Let f (x) = x3 + 6x2 + 11x + 6 be the given polynomial.
The constant term in f (x) is 6 and factors of 6 are +1,+2,+3 and +6
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)