the
left hand side of the equation (1), then it can be written as the difference
of the squares of the following polynomials:
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:
,
given by (3) we are going substitute in (2). We will find the same solutions
for (1).
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:
