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Roots of polynomial

by rock.freak667
Tags: polynomial, roots
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rock.freak667
#1
Nov29-07, 08:46 PM
HW Helper
P: 6,202
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?
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Ben Niehoff
#2
Nov29-07, 09:17 PM
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P: 1,593
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.
rock.freak667
#3
Nov29-07, 09:38 PM
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P: 6,202
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]

Xevarion
#4
Nov29-07, 11:31 PM
P: 75
Roots of polynomial

Something is wrong in your expansion. Try:
[tex](x-a)(x-b)(x-c)(x-d) = (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.
Ben Niehoff
#5
Nov30-07, 03:47 PM
Sci Advisor
P: 1,593
Quote Quote by rock.freak667 View Post
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]
The very first line is your mistake. This should be

[tex](x-\alpha)(x-\beta)(x-\gamma)(x-\delta)

=(x^2-(\alpha+\beta)x+\alpha\beta)(x^2-(\gamma+ \delta)x+ \gamma\delta)[/tex]
rock.freak667
#6
Nov30-07, 06:33 PM
HW Helper
P: 6,202
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?
Xevarion
#7
Nov30-07, 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.
Integral
#8
Nov30-07, 08:27 PM
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Quote Quote by rock.freak667 View Post
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?
I have to admit that I do not understand your notation. If the 3 roots are [itex] \alpha, \beta, \gamma [/itex] Then what do your sums mean?
rock.freak667
#9
Nov30-07, 08:37 PM
HW Helper
P: 6,202
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
Xevarion
#10
Nov30-07, 11:38 PM
P: 75
So, in the notation of that link, [tex]a_{n-k}[/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.
rock.freak667
#11
Dec1-07, 01:00 PM
HW Helper
P: 6,202
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] ?
Xevarion
#12
Dec1-07, 01:37 PM
P: 75
right
rock.freak667
#13
Dec1-07, 02:10 PM
HW Helper
P: 6,202
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_{n-1}+cS_{n-2}+...+ n[/tex]*(The term independent of x in polynomial)


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