This under Amaths Remainder and Factor Theorm.
The A B C D are given in the question I suppose.
Step1, Trial and error.
Let this equation be f(x)=Ax^3 +Bx^2 +Cx+D
When x= ?, f(?)=0 <---? can be any no. but most of the time is between -4<4.
Hence, (x - ?) is a factor of the equation.
Step 2, find the other two factors by comparing coefficient.(Left side and right side)
Ax^3 +Bx^2 +Cx+D=(x-?)(ax^2+bx+c)
Comparing coefficient of x^3,
A(know no.)= a
a=A//
Comparing coefficient of x^2,
B=b+a?
b=B-a?//
Comparing constants,
D=c?
c=D/?
Therefore,Ax^3 +Bx^2 +Cx+D=0
(x-?)(ax^2+bx+c)=0 a b c d all known
factorise (ax^2+bx+c),
(x-?)(x-??)(x-???)=0
So, the answer is, The factors or solution for this equation is ?, ?? and ???