If ax3+bx2+cx+d is divisible by ax2+c, then a,b,c,d are _ in

Since ax3+bx2+cx+d is divisible by ax2+c, therefore, when ax3+bx2+cx+d is divided by ax2+c the remainder should be zero, Now when ax3+bx2+cx+d is divided by ax2+c, then the remainder is bca−d∴bca−d=0⇒bc=ad⇒ba=dc Hence, from this a,b,c,d are not necessarily in G.P.