Q. 1

# Using factor theo

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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