Answer :

Let f (x) = x3+6x2+11x+6 be the given polynomial.

The constant term in f (x) is 6 and factors of 6 are


Putting x = - 1 in f (x) we have,


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


= -1 + 6 – 11 + 6


= 0


Therefore, (x + 1) is a factor of f (x)


Similarly, (x + 2) and (x + 3) are factors of f (x).


Since, f (x) is a polynomial of degree 3. So, it cannot have more than three linear factors.


Therefore, f (x) = k (x + 1) (x + 2) (x + 3)


x3+6x2+11x+6 = k (x + 1) (x + 2) (x + 3)


Putting x = 0, on both sides we get,


0 + 0 + 0 + 6 = k (0 + 1) (0 + 2) (0 + 3)


6 = 6k


k = 1


Putting k = 1 in f (x) = k (x + 1) (x + 2) (x + 3), we get


f (x) = (x + 1) (x + 2) (x + 3)


Hence,


x3+6x2+11x+6 = (x + 1) (x + 2) (x + 3)


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