If the polynomials

Let p1(z) = az3 +4z2 + 3z-4 and p2(z) = z3-4z + o

When we divide p1(z) by z – 3, then we get the remainder p,(3).

Now, p1(3) = a(3)3 + 4(3)2 + 3(3) – 4

= 27a+ 36+ 9-4= 27a+ 41 When we divide p2(z) by z-3 then we get the remainder p2(3).

Now, p2(3) = (3)3-4(3)+a

= 27-12 + a = 15+a According to’ the question, both the remainders are same.

p1(3)= p2(3)

27a+41 = 15+a

27a-a = 15-41 .

26a = 26 a = -1