Finding roots of a polynomial

Question

LukeO 2016-11-10

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/311631-finding-roots-of-a-polynomial

编辑： James Tursa 2016-11-10

I've carried out manual calculations of newton raphson and have proven it in MATLAB with the following code in the command window:

>> f= 0.5*x.^3 + x.^2 - 10*x+14.037;

>> roots ([0.5 1 -10 14.037])

ans =

-6.0627 + 0.0000i
 2.0313 + 0.7101i
 2.0313 - 0.7101i

I know the -6.0627 is correct as this is the same as my manual calculation, but I don't understand the following 2 answers MATLAB has provided. Could someone explain this in reasonably simple terms what they mean?

Thanks.

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Answer 1

James Tursa 2016-11-10

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/311631-finding-roots-of-a-polynomial#answer_242815

编辑：James Tursa 2016-11-10

在 MATLAB Online 中打开

An n'th order polynomial will have n roots, some of which may be repeated or complex. You have a 3rd order polynomial which will have 3 roots. Turns out one of the roots is real and the other two are complex (meaning a plot of the polynomial will only cross the x-axis once). If you plug the roots into the polynomial you can see that all three results are very close to zero as expected:

>> f = @(x) 0.5*x.^3 + x.^2 - 10*x + 14.037
f = 
    @(x)0.5*x.^3+x.^2-10*x+14.037
>> r = roots ([0.5 1 -10 14.037])
r =
 -6.062670505918885                     
  2.031335252959442 + 0.710147689980882i
  2.031335252959442 - 0.710147689980882i
>> f(r)
ans =
  1.0e-013 *
  0.230926389122033                     
  0.035527136788005 - 0.062172489379009i
  0.035527136788005 + 0.062172489379009i

Look at it this way, if you take your real root and form (x+6.062670505918885) and factor that out of your polynomial, you will be left with a 2nd order polynomial that has no real roots.