Answer :

We have,

f(x) = x3+6x2+11x+6

Clearly, f (x) is a polynomial with integer coefficient and the coefficient of the highest degree term i.e., the leading coefficient is 1.

Therefore, integer root of f (x) are limited to the integer factors of 6, which are:

We observe that 

f (-1) = (-1)3 + 6 (-1)2 + 11 (-1) + 6

= -1 + 6 -11 + 6

= 0

f (-2) = (-2)3 + 6 (-2)2 + 11 (-2) + 6

= -8 + 24 – 22 + 6

= 0

f (-3) = (-3)3 + 6 (-3)2 + 11 (-3) + 6

= -27 + 54 – 33 + 6

= 0

Therefore, integral roots of f (x) are -1, -2, -3.

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