# Roots of polynomial

by rock.freak667
Tags: polynomial, roots
 HW Helper P: 6,202 Considering the roots of a cubic polynomial($ax^3+bx^2+cx+d$),$\alpha,\beta,\gamma$ $$\sum \alpha=\frac{-b}{a}$$ $$\sum \alpha\beta=\frac{c}{a}$$ $$\sum \alpha\beta\gamma=\frac{-d}{a}$$ If I have those sums of roots..and I am told to find $\alpha^9+\beta^9+\gamma^9[/tex] is there any easy way to find this without having to expand? and also for a quartic polynomial when I expand [itex](x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$ I get: $$x^4-(\alpha+\beta+\alpha\gamma+\beta\gamma+\alpha\delta+\beta\gamma)x^3+(\a lpha\beta+\gamma\delta+\alpha\gamma+\beta\gamma+\alpha\delta+\beta\delt a)x^2 -(\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\gamma\delta\be ta)x+\alpha\beta\gamma\delta$$ for -x^3 I am supposed to get the sum of the roots...yet I expanded correctly, where did i go wrong?
 Sci Advisor PF Gold P: 1,594 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.
 HW Helper P: 6,202 Oh I thought I typed it out well this is it $$(x-\alpha)(x-\beta)(x-\gamma)(x-\delta) =(x^2-(\alpha+\beta)+)(x^2-(\gamma\delta)+(\gamma\delta)$$ =$$x^4-(\alpha+\beta)x^3+\alpha\beta x^2 -(\alpha+\beta)(\gamma+\delta)x^3+(\alpha+\beta)(\gamma+\delta)x^2-\alpha\beta(\gamma+\delta)x +\alpha\gamma x^2-\gamma\delta(\alpha+\beta)x+\alpha\beta\gamma\delta$$ = $$x^4-(\alpha+\beta+\alpha\gamma+\beta\gamma+\alpha\delt a+\beta\gamma)x^3+(\alpha\beta+\gamma\delta+\alpha \gamma+\beta\gamma+\alpha\delta+\beta\delta)x^2 -(\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\ delta+\gamma\delta\beta)x+\alpha\beta\gamma\delta$$
 P: 75 Roots of polynomial Something is wrong in your expansion. Try: $$(x-a)(x-b)(x-c)(x-d) = (x^2 - (a+b)x + ab)(x^2 - (c+d)x + cd)$$ $$= x^4 - (c+d)x^3 + cdx^2 - (a+b)x^3 + (a+b)(c+d)x^2 - cd(a+b)x + abx^2 - ab(c+d)x + abcd$$ $$= x^4 - (a+b+c+d)x^3 + (ab + cd + ac + ad + bc + bd)x^2 - (acd + bcd + abc + abd)x + abcd$$ which is what you'd expect. For the first problem, try using the Newton's sums trick.
PF Gold
P: 1,594
 Quote by rock.freak667 Oh I thought I typed it out well this is it $$(x-\alpha)(x-\beta)(x-\gamma)(x-\delta) =(x^2-(\alpha+\beta)+)(x^2-(\gamma\delta)+(\gamma\delta)$$
The very first line is your mistake. This should be

$$(x-\alpha)(x-\beta)(x-\gamma)(x-\delta) =(x^2-(\alpha+\beta)x+\alpha\beta)(x^2-(\gamma+ \delta)x+ \gamma\delta)$$
 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?
 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.
Emeritus
 Quote by rock.freak667 Considering the roots of a cubic polynomial($ax^3+bx^2+cx+d$),$\alpha,\beta,\gamma$ $$\sum \alpha=\frac{-b}{a}$$ $$\sum \alpha\beta=\frac{c}{a}$$ $$\sum \alpha\beta\gamma=\frac{-d}{a}$$ If I have those sums of roots..and I am told to find $\alpha^9+\beta^9+\gamma^9[/tex] is there any easy way to find this without having to expand? and also for a quartic polynomial when I expand [itex](x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$ I get: $$x^4-(\alpha+\beta+\alpha\gamma+\beta\gamma+\alpha\delta+\beta\gamma)x^3+(\a lpha\beta+\gamma\delta+\alpha\gamma+\beta\gamma+\alpha\delta+\beta\delt a)x^2 -(\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\gamma\delta\be ta)x+\alpha\beta\gamma\delta$$ 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 $\alpha, \beta, \gamma$ Then what do your sums mean?
 HW Helper P: 6,202 Oh well... $\sum \alpha$ is simply the sum of the roots taking one at a time, i.e.$\alpha+\beta+\gamma$ and well $\sum \alpha\beta$ is the sum of the roots taking two at a time, i.e. $\alpha\beta+\alpha\gamma+\beta\gamma$ 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$_9$ or for 4 and higher
 P: 75 So, in the notation of that link, $$a_{n-k}$$ is the sum of the products of roots taking $$k$$ at a time. In your cubic equation, $$a_3 = a, a_2 = b, a_1 = c, a_0 = d$$. Using the Newton sum equations, you can find $$S_1, S_2$$ and so on, up through $$S_9$$, which is what you asked for. $$aS_1 + b = 0$$ $$aS_2 + bS_1 + 2c = 0$$ $$aS_3 + bS_2 + cS_1 + 3d = 0$$ $$aS_4 + bS_3 + cS_2 + dS_1 = 0$$ (there's nothing after $$d$$) $$aS_5 + bS_4 + cS_3 + dS_2 = 0$$ .... So you should be able to get all the way to $$S_9$$ on your own this way.
 HW Helper P: 6,202 ah ok...but if there was something after d it would be $$aS_5 + bS_4 + cS_3 + dS_2 + eS_1 = 0$$ ?
 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 $$aS_n + bS_{n-1}+cS_{n-2}+...+ n$$*(The term independent of x in polynomial)