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

(1)

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.

[…] 2. Let be a polynomial with real coefficients, and let be a complex root of Show that as well. Conclude that if the degree of is odd and the coefficients of are real, then has at least one real root. (You may use the fundamental theorem of algebra, if needed.) Conclude also that if is of degree four and has real coefficients, then can be factor as the product of two quadratic polynomials with real coefficients. (Does this follow “directly” from the argument described in lecture?) […]

[…] formally (for example, they are needed to even make sense of the formulas we found in the previous lectures) what was not clear was that they made sense. Perhaps indiscriminate use of them would lead to […]

[…] Proof: This is similar to the previous case, but now we use Ferrari’s method, see lecture 2.2. […]

[…] The trick in this case is to see that this equation can be written as a difference of squares, which leads to its factorization as the product of two quadratics, and can therefore be solved. In order to find the squares that lead to this factorization, a parameter is introduced, and one checks that the restrictions it must satisfy lead to a cubic equation. The details will be presented next lecture. […]

[…] Proof: This is similar to the previous case, but now we use Ferrari’s method, see lecture 2.2. […]