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Question

Let f(x)=5x3+px+q, where p and q are real numbers. When f(x) is divided by x2+x+1, the remainder is 0. Then the value of pq is
  1. 0
  2. 5
  3. 5
  4. 1


Solution

The correct option is B 5
f(x)=5x3+px+q
We know that 
x2+x+1=0 has roots ω,ω2 where ω is the cube root of unity.

So, f(ω)=5ω3+pω+q=0
5+pω+q=0     (1)f(ω2)=5ω6+pω2+q=05+pω+q=0     (2)
Using equation (1) and (2), we get
p=0
q=5pq=5

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