[Friday’s lecture was covered by Marion Scheepers. Many thanks!]
Let’s return to the problem of solving quartic polynomials. In the first lecture on this topic, we reduced the problem of solving an equation like
to solving the similar problem
in which the coefficient of is zero. This is achieved by a simple translation, simply set . This was motivated and explained in that lecture. Let us now see how we can approach this problem.
The key idea is that a polynomial without a cubic term somewhat resembles a square . Following the book’s notation, write instead of . We have:
We need to deal with the remaining polynomial Being of degree 2, perhaps it could be a square as well. What we try then is a bit more ambitious: We look for values of and such that
If we succeed, then we would have a representation of the original polynomial in as a difference of squares:
and the quartic is then solved by solving two quadratic equations.
We are left with the task of looking for The first observation we need to make is that there is really not much leeway on what we can try. From the displayed equation (1), we see we need the following equations to hold:
- , and
We solve this system in two steps: First, we use these equations to find , and of course once we have the value of , the equations give us the values of and as well. To find , multiply the first two equations. We get:
Comparing with the square of the third equation, we now have:
This last equation is a cubic in . We can then solve it with the method explained in the previous lecture. Once we have , the first and second equations determine the values of and , up to sign. Once a value of is chosen, the corresponding sign for is determined by the third equation .
Once are known, we can use them to factor the quartic in as a difference of squares. This gives us two quadratic polynomials, that we can now solve. Translating the roots back to , we obtain the desired solutions.
An important remark needs to be made:
There are three possible values of found this way. Each gives rise to two possible values of (the value of is completely determined then by the sign of the third equation above). This gives us six possible factorizations of the original polynomial, each with four possibly different roots. Of course, there can only be four roots of a quartic polynomial, which means that the different values of and must result in the same roots, perhaps after a permutation of the order in which they appear. Check that this is indeed the case.
The book presents several examples of these methods in varying degrees of detail. Make sure you work through them to familiarize yourself with the algebra involved.