A possibility is to start with A, B, D
and ask for the roots of
D
, condidered as a function of degree two in M
. The code is:
var( 'p,M' )
A = 18*p^10  54*p^9 + 59*p^8 + 130*p^7  209*p^6  98*p^5 + 407*p^4 + 362*p^3 + 49*p^2  16*p + 8
B = 9 \
* (p + 1)^2 \
* (p^4  2*p^3 + 2*p^2 + 2*p + 1) \
* (4*p^8  52*p^7 + 373*p^6 + 68*p^5  445*p^4 + 72*p^3 + 163*p^2  48*p+ 9)
D = M^2  A/2/p^4*M + B/16/p^4
solutions = solve( D==0, M )
for sol in solutions:
print sol
This gives:
M == 1/4*(18*p^10  54*p^9 + 59*p^8 + 130*p^7  209*p^6  98*p^5 + 407*p^4 + 362*p^3 + 49*p^2

2*(9*p^8  18*p^7  7*p^6 + 45*p^5  21*p^4  74*p^3  18*p^2 + 6*p  2)
*sqrt(p^4  2*p^3 + 5*p^2 + 8*p + 4)  16*p + 8)/p^4
M == 1/4*(18*p^10  54*p^9 + 59*p^8 + 130*p^7  209*p^6  98*p^5 + 407*p^4 + 362*p^3 + 49*p^2
+
2*(9*p^8  18*p^7  7*p^6 + 45*p^5  21*p^4  74*p^3  18*p^2 + 6*p  2)
*sqrt(p^4  2*p^3 + 5*p^2 + 8*p + 4)  16*p + 8)/p^4
(Result was manually broken.)
Alternatively, one can ask for the factorization of the discriminant of D
, seen as a
polynomial in M
. The rest is applying the formula for the roots of an equation
of degree two in the variable M
. A possible code for this path is:
R0.<p> = PolynomialRing( QQ )
F0 = R0.fraction_field()
R.<M> = PolynomialRing( F0 )
A = 18*p^10  54*p^9 + 59*p^8 + 130*p^7  209*p^6  98*p^5 + 407*p^4 + 362*p^3 + 49*p^2  16*p + 8
B = 9 \
* (p + 1)^2 \
* (p^4  2*p^3 + 2*p^2 + 2*p + 1) \
* (4*p^8  52*p^7 + 373*p^6 + 68*p^5  445*p^4 + 72*p^3 + 163*p^2  48*p+ 9)
D = M^2  A/2/p^4*M + B/16/p^4
print D.discriminant().factor()
latex( D.discriminant().factor() )
This gives:
(81) * p^8 * (p^4  2*p^3 + 5*p^2 + 8*p + 4) * (p^8  2*p^7  7/9*p^6 + 5*p^5  7/3*p^4  74/9*p^3  2*p^2 + 2/3*p  2/9)^2
The discriminant has thus in its factorization the posted factors
v
, simple factor,  and
F
, with the power two.
The latex print for the factorized discriminant is displayed here, also after some small change, as follows:
$$\left(81\right) \cdot p^{8} \cdot (p^{4}  2 p^{3} + 5 p^{2} + 8 p + 4) \cdot \left(p^{8}  2 p^{7}  \frac{7}{9} p^{6} + 5 p^{5}  \frac{7}{3} p^{4}  \frac{74}{9} p^{3}  2 p^{2} + \frac{2}{3} p  \frac{2}{9} \right)^{2} \ .$$
(Use the $81=9^2$ for the squared factor to get the posted F
.)
Could you fix the formatting of your question (i. e. closing your displayed math by another pair of dollars) ? It is difficult to read you...
What is $M$ ?
Could you clarify your question ?
@emmanuel sorry for the poor display. I have corrected it. M is a rational. My question is that I want to factorize D using v so that I get
[M((A+2*F*v)/(4*p^4))]*[M((A2*F*v)/(4*p^4))]