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It can similarly be shown that if + + = + , x = −1 is a root. In either case the full quartic can then be divided by the factor (x − 1) or (x + 1) respectively yielding a new cubic polynomial, which can be solved to find the quartic's other roots.
There is an alternative solution using algebraic geometry [18] In brief, one interprets the roots as the intersection of two quadratic curves, then finds the three reducible quadratic curves (pairs of lines) that pass through these points (this corresponds to the resolvent cubic, the pairs of lines being the Lagrange resolvents), and then use ...
Finding the roots (zeros) of a given polynomial has been a prominent mathematical problem.. Solving linear, quadratic, cubic and quartic equations in terms of radicals and elementary arithmetic operations on the coefficients can always be done, no matter whether the roots are rational or irrational, real or complex; there are formulas that yield the required solutions.
A quadratic with two real roots, for example, will have exactly two angles that satisfy the above conditions. For complex roots, one also needs to find a series of similar triangles, but with the vertices of the root path displaced from the polynomial path by a distance equal to the imaginary part of the root. In this case the root path will ...
Finding the root of a linear polynomial (degree one) is easy and needs only one division: the general equation + = has solution = /. For quadratic polynomials (degree two), the quadratic formula produces a solution, but its numerical evaluation may require some care for ensuring numerical stability.
The roots , of the quadratic polynomial () = + + satisfy + =, =. The first of these equations can be used to find the minimum (or maximum) of P ; see Quadratic equation § Vieta's formulas .
This explains the existence of the quadratic, cubic, and quartic formulas, since a major result of Galois theory is that a polynomial equation has a solution in radicals if and only if its Galois group is solvable (the term "solvable group" takes its origin from this theorem).
The rational root test allows finding q and p by examining a finite number of cases (because q must be a divisor of a, and p must be a divisor of d). Thus, one root is =, and the other roots are the roots of the other factor, which can be found by polynomial long division.