Roots of Cubic & Quartic Polynomials - Finding Sums & Expanding

In summary, the conversation discusses the sums of roots in a cubic polynomial and a quartic polynomial, as well as a method for finding the sum of the 9th powers of the roots without having to expand. The conversation also introduces the concept of Newton sums, which can be used to find the sums of products of roots in a polynomial using a set of equations. These equations can be applied to find the sums of higher powers of the roots as well.
  • #1
rock.freak667
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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+(\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]
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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  • #2
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
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
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 http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums" trick.
 
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  • #5
rock.freak667 said:
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]
 
  • #6
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
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
rock.freak667 said:
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+(\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]
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?
 
  • #9
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 don't get how I would find an expression to find S[itex]_9[/itex] or for 4 and higher
 
  • #10
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.
 
  • #11
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
right
 
  • #13
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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FAQ: Roots of Cubic & Quartic Polynomials - Finding Sums & Expanding

What are the roots of a cubic polynomial?

The roots of a cubic polynomial are the values of the variable that make the polynomial equal to zero. A cubic polynomial has three roots, which can be real or complex numbers.

How do you find the sum of the roots of a cubic polynomial?

The sum of the roots of a cubic polynomial can be found by dividing the second coefficient by the first coefficient. This will give you the negative sum of the roots.

Can a cubic polynomial have more than three roots?

No, a cubic polynomial can only have three roots. This is because a cubic polynomial is a third-degree polynomial, meaning it has a maximum of three terms.

What is the difference between expanding and factoring a polynomial?

Expanding a polynomial means to multiply out all the terms and simplify, while factoring means to break down a polynomial into its factors. Expanding is the reverse process of factoring.

How do you expand a quartic polynomial?

To expand a quartic polynomial, you must multiply each term by every other term, resulting in a polynomial with four terms. Then, you can simplify the terms and combine like terms to get the expanded form of the polynomial.

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