
#1
Nov2812, 10:33 PM

P: 57

Hi.I'd like to ask a question about polynomal equations.
http://www.intmath.com/equationsof...functions.php here is the solution polynomial function of degree 4. graphically.But I don't understand the method ? How we can use graphic method to solve? or can we use? for example ; x^3+6x^2+15x+18 Can we solve this withous using the factorisation method. 



#2
Nov2812, 10:51 PM

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Step 1: Use a computer program to plot the graph (precomputers, we'd get good at guessing by plotting a few points.)
Step 2: using the mk1 eyeball, get approximate positions for the roots Step 3: use a computer program to refine these approximations to some prespecified accuracy. the computer program will typically use the NewtonRaphson algorithm to refine the guess. It is not really a graphical method  it is a numerical method. 



#3
Nov2912, 04:24 AM

P: 57

Understood :) I thought that they're usign the first and second derivative of a function to find the roots from graph.




#4
Nov2912, 04:30 AM

P: 57

Polynoms Graphically?
Understood :) I thought that they're usign the first and second derivative of a function to find the roots from graph.
I mean ; http://en.wikipedia.org/wiki/Maxima_and_minima 



#5
Nov2912, 08:41 PM

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NewtonRaphson uses the first derivative to refine guesses.
Graphically, you guess some value, take the tangent to f(x) at that value: where the tangent intercepts the x axis will be closer to the root than the initial guess. Now take the tangent for that value: wash and repeat. For instance  say we wanted to find the roots of ##y=(x1)^2## by this method  lets illustrate the method by making a silly guess of x=4 That means y=9 ... no: not a root. The slope of the tangent is y'(4)=6 and the equation of the tangent is y=6x15 so the intercept with the x axis is x=15/6 which is closer to the actual root (x=1) than the initial guess. The derivatives are only indirectly useful for finding the roots  you'll see the article you have referenced shows how to find and characterize the critical points. See: http://en.wikipedia.org/wiki/Newton's_method 



#6
Nov3012, 07:15 AM

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You seem to be confusing a number of concepts. You cannot "solve" a polynomial such as [itex]x^3+ 6x^2+ 15x+ 18[/itex]. You can solve a polynomial equation such as [itex]x^3+ 6x^2+ 15x+ 18= 0[/itex]
There is a formula for solving cubic equations: look at the Cardano formula http://www.sosmath.com/algebra/factor/fac11/fac11.html There is a related formula for solving quartic (fourth degree) equations. It can be shown that there cannot be a similar formula for equations of degree 5 or higher because they can have roots that cannot be written in terms of roots and the other arithmetic operations. 



#7
Nov3012, 08:46 AM

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I think he's just being sloppy about language.
The problem he wants to solve is of form "find the roots of..."  without factorizing. <grizzle> Did you see the webpage he linked to? ... it calls the factorization approach "the dinosaur method" and leaves it as a "if you really must"... favoring numerical methods instead. OK it's chatty but how many people using the site have decided they don't need to be dinosaurs? </grizzle> I suppose, for completeness: ##y=x^3+6x^2+15x+18## has one real root at ##x=3## Hopefully weve covered all the bases  I'm sure Erbil can figure out how to use regression on complex numbers ... though this one is probably best done by quadratic equation now we know one root. 



#8
Dec112, 05:03 AM

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#9
Dec112, 05:16 AM

P: 57

Ok I have confused something. =) The web page that I linked is about polynomal functions.So functions can be graphed.But in my equation,only root can be graph.Because there's no depended variable to x.
There's the difference; http://www.wolframalpha.com/input/?i...2B15x%2B18%3D0 for roots, http://www.wolframalpha.com/input/?i...5E2%2B15x%2B18 for function. 



#10
Dec112, 05:20 AM

P: 57

ohh=) but in intmath.com he wrotes;
Finding the roots of an equation, for example x4 − x3 − 19x2 − 11x + 31 = 0, f(x) = x4 − x3 − 19x2 − 11x + 31 We see that there are 4 roots, at approximately x = 3, x = 2, x = 1, x = 5. 



#11
Dec112, 08:29 AM

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