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 p−q is05−51

Solution

## The correct option is B 5f(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=0⇒5+pω+q=0     ⋯(2) Using equation (1) and (2), we get p=0 ⇒q=−5∴p−q=5

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