# Alg 2

Is there another way to find the roots of 3rd degree and higher polynomials instead of using trial and error (using synthetic division)?

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1. One way it helps to look at roots of 3rd degree is that it's the same as the what is under the root to the power of (1/3).

Ex:
(3√x)= x^(1/3)

I would then do this with a higher polynomial like
(3√x^5)= (x^5)^(1/3)
= x^(5/3)

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2. I use Newton's method, which involves repeated iterations, where you start with some reasonable guess and within a few steps it will "Zoom in" on the answer

in general it says
if for some function f(x) , starting with a guess of x
new x = x - f(x)/derivative of f(x)

so for a cubic
f(x) = ax^3 + bx^2 + cx + d

then
newx = (2ax^3 + bx^2 - d)/(3ax^2 + 2bx + c)

the result of newx then becomes the input for x in the second round.
Your newx will get closer and closer to your input value of x
You will be amazed how fast it is.

BTW, no matter how 'weird' your first guess is, eventually it will get you there, it just takes longer.

BTW#2
for square roots, eg. √5 it would work this way

newx = (x^2+5)/(2x)

guess at x = 2

newx - x
---- 2.25
2.25 ---2.236111
2.236111 -- 2.2360679
2.2360679 -- 2.2360679 same as input

4 steps to get 7 decimal accuracy !!!

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3. For polynomials of third and higher degree, it matters whether you are looking for an exact solution (factorable) or contented with a numerical answer.

For factorable polynomials, apart from testing by trial and error for integer roots, you can read up on Descartes rule of signs to help you limit the number of trials.
See:
http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs

There are many tools for numerical solutions. One of them is graphing, another is by iteration. One of the well-known iteration mthods is Newton's method. See for example:
http://en.wikipedia.org/wiki/Newton%27s_method

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4. There is an analytical way of getting a exact closed-form solutions to cubic equations, but it fills two pages. For higher order polynomials, I am not aware of any exact methods.

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