Can anyone solve this question of trinomial?
Solve: 7x^5+7x-63
:) :p :confused:
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Can anyone solve this question of trinomial?
Solve: 7x^5+7x-63
:) :p :confused:
The polynomial is essentially x^5 + x - 9. It cannot be solved by radicals, which follows from Galois theory:
This polynomial is an irreducible quintic, so its Galois group is a transitive subgroup of the symmetric group on 5 letters, S5. The only such subgroups are (1) S5 itself; (2) the alternating group on 5 letters, A5; (3) the general affine group GA(1, 5) of order 20; (4) the cyclic group of order 5, Z5; and (5) the dihedral group of order 10, D10. The polynomial discriminant is 20503381 = 61 * 336121 [these are prime]. Since the square root is not rational, the Galois group is not a subgroup of the alternating group A5, eliminating possibilities (2), (4), and (5), leaving only (1) and (3).
Over the field on two elements, x^5 + x - 9 becomes (x^2 + x + 1)(x^3 + x^2 + 1). The cubic is irreducible over the field on four elements, so the splitting field of the cubic is an extension of order divisible by 3. The polynomial is also separable, so the splitting field is Galois, and the Galois group has order divisible by 3. Since the Galois group of the original polynomial in this setting has the reduced Galois group as a subgroup, from Lagrange's theorem the Galois group of x^5 + x - 9 has order divisible by 3. Since (3) has order 20 which is not divisible by 3, we're left with (1). That is, the Galois group of x^5 + x - 9 over the rationals is the full group S5. This is not solvable, so the polynomial cannot be solved by radicals.
Hi jemidiah,
I am Mohit S.Jain. You are right . This is not solvable .
Unsolvable, indeed, unless you'd like to find the real roots by numerical methods -e.g. Newton's method.Quote:
Originally Posted by ak28it