Let me please also join the party.
"Short" answer, which gives sage code for the roots, manually computed assisted by sage:
R.<x> = PolynomialRing( QQ )
f = x^5 + x^4 - 8*x^3 + 11*x^2 - 15*x + 2
f.factor()
# the factor (x-2) is simple, the other one...
g = x^4 + 3*x^3 - 2*x^2 + 7*x - 1
e,d,c,b,a = g.coefficients()
p = ( 8*a*c - 3*b^2 ) / ( 8*a^2 )
q = ( b^3 - 4*a*b*c + 8*a^2*d ) / ( 8*a^3 )
U = 649/27 + 2*sqrt(7214/27)
V = 649/27 - 2*sqrt(7214/27)
U3 = U ^(1/3)
V3 = -(-V)^(1/3) # since V^(1/3) is not real
S = (1/2) * sqrt( 43/12 + U3 + V3 )
signs = [ -1, 1 ]
myroots = [ -b/4/a + s1*S + s2*sqrt( - S^2 - p/2 - s1*q/4/S )
for s1 in signs
for s2 in signs ]
for myroot in myroots:
print myroot
The printed roots are somehow parallely digestible in this form.
The first printed one is after manual rearrangement numerically
sage: (
....: -1/2*sqrt( (2/3*sqrt(7214/3) + 649/27)^(1/3)
....: - (2/3*sqrt(7214/3) - 649/27)^(1/3) + 43/12)
....: - sqrt( - 1/4*(2/3*sqrt(7214/3) + 649/27)^(1/3)
....: + 1/4*(2/3*sqrt(7214/3) - 649/27)^(1/3)
....: + 107/16/sqrt( (2/3*sqrt(7214/3) + 649/27)^(1/3)
....: - (2/3*sqrt(7214/3) - 649/27)^(1/3) + 43/12)
....: + 43/24)
....: - 3/4
....: ).n()
....:
-3.96552349222940
sage: g.roots( ring=QQbar, multiplicities=False )[0]
-3.965523492229398?
and it coincides with the first root, computed by sage in QQbar.
Longer answer
I appologize in advance for this long answer,
did the work for myself, to make it once for all times clear
how to figure out the four roots of g = x^4 + 3*x^3 - 2*x^2 + 7*x - 1 in a structural way.
Here, very quickly, the problem translates in finding the roots of the
factor g of degree four in
sage: R.<x> = PolynomialRing( QQ )
sage: f = x^5 + x^4 - 8*x^3 + 11*x^2 - 15*x + 2
sage: f.factor()
(x - 2) * (x^4 + 3*x^3 - 2*x^2 + 7*x - 1)
sage: g = x^4 + 3*x^3 - 2*x^2 + 7*x - 1
and to give their description using only radicals.
We can build
sage: K.<a> = NumberField( g )
sage: L.<b> = K.galois_closure()
sage: L.degree()
24
sage: L
Number Field in b with defining polynomial x^24 + 24*x^23 + 334*x^22 + 8432*x^21 + ...
and so on, result was truncated before coefficients of L.polynomial() get to big.
There are many examples in many answers on this site, where we compute in
one of the "big" fields AA, or QQbar, or SR.
Let us work here in K and L.
(I appologize again, there will be a dry exposition till the point with the direct connection to the question,
which is bold face marked in the sequel. The reader in hurry should please scroll down.)
The answer applies only for (cubic, and) quartic equations. This particular polynom g is a good example
to understand the Galois group of the Galois closure of the roots of the quartic. So let us use it.
Note that we have:
sage: g.discriminant().factor()
-1 * 2^5 * 3607
and also, looking into some subfields:
for M, mapM, _ in L.subfields( degree=2 ):
print "DISC = %s for %s" % ( M.discriminant().factor(), M )
This gives the one field of degree two,
DISC = -1 * 2^3 * 3607 for Number Field in b0 with defining polynomial x^2 + 15072*x + 73412352
and the same, using degree=3 as parameter above gives:
DISC = -1 * 2^3 * 3607 for Number Field in b0 with defining polynomial x^3 + 92*x^2 - 24064*x - 6486016
DISC = -1 * 2^3 * 3607 for Number Field in b1 with defining polynomial x^3 + 92*x^2 + 3896*x + 223904
DISC = -1 * 2^3 * 3607 for Number Field in b2 with defining polynomial x^3 + 92*x^2 - 18200*x + 1414048
The solution of the quartic
is given for instance here:
wiki/Quartic_function#General_formula_for_roots
The main idea is to find the roots in the form:
$$ A\pm_1 B\pm_2\sqrt{C\mp_1D} $$ where the signs $\pm_1 $ and $\mp_1 $ do correspond, their product is $-1$,
and the sign $\pm_2$ can be chosen independently.
A simple computation
var( 'A,B,C,D,X' );
signs = [ -1, 1 ]
prod( [ X - ( A+s1*B + s2*sqrt( C+s1*D ) ) for s1 in signs for s2 in signs ] ).expand()
shows how to search for A,B,C,D, when the equation is given.
We usually clear first the factor in $x^3$ by a linear substitution, so that after this step $A=0$.
The value of $A$ is simple. The main point is to find the value of $B$, which becomes an $S$ in the sequel.
More precisely now. The wiki article gives for the equation
$$ ax^4 + bx^3 + cx^2 + dx + e = 0 $$
the solution
$$ -\frac b{4a}\pm_1 S\pm_2 \sqrt{ -S^2 -\frac p2 \mp_1 \frac q{4S}} \ , $$
and the main complication is getting $S$. Above, $p,q$ are simple rational expressions involving
$a,b,c,d$. If we give weights $0,1,2,3,4$ to the coefficients $a,b,c,d,e$,
then $S$ is of weight $1$, as $b/(4a)$, $p$ of weight $2$, as $-S^2$, and $q$ of weight three.
So the particular case extracted from the question, is asking for $S$, if we have $S$,
we have basicly solved and understood the situation.
Let us ask sage for the algebraic equation satisfied by the
sum of two roots $x_0+x_1$, for some choice of the order of the roots.
Note that considering all choices...
g_roots = g.roots( ring=L, multiplicities=False )
for k in [0..3]:
for kk in [k+1..3]:
xk = g_roots[ k]
xkk = g_roots[kk]
print "x%s + x%s has minpoly = %s" % ( k, kk, (xk+xkk).minpoly() )
is given:
x0 + x1 has minpoly = x^6 + 9*x^5 + 23*x^4 + 3*x^3 - 7*x^2 + 87*x - 82
x0 + x2 has minpoly = x^6 + 9*x^5 + 23*x^4 + 3*x^3 - 7*x^2 + 87*x - 82
x0 + x3 has minpoly = x^6 + 9*x^5 + 23*x^4 + 3*x^3 - 7*x^2 + 87*x - 82
x1 + x2 has minpoly = x^6 + 9*x^5 + 23*x^4 + 3*x^3 - 7*x^2 + 87*x - 82
x1 + x3 has minpoly = x^6 + 9*x^5 + 23*x^4 + 3*x^3 - 7*x^2 + 87*x - 82
x2 + x3 has minpoly = x^6 + 9*x^5 + 23*x^4 + 3*x^3 - 7*x^2 + 87*x - 82
and the choice of the two roots is rather immaterial.
Alternatively, if we work in L:
sage: K.<a> = NumberField( g )
sage: L.<b> = K.galois_closure()
sage: maps = K.embeddings(L)
sage: len( maps )
4
sage: for m1, m2 in Combinations( maps, 2 ):
....: print "minpoly =", ( m1(a) + m2(a) ).minpoly()
....:
minpoly = x^6 + 9*x^5 + 23*x^4 + 3*x^3 - 7*x^2 + 87*x - 82
minpoly = x^6 + 9*x^5 + 23*x^4 + 3*x^3 - 7*x^2 + 87*x - 82
minpoly = x^6 + 9*x^5 + 23*x^4 + 3*x^3 - 7*x^2 + 87*x - 82
minpoly = x^6 + 9*x^5 + 23*x^4 + 3*x^3 - 7*x^2 + 87*x - 82
minpoly = x^6 + 9*x^5 + 23*x^4 + 3*x^3 - 7*x^2 + 87*x - 82
minpoly = x^6 + 9*x^5 + 23*x^4 + 3*x^3 - 7*x^2 + 87*x - 82
Now we can try:
x0, x1, x2, x3 = g_roots
( x0 + x1 + 9/6 ).minpoly() # this is (2S).minpoly()
(Above, $9/6=3/2=2\cdot b/(4a)$ is clearing the monomial in $x^5$,
and it corresponds to the term $-b/(4a)$ from the wiki formula.)
This gives:
x^6 - 43/4*x^4 + 995/16*x^2 - 11449/64
Which is a cubic equation in $x^2$. We can further ask for...
sage: ( ( x0 + x1 + 9/6 )^2 - 43/12 ).minpoly()
x^3 + 71/3*x - 1298/27
sage: _.discriminant().factor()
-1 * 2^5 * 3607
sage: ( ( x0 + x1 + 9/6 )^2 - 43/12 ).minpoly().roots( ring=QQbar, multiplicities=False )
[1.789260758493839?,
-0.8946303792469193? - 5.105659331867054?*I,
-0.8946303792469193? + 5.105659331867054?*I]
sage: for r in _:
....: print r, '=', r.radical_expression()
1.789260758493839? = (2/9*sqrt(7214)*sqrt(3) + 649/27)^(1/3) - 71/9/(2/9*sqrt(7214)*sqrt(3) + 649/27)^(1/3)
and two further solutions. All three are of the form
$$ \epsilon \left( \frac{649}{27} +\frac 29\sqrt{7214\cdot 3} \right)^{1/3}
+ \epsilon^2\left( \frac{649}{27} -\frac 29\sqrt{7214\cdot 3} \right)^{1/3}\ , $$
where $\epsilon$ is one of the three roots of unity, either $1$, or one of the two primitive roots,
and the two radicals are taken so that their products is the real number
$71/9$:
( ( 649/27 + 2/9*sqrt(7214)*sqrt(3) ) * \
( 649/27 - 2/9*sqrt(7214)*sqrt(3) ) ).expand()^(1/3)
which gives:
71/9*(-1)^(1/3)
We can now write simple sage code, using radicals, which generates the four roots of g.
coeffs = g.coefficients()
coeffs . reverse()
a,b,c,d,e = coeffs
p = ( 8*a*c - 3*b^2 ) / ( 8*a^2 )
q = ( b^3 - 4*a*b*c + 8*a^2*d ) / ( 8*a^3 )
U = 649/27 + 2*sqrt(7214/27)
V = 649/27 - 2*sqrt(7214/27)
U3 = U ^(1/3)
V3 = -(-V)^(1/3) # since V^(1/3) is not real
S = (1/2) * sqrt( 43/12 + U3 + V3 )
signs = [ -1, 1 ]
myroots = [ -b/4/a + s1*S + s2*sqrt( - S^2 - p/2 - s1*q/4/S )
for s1 in signs
for s2 in signs ]
print "The roots computed with bare hands starting from the sage value\nS = %s" % S
print "and setting root = %s (+/-) S + [+/-] sqrt( -S^2 - %s (-/+) %s/S )" % ( -b/4/a, -p/2, q/4 )
for r in myroots:
print r.n()
print
print "The roots computed with sage in QQbar"
for rr in g.roots( ring=QQbar, multiplicities=False ):
print rr
(The code in the above block is basicly the short answer.)
Results:
The roots computed with bare hands starting from the sage value
S = 1/2*sqrt((2/3*sqrt(7214/3) + 649/27)^(1/3) - (2/3*sqrt(7214/3) - 649/27)^(1/3) + 43/12)
and setting root = -3/4 (+/-) S + [+/-] sqrt( -S^2 - 43/16 (-/+) 107/32/S )
-3.96552349222940
0.147637797932293
0.408942847148552 - 1.24129810909174*I
0.408942847148552 + 1.24129810909174*I
The roots computed with sage in QQbar
-3.965523492229398?
0.1476377979322930?
0.4089428471485524? - 1.241298109091741?*I
0.4089428471485524? + 1.241298109091741?*I
The explicit computation shows also how the Galois group of L acts on the
roots (and on "similar" expressions).
sage: G.<v> = K.galois_group()
sage: G.order()
24
sage: G.gens()
[(1,12)(2,19)(3,17)(4,22)(5,11)(6,8)(7,16)(9,10)(13,14)(15,23)(18,21)(20,24),
(1,18)(2,21)(3,11)(4,16)(5,20)(6,7)(8,23)(9,17)(10,24)(12,14)(13,19)(15,22),
(1,19)(2,12)(3,24)(4,8)(5,9)(6,22)(7,15)(10,11)(13,18)(14,21)(16,23)(17,20),
(1,20,7)(2,4,11)(3,14,8)(5,22,19)(6,13,17)(9,15,18)(10,21,23)(12,16,24)]
sage: G.is_solvable()
True
sage: G.is_polycyclic()
True
sage: G.is_supersolvable()
False
sage: P = PermutationGroup( G.gens() )
sage: for N in P.normal_subgroups():
....: print "Normal subgroup of order %s" % N.order()
....:
Normal subgroup of order 1
Normal subgroup of order 4
Normal subgroup of order 12
Normal subgroup of order 24
sage: N = [ N for N in P.normal_subgroups() if N.order() == 4 ][0]
sage: PQ = P.quotient(N)
sage: N.gens()
[(1,18)(2,21)(3,11)(4,16)(5,20)(6,7)(8,23)(9,17)(10,24)(12,14)(13,19)(15,22),
(1,19)(2,12)(3,24)(4,8)(5,9)(6,22)(7,15)(10,11)(13,18)(14,21)(16,23)(17,20)]
sage: PQ.order()
6
sage: PQ.is_commutative()
False
So $G=Gal(L:\mathbb Q)$ has a normal subgroup
$\cong (Z/2)^2$ and it is a twisted product of it with the symmetric group with $3!$ elements.
The one element of order 3 acts by replacing U3, V3 above by an other
choice for the roots of order three of U, V, constrained to have a fixed product.
The elements of order two act by changing signs in the places where we
used $\pm$...
The key is to use
radical_expressionfor algebraic numbers. See the answers to similar questions: https://ask.sagemath.org/questions/query:radical_expression/ .