If a, b, c and d are in GP show that ax^3 + bx^2 +cx + d is divisible by - Maths - Sequences and Series

Question

If a, b, c and d are in GP show that ax^3 + bx^2 +cx + d is
divisible by ax^2 +c

Neha Sethi · Accepted Answer

Dear student
Given: a,b,c and d are in GP⇒a=a,b=ar,c=ar2,d=ar3 with first term=a and common ratio=rConsider,ax3+bx2+cx+d=ax3+arx2+ar2x+ar3=ax3+rx2+r2x+r3and ax2+c=ax2+ar2=ax2+r2Now consider,ax3+bx2+cx+dax2+c=ax3+rx2+r2x+r3ax2+r2=x3+rx2+r2x+r3x2+r2

=x2x+r+r2x+rx2+r2=x2+r2x+rx2+r2=x+r⇒ax3+bx2+cx+d is divisible by ax2+c
Regards

If a, b, c and d are in GP show that ax^3 + bx^2 +cx + d is divisible by ax^2 +c.