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58 CHAPTER 3. POLYNOMIAL EQUATIONS
3.1.3 Quartic Equations
Here the equation would be x 4 + ax 3 + bx 2 + cx + d, but after the Tschirnhaus
transformation x → x− a 4
, analogous to that which took equation (3.6) to (3.7),
we can assume that a = 0. A truly marvellous solution then looks as follows
(but the page is too small to contain it!).
√
6
12
√
√−4 b 3√ −288 db + 108 c 2 + 8 b 3 + 12 √ −768 d 3 + 384 d 2 b 2 − 48 db 4 − 432 dbc 2 + 81 c 4 + 12 c 2 b 3 + ( −
√
3
−288 db + 108 c2 + 8 b 3 + 12 √ −768 d 3 +
(3.11)
We can adopt the same formulation as in figure 3.1, as shown in figure 3.2. Here
return
S := √ −768 d 3 + 384 d 2 b 2 − 48 db 4 − 432 dbc 2 + 81 c 4 + 12 c 2 b 3
T := 3√ −288 db + 108 c 2 + 8 b 3 + 12 S
√
−4 bT + T
2
+ 48 d + 4 b 2
U :=
T
√
√ √
6 6
12 U + − ( 8 bT U + UT 2 + 48 Ud + 4 Ub 2 + 12 c √ 6T )
12
T U
Figure 3.2: Program for computing solutions to a quartic
the problem of multiple choices is even more apparent, but in this formulation
it turns out that choices cancel, much as in the case of the cubic. We have
the same problem as in the case of the cubic, that real solutions can arise from
complex intermediates, but also that the answer apparently involves √ 6, even
though it clearly need not do so in reality. For example, with x 4 − 5x 2 + 4,
whose solutions are ±1, ±2, we can evaluate
S := 72 √ −3; T := 17 + √ −3; U := 3 √ 6; return 2; (3.12)
taking the other square root at the end gives 1, and taking the other square root
when computing U gives −1 or −2. We should also note that T was evaluated
as 3√ 4760 + 864 √ −3: not entirely obvious.
3.1.4 Higher Degree Equations
When it comes to higher degree equations, the situation is very different.
Theorem 8 (Abel, Galois [Gal79]) The general polynomial equation of degree
5 or more is not soluble in radicals (i.e. in terms of k-th roots).