Question:
If the polynomials az3 +4z2 + 3z-4 and z3-4z + o leave the same remainder when divided by z – 3, find the value of a.
Solution:
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