. So x^4 - 4x^3 + 16x - 16 is
(x+2)(x-2)^3 with a single root at -2 and a triple root at 2.
Dan
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| OK, but if it *hadn't* been that easy, how *do* you solve a general
quartic? I've grubbed together an algorithm to do cubics, from various
assorted sources, but haven't been able to find any information on
quartics. It's possible though, isn't it? Anyone?
Simon
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| This has probably been given elsewhere in the conference, but it won't hurt to
repeat it. I have been told that this method (the classic one) isn't very good
numerically, which claim I haven't checked.
Starting with x^4 + ax^3 + bx^2 + cx + d = 0, let z = x + a/4, so x = z - a/4;
you will get f(z) = z^4 + uz^2 + vz + w = 0 (you have eliminated the cubic term).
Write f(z) = (z^2 + px + q)(z^2 - px + w/q) and solve for p and q. You will
find that you end up with a cubic in p, or p^2, or something. So solving
quartics is reducible to solving cubics (and quadratics).
Of course I've omitted some details. :-) If you get stuck, I'm willing to plow
through the details myself and post them.
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