Answer :

Let f(x) = x3 – x2 – (3 –√3 ) x + √3


By Factor Theorem, we know that,


If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0


For checking (x+1) to be a factor, we will find f(–1)


f(–1) = (–1)3 – (–1)2 – (3 –√3)(–1) + √3


f(–1) = –1 – 1 + 3 – √3 + √3


f(–1) = 1


As, f(–1) is not equal to zero, therefore (x+1) is not a factor x3 – x2 –(3 –√3) x + √3


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