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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.
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]
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 .
If two or more factors of a polynomial are identical, then the polynomial is a multiple of the square of this factor. The multiple factor is also a factor of the polynomial's derivative (with respect to any of the variables, if several). For univariate polynomials, multiple factors are equivalent to multiple roots (over a suitable extension field).
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 ...
The symmetries in this solution are easy to see. There are three roots of the cubic, corresponding to the three ways that a quartic can be factored into two quadratics, and choosing positive or negative values of for the square root of merely exchanges the two quadratics with one another.
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