Find Polynomial Roots: x^4+2x^3-8x^2-18x-9=0

[Mentallic]

Homework Statement

If the sum of two roots of:
x^4+2x^3-8x^2-18x-9=0
are equal to 0, find the roots of the equation.

Homework Equations

For this quartic, P(x)=ax^4+bx^3+cx^2+dx+e

Let the roots be \alpha, \beta, \gamma, \delta

\alpha +\beta +\gamma +\delta=\frac{-b}{a}

\alpha\beta +\alpha\gamma +\alpha\delta +\beta\gamma +\beta\delta +\gamma\delta=\frac{c}{a}

\alpha\beta\gamma +\alpha\beta\delta +\beta\gamma\delta=\frac{-d}{a}

\alpha\beta\gamma\delta=\frac{e}{a}

The Attempt at a Solution

Since 2 roots are equal to 0, i.e. \alpha +\beta=0 therefore, \beta =-\alpha
So I began to rearrange the equations, trying to find values for each root by making it the subject. When I end up 'simplifying' the equations down to 2 simultaneous equations, with 2 roots, trying to make the subject for one becomes chaotic.

Let \beta =-\alpha

Therefore, by the sum of the roots one at a time:

\alpha -\alpha +\beta +\delta=-2

Thus, \beta = -(2+\delta )

Now the roots are: \alpha ,-\alpha ,\delta ,-(\delta +2)

By multiplying roots two at a time:

-\alpha ^2 + \alpha \delta -\alpha (\delta +2) -\alpha\delta +\alpha(\delta +2)-\delta (\delta +2)=-8

Simplified: \alpha ^2+\delta ^2-2\delta =8 (1)

By multiplying three at a time:

-\alpha ^2\delta +\alpha ^2(\delta +2) +\alpha \delta (\delta +2)=18

Simplified: \alpha (2\alpha +\delta ^2+2\delta )=18 (2)

By multiplying all at a time:

\alpha ^2\delta (\delta +2)=-9 (3)

Now with equations (1), (2) and (3). It is possible to find \alpha or \delta by making them the subject through simultaneous equations, but this becomes ridiculously complicated which just looks like another polynomial.
Is there another method to solve this problem, or is the answer to the roots easier than meets the eye?
p.s. I know I can just graph the polynomial to find the roots that way, since they're integral, but I *think* it's meant to be solved somewhat along these lines.

 

[epenguin]

Something like that, you are just making a bit heavy weather of it.

You are looking for integer roots. Their product is -9. This has got to come from either 3, 3, 1, 1 or possibly 9, 1, 1, 1 with pluses and minuses.

Then your \alpha +\beta +\gamma +\delta=\frac{-b}{a}
is

\alpha +\beta +\gamma +\delta= -2

Not many ways you can get -2 out of the above set of numbers, I did it quickly.

 

[gabbagabbahey]

Ewww!

Try assuming that \pm\alpha are your two roots that sum to zero; that tells you that (x\pm\alpha) are factors of your polynomial, and that if you were to factor both of them out, you'd be left with some quadratic equation. That tells you that you can write your polynomial like the following:

(x+\alpha)(x-\alpha)(ax^2+bx+c)=x^4+2x^3-8x^2-18x-9=0

But what is (x+\alpha)(x-\alpha)? Expand your assumed polynomial form and compare the coefficients of each power of x. You should be able to easily determine a,b,c and alpha.

 

[Mentallic]

Something like that, you are just making a bit heavy weather of it.

You are looking for integer roots. Their product is -9. This has got to come from either 3, 3, 1, 1 or possibly 9, 1, 1, 1 with pluses and minuses.

Then your \alpha +\beta +\gamma +\delta=\frac{-b}{a}
is

\alpha +\beta +\gamma +\delta= -2

Not many ways you can get -2 out of the above set of numbers, I did it quickly.

The problem with testing points is that they're not always integer roots. There might be irrational roots which I must find from the resultant quadratic. Of course testing points would work if they're all integral, but wastes valuable time in the test if they're not.

Ewww!

Try assuming that \pm\alpha are your two roots that sum to zero; that tells you that (x\pm\alpha) are factors of your polynomial, and that if you were to factor both of them out, you'd be left with some quadratic equation. That tells you that you can write your polynomial like the following:

(x+\alpha)(x-\alpha)(ax^2+bx+c)=x^4+2x^3-8x^2-18x-9=0

But what is (x+\alpha)(x-\alpha)? Expand your assumed polynomial form and compare the coefficients of each power of x. You should be able to easily determine a,b,c and alpha.

Aha how could I miss this lol.

Transforming the equation into:

(x+\alpha)(x-\alpha)(Ax^2+Bx+C) \equiv x^4+2x^3-8x^2-18x-9=0

thus, \equiv Ax^4+Bx^3+(C-A\alpha ^2)x^2-B\alpha ^2x-C\alpha ^2

A=1,B=2,C=1,\alpha =\pm 3

Thus I end up with the factors: P(x)=(x-3)(x+3)(x+1)^2

Thanks for all the help!

 

[Mark44]

The problem with testing points is that they're not always integer roots. There might be irrational roots which I must find from the resultant quadratic. Of course testing points would work if they're all integral, but wastes valuable time in the test if they're not.

True, the roots of a polynomial are not always integral, but third-degree or higher polynomials can be difficult or even impossible to factor, so you want to eliminate the easy possible roots first.

There is a theorem that is useful, called the Rational Roots Theorem, IIRC. It says that in a polynomial equation a_n * x^n + a_(n - 1) *x^(n - 1) + ... + a_1 * x + a_0 = 0, if p/q is a root of the equation, then p must divide a_0 and q must divide a_n.

In your equation, a_4 = 1 and a_0 = -9. The numerator of any possible rational root has to divide -9, and the denominator of the root has to divide 1, so the possible rational roots are +/-1, +/- 3, and +/-9. You can check these 6 possible roots fairly quickly, especially if you use the technique of synthetic division, which I won't explain here.

You mentioned not wanting to test integral roots (the technique above works for rational roots as well) because it might waste time during a test. If you don't eliminate the easy possibilities right off the bat, the time you save will probably be wasted on trying to get irrational roots by brute force.

Aha how could I miss this lol.

Transforming the equation into:

(x+\alpha)(x-\alpha)(Ax^2+Bx+C) \equiv x^4+2x^3-8x^2-18x-9=0

thus, \equiv Ax^4+Bx^3+(C-A\alpha ^2)x^2-B\alpha ^2x-C\alpha ^2

A=1,B=2,C=1,\alpha =\pm 3

Thus I end up with the factors: P(x)=(x-3)(x+3)(x+1)^2

Thanks for all the help!

 

[Mentallic]

True, the roots of a polynomial are not always integral, but third-degree or higher polynomials can be difficult or even impossible to factor, so you want to eliminate the easy possible roots first.
...
You can check these 6 possible roots fairly quickly, especially if you use the technique of synthetic division, which I won't explain here.

You mentioned not wanting to test integral roots (the technique above works for rational roots as well) because it might waste time during a test. If you don't eliminate the easy possibilities right off the bat, the time you save will probably be wasted on trying to get irrational roots by brute force.

Thanks a lot for your advice on the topic mark44! I guess all the 3rd degree or higher polynomials given in a test would have rational roots, so testing points either by finding P(a) or by this synthetic division you mentioned (I'll read up on it) would be faster.