1. The problem statement, all variables and given/known data Find x. \(x^4+4x^3+3x^2-20x-20=0\) 2. Relevant equations 3. The attempt at a solution \(x^4+4x^3+3x^2-20x-20=0\)/:\(x^2\) \(x^2+4x+3-20\frac{1}{x}-20\frac{1}{x^2}=0\) \((x^2-20\frac{1}{x^2})+4(x-5\frac{1}{x})+3=0\) \(x-5\frac{1}{x}=y\) \(x^2-10+25\frac{1}{x^2}=y^2\) \(x^2+25\frac{1}{x^2}=y^2+10\) ???? I can't substitute for \(x^2-20\frac{1}{x^2}\)

That is not the way to solve this problem. It is just luck that this works out here. Where a derivative of a function reaches zero (or the second derivative) has absolutely nothing to do with where the original function reaches zero. Think about it this way: if the derivative of some function \(f(x)\) is some other function \(g(x)\), the derivative of any function that differs from \(f(x)\) by a fixed constant is also \(g(x)\).

Dragon, your method is wrong. You're finding the points of inflection, not the roots. If P(x) = 0, that doesn't mean that \(\frac{d^{n}}{dx^{n}}P(x) = 0\) will provide you with roots. That's only true in general if the multiplicity of a particular root is n (ie \(P(x) = (x-a)^{n}Q(x)\). In this case it's a lucky coincidence, not the right method. For instance \(P(x) = x^{3}+1\). The only real root is x=-1. However, your method would claim that P''(x) = 6x = 0 provides you with a root, x=0. Obviously this is not true. There is a general formula for quartic polynomials but it's horrific and vast and completely unmemorisable. Instead, you have to employ other methods. Short of learning how to do numerous transformations to reduce quartics down to a few particular cases (which amounts to semi-deriving the quartic formula anyway) you should notice things like the coefficents of the polynomial in question have a nice symmetry. \(P(x) = 1x^{4}+4x^{3}+3x^{3}-20x-20\) The fact that 1-4+3 = 0 and 20-20=0 suggests that if you can later the signs of the coefficents just right, you'll sum to them zero. Putting x=-1 does this, so \(P(-1) = 1-4+3+20-20 = 0\). Thus x=-1 is a root. This then gives you that \(P(x) = (x+1)Q(x)\). Doing some long division or just being able to do it by inspection gives you \(P(x) = (x+1)(x^{3}+3x^{2}- 20)\) Now to find the roots of \(Q(x)=x^{3}+3x^{2}- 20\). It has a real root, since it's a cubic, so let's find that. Q(-5) < 0. Q(0) < 0. Q(3) > 0. So it's crossed somewhere between x=0 and x=3. A moments thought tells you it's x=2. Thus \(P(x) = (x+1)(x-2)(x^{2}+5x+10)\) Now you just use the quadratic formula to find those, complex, roots. Homework questions like this are inevitably constructed to have enough nice roots to get you down to a quadratic, which you can then solve by the formula. Obviously in physics things are not that nice but if you were doing a real system you'd have inevitably computerised it. That or you can use something like algebraic geometry to factorise it for you, algebraicly. /edit DH beat me to the correction.

Are you forced to solve algebraically, or are you given the option of a TI? As Ben said, I'd suggest synthetic division. Of course, you'll have to learn the Rational Root Theorem before that, though.

There is a messy procedure for solving 3rd & 4th order polynomials. Try a search for Cardano or Tartaglia who published a method for solving 3rd order polynomials at least a century ago. I do not know a search key for the 4th order polynomial solution method. There is much donkey work involved. I would not consider using anything other than the Newton numerical solution, which is a sucessive approximations solution. The following indicates the method. NextGuess = CurrentGuess - Polynomial(CurrentGuess)/Derivative(CurrentGuess) To illustrate the above: Find cube root of 1000 starting with a lousy Guess[sub]0[/sub] = 20 The pertinent 3rd order equation to be solved is: X[sup]3[/sup] - 1000 = 0 Guess[sub]1[/sub] = 20 - (20[sup]3[/sup] - 1000)/3*20[sup]2[/sup] --->> Guess[sub]1[/sub] = 14.166 667 Guess[sub]2[/sub] = 14.166 667 - (14.166 667[sup]3[/sup] - 1000)/3*14.166 667[sup]2[/sup] --->> Guess[sub]2[/sub] = 11.105 344 Guess[sub]3[/sub] = 11.105 344 - (11.105 344[sup]3[/sup] - 1000)/3*11.105 344[sup]2[/sup] --->> Guess[sub]3[/sub] = 10.106 368 Guess[sub]4[/sub] = 10.106 368 - (10.106 368[sup]3[/sup] - 1000)/3*10.106 368[sup]2[/sup] --->> Guess[sub]4[/sub] = 10.001 116 Guess[sub]5[/sub] = 10.001 116 - (10.001 116[sup]3[/sup] - 1000)/3*10.001 116[sup]2[/sup] --->> Guess[sub]5[/sub] = 10.000 000 It is not difficult to write a PC program to solve polynomials. I wrote one using Visual Basic. I do not know the limits of that program. I tested it on 50th order polynomials. It is a bit of work to solve polynomials using a hand calculator to do the Newton method, although it is easy to use a calculator for nth roots, assuming that it will not do it directly. My HP 48G+ & 48GX can solve polynomials, & I think it will solve other types of equations (I have not used it for anything but polynomials). I assume that the better TI calculators can do like wise.

Though it's not relevent here, if any of you ever have to find the common roots for large quantities of polynomials, even in multiple variables, I highly recommend learning a bit of algebraic geometry. I came across it last year when someone published it's applications to string theory. It's amazing at finding common roots to things. Even when working with hundreds of polynomials in dozens of variables, it sifts through and gives the list of essential constraints and roots so quickly.