Every polynomial equation with degree 4, as it can be easily seen, can take the form of (1).
If we call the left hand side of the equation (1), then it can be written as the difference of the squares of the following polynomials:
where are proper complex numbers, which we have to specify.
We write:
and after the relevant operations we have:
In order the right hand side of (2) to
be a perfect square, we must find a proper such
that its discriminant
D to be equal to zero.
We conclude that:
Equation (3) has degree 3 and can be solved
as described in paragraph 8.1 of this chapter.
With one of the values of ,
that we get from (3) we find the value of
from (2) and then
(1) because of the following equation:
is equivalent to:
which is reduced to two polynomial equations
with degree 2.
Observations:
Solve the equation :
Solution
It is a=1, b=2, c=3,
d=27
The constant term
of (3) is :
and the coefficient
of the third degree term is :
Hence we have the equation: .
Solving the above equation as described in paragraph 8.1 we get :
.
For we get from (M)
and hence from the equation
we find that
The equations A(x) + B(x) = 0, A(x) - B(x) = 0 that equation (4) gives become:
and from these we get the roots: