SOLUTION: Please help me solve the question if ax^3+bx^2+cx+d is divisible by ax^2+c then a,b,c,d are in ap or gp or hp or none?

Question 1000064: Please help me solve the question
if ax^3+bx^2+cx+d is divisible by ax^2+c then a,b,c,d are in ap or gp or hp or none?
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
In general, the answer is NONE.
Adding some requirements to the coefficients a, b, c, and d,
we could be sure to have a gp,
or even a gp that is also an ap, and a hp.

If there are no specifications about a,b,c,and d,
could be divisible by if ,
and then a,b,c,d would be neither an ap, nor a gp, nor a hp.

If it were specified that none of the coefficients a, b,c,and d is zero,
when we divide by we find
a quotient of and
a remainder of .
In that case,
if is divisible by , then
--->---> .
Conversely, if <---><---><---> ,
then the remainder is zero, and is divisible by .
For <---> to happen,
it is not necessary that a,b,c,d form an ap, an gp, or an hp.
Sometimes a,b,c,d will be an ap, a gp, and a hp,
sometimes it will only be a gp,
but other times it will be none of those.
For example, 1,2,3,6 is neither an ap, nor a gp, nor a hp.

With <---> , and <---> ,
a,b,c,d is a gp with common ratio .
1,2,4,8 is an example.
A special case of that is when , which makes a,b,c,d
a gp, an ap, and a hp.
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