# Multiplying Roots of a Polynomial

Hello - I'm new to SAGE, so trying to get to grips with the basics!

I have a polynomial $f = x^{2}((x+\frac{1}{x})^{2} - 1)$. I would like to be able to do to things:

1) See $f$ in the form $\beta(x - \alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)$

2) Know the value of $\beta (\prod_i \alpha_i)$

Using

f.factor

f.roots

Does not give me the linear factorisation, nor can I see a simple way of getting the product of the roots.

In the long term I'm hoping to do this for polynomials of larger degree, so any advice would be great!

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You can also get exact results by using the Algebraic Field QQbar, which avoids the need of manually constructing a splitting field (like rburing shows in his answer).

In Sage 8.4, I'd also suggest setting QQbar's display option to radical instead of decimal (the default). Try both and see the difference; it's the simplest way I know how to explain it!

sage: QQbar.options.display_format = 'radical'
sage: R.<x> = PolynomialRing(QQbar)
sage: f = x^2*((x+1/x)^2-1)
sage: f.factor()
(x - 1/2*I*sqrt(3) - 1/2) * (x + 1/2*I*sqrt(3) - 1/2) * (x - 1/2*I*sqrt(3) + 1/2) * (x + 1/2*I*sqrt(3) + 1/2)
sage: mul([p.constant_coefficient() for p,m in f.factor()])
1

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Let's see...

sage: f=x^2*((x+1/x)^2-1)


The roots are :

sage: f.roots(multiplicities=False)
[-sqrt(1/2*I*sqrt(3) - 1/2),
sqrt(1/2*I*sqrt(3) - 1/2),
-sqrt(-1/2*I*sqrt(3) - 1/2),
sqrt(-1/2*I*sqrt(3) - 1/2)]


Therefore, the factorized polynom is :

sage: prod([x-u for u in f.roots(multiplicities=False)])
(x + sqrt(1/2*I*sqrt(3) - 1/2))*(x - sqrt(1/2*I*sqrt(3) - 1/2))*(x + sqrt(-1/2*I*sqrt(3) - 1/2))*(x - sqrt(-1/2*I*sqrt(3) - 1/2))


And the roots' products is:

sage: prod(f.roots(multiplicities=False)).expand()
1


Whereas moving to a polynomial ring, more specialized, is often useful, it is not necessary here...

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I would argue polynomial rings are simpler than the symbolic ring. It depends on your point of view. In any case, it is good to see both approaches.

( 2019-01-17 16:39:49 -0500 )edit

Thank you - this is perfect for what I'm after!

( 2019-01-18 03:11:45 -0500 )edit

For fun, you can check what happens for f = x^5 - x - 1 in both answers.

( 2019-01-18 03:36:23 -0500 )edit

Do you want exact results or approximations? For exact results, you can define

R.<x> = PolynomialRing(QQ)
f = R(x^2*((x+1/x)^2-1))
K.<a> = f.splitting_field()


so you can do:

sage: f.change_ring(K).factor()
(x - a) * (x + a - 1) * (x - a + 1) * (x + a)
sage: f.change_ring(K).roots()
[(a, 1), (-a + 1, 1), (a - 1, 1), (-a, 1)]
sage: myprod = beta*prod(r[0]^r[1] for r in f.change_ring(K).roots()); myprod
1


By Vieta's formulas what you get is just $(-1)^n f_0$ where $n = \deg(f)$ and $f_0$ is the constant coefficient of $f$.

sage: myprod == (-1)^f.degree()*f.constant_coefficient()
True


If you are wondering what $a$ is in the discussion above: it is a root of

sage: a.minpoly()
x^2 - x + 1


As for numerics, you can do e.g.:

sage: f.change_ring(QQbar).roots()
[(-0.500000000000000? - 0.866025403784439?*I, 1),
(-0.500000000000000? + 0.866025403784439?*I, 1),
(0.500000000000000? - 0.866025403784439?*I, 1),
(0.500000000000000? + 0.866025403784439?*I, 1)]
sage: beta*prod(r[0]^r[1] for r in f.change_ring(QQbar).roots())
1.000000000000000? + 0.?e-18*I

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