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Roots of polynomial 
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#1
Nov2907, 08:46 PM

HW Helper
P: 6,204

Considering the roots of a cubic polynomial([itex]ax^3+bx^2+cx+d[/itex]),[itex]\alpha,\beta,\gamma[/itex]
[tex]\sum \alpha=\frac{b}{a}[/tex] [tex]\sum \alpha\beta=\frac{c}{a}[/tex] [tex]\sum \alpha\beta\gamma=\frac{d}{a}[/tex] If I have those sums of roots..and I am told to find [itex]\alpha^9+\beta^9+\gamma^9[/tex] is there any easy way to find this without having to expand? and also for a quartic polynomial when I expand [itex](x\alpha)(x\beta)(x\gamma)(x\delta)[/itex] I get: [tex]x^4(\alpha+\beta+\alpha\gamma+\beta\gamma+\alpha\delta+\beta\gamma)x^3+(\a lpha\beta+\gamma\delta+\alpha\gamma+\beta\gamma+\alpha\delta+\beta\delt a)x^2 (\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\gamma\delta\be ta)x+\alpha\beta\gamma\delta[/tex] for x^3 I am supposed to get the sum of the roots...yet I expanded correctly, where did i go wrong? 


#2
Nov2907, 09:17 PM

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P: 1,588

On the last bit, you obviously didn't expand correctly, but with no intermediate steps, I don't see how one could say where you went wrong, exactly.



#3
Nov2907, 09:38 PM

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P: 6,204

Oh I thought I typed it out well this is it
[tex](x\alpha)(x\beta)(x\gamma)(x\delta) =(x^2(\alpha+\beta)+)(x^2(\gamma\delta)+(\gamma\delta)[/tex] =[tex]x^4(\alpha+\beta)x^3+\alpha\beta x^2 (\alpha+\beta)(\gamma+\delta)x^3+(\alpha+\beta)(\gamma+\delta)x^2\alpha\beta(\gamma+\delta)x +\alpha\gamma x^2\gamma\delta(\alpha+\beta)x+\alpha\beta\gamma\delta[/tex] = [tex]x^4(\alpha+\beta+\alpha\gamma+\beta\gamma+\alpha\delt a+\beta\gamma)x^3+(\alpha\beta+\gamma\delta+\alpha \gamma+\beta\gamma+\alpha\delta+\beta\delta)x^2 (\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\ delta+\gamma\delta\beta)x+\alpha\beta\gamma\delta[/tex] 


#4
Nov2907, 11:31 PM

P: 75

Roots of polynomial
Something is wrong in your expansion. Try:
[tex](xa)(xb)(xc)(xd) = (x^2  (a+b)x + ab)(x^2  (c+d)x + cd) [/tex] [tex] = x^4  (c+d)x^3 + cdx^2  (a+b)x^3 + (a+b)(c+d)x^2  cd(a+b)x + abx^2  ab(c+d)x + abcd [/tex] [tex] = x^4  (a+b+c+d)x^3 + (ab + cd + ac + ad + bc + bd)x^2  (acd + bcd + abc + abd)x + abcd[/tex] which is what you'd expect. For the first problem, try using the Newton's sums trick. 


#5
Nov3007, 03:47 PM

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P: 1,588

[tex](x\alpha)(x\beta)(x\gamma)(x\delta) =(x^2(\alpha+\beta)x+\alpha\beta)(x^2(\gamma+ \delta)x+ \gamma\delta)[/tex] 


#6
Nov3007, 06:33 PM

HW Helper
P: 6,204

No, I typed it wrongly, on paper I expanded it and found my error...so thanks...
but is there any general formula that will give me the sums of the roots in a form that I need rather than having to expand it? 


#7
Nov3007, 08:04 PM

P: 75

Read the page I linked to about Newton sums. There's nothing faster than that, I think; that method allows you to calculate that kind of stuff pretty quickly though.



#8
Nov3007, 08:27 PM

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#9
Nov3007, 08:37 PM

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P: 6,204

Oh well...
[itex]\sum \alpha[/itex] is simply the sum of the roots taking one at a time, i.e.[itex]\alpha+\beta+\gamma[/itex] and well [itex]\sum \alpha\beta[/itex] is the sum of the roots taking two at a time, i.e. [itex]\alpha\beta+\alpha\gamma+\beta\gamma[/itex] and for newton's sums I get up to the 3rd sum formula but I dont get how I would find an expression to find S[itex]_9[/itex] or for 4 and higher 


#10
Nov3007, 11:38 PM

P: 75

So, in the notation of that link, [tex]a_{nk}[/tex] is the sum of the products of roots taking [tex]k[/tex] at a time. In your cubic equation, [tex]a_3 = a, a_2 = b, a_1 = c, a_0 = d[/tex]. Using the Newton sum equations, you can find [tex]S_1, S_2[/tex] and so on, up through [tex]S_9[/tex], which is what you asked for.
[tex]aS_1 + b = 0[/tex] [tex]aS_2 + bS_1 + 2c = 0[/tex] [tex]aS_3 + bS_2 + cS_1 + 3d = 0[/tex] [tex]aS_4 + bS_3 + cS_2 + dS_1 = 0[/tex] (there's nothing after [tex]d[/tex]) [tex]aS_5 + bS_4 + cS_3 + dS_2 = 0[/tex] .... So you should be able to get all the way to [tex]S_9[/tex] on your own this way. 


#11
Dec107, 01:00 PM

HW Helper
P: 6,204

ah ok...but if there was something after d it would be
[tex]aS_5 + bS_4 + cS_3 + dS_2 + eS_1 = 0[/tex] ? 


#12
Dec107, 01:37 PM

P: 75

right



#13
Dec107, 02:10 PM

HW Helper
P: 6,204

oh thanks, then...this is a real help...Now i can do my roots of polynomials questions even faster now
Edit: so in general the sums would be like this [tex]aS_n + bS_{n1}+cS_{n2}+...+ n[/tex]*(The term independent of x in polynomial) 


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