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Given a quadratic polynomial of the form + + it is possible to factor out the coefficient a, and then complete the square for the resulting monic polynomial. Example: + + = [+ +] = [(+) +] = (+) + = (+) + This process of factoring out the coefficient a can further be simplified by only factorising it out of the first 2 terms.
The non-real factors come in pairs which when multiplied give quadratic polynomials with real coefficients. Since every polynomial with complex coefficients can be factored into 1st-degree factors (that is one way of stating the fundamental theorem of algebra ), it follows that every polynomial with real coefficients can be factored into ...
Let m be the degree over k of the splitting field of the resolvent cubic (it can be either R 4 (y) or R 5 (y); they have the same splitting field). Then the group G is a subgroup of the symmetric group S 4. More precisely: [4] If m = 1 (that is, if the resolvent cubic factors into linear factors in k), then G is the group {e, (12)(34), (13)(24 ...
The discriminant Δ of the cubic is the square of = () (), where a is the leading coefficient of the cubic, and r 1, r 2 and r 3 are the three roots of the cubic. As Δ {\displaystyle {\sqrt {\Delta }}} changes of sign if two roots are exchanged, Δ {\displaystyle {\sqrt {\Delta }}} is fixed by the Galois group only if the Galois group is A 3 .
In the case of a cubic equation, this resolvent is sometimes called the quadratic resolvent; its roots appear explicitly in the formulas for the roots of a cubic equation. The cubic resolvent of a quartic equation, which is a resolvent for the dihedral group of 8 elements.
A similar but more complicated method works for cubic equations, which have three resolvents and a quadratic equation (the "resolving polynomial") relating and , which one can solve by the quadratic equation, and similarly for a quartic equation (degree 4), whose resolving polynomial is a cubic, which can in turn be solved. [14]
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