Option 2 : (x + 1)(x - 2)(x + 3)

**Concept:**

*Integral Root theorem:*If f(x) is a polynomial with integral coefficients and leading coefficient 1, then any integer root of f(x) is a factor of the constant term.-
*Factor Theorem:*Let p(x) be a polynomial of degree greater than or equal to 1 and 'a' be a real number such that p(a) = 0, then (x - a) is a factor of p(x).Conversely, if (x - a) is a factor of p(x), then p(a) = 0. -
A polynomial function cannot have more real zeroes than its degree.

**Calculation:**

Let f(x) = x3 + 2x2 - 5x - 6.

The constant term in f(x) is equal to -6 and factors of -6 are ± 1, ± 2, ± 3, ± 6.

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

f(-1) = (-1)3 + 2(-1)2 - 5(-1) - 6

⇒ f(-1) = -1 + 2 + 5 - 6

⇒ f(-1) = -7 + 7

⇒ f(-1) = 0

∴ (x + 1) is a factor of f(x)

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

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

∴ f(x) = k(x + 1)(x - 2)(x + 3) ----(1)

⇒ x3 + 2x2 - 5x - 6 = k(x + 1)(x - 2)(x + 3)

Putting x = 0 on both sides, we get

-6 = k(1)(-2)(3)

⇒ k = 1

Putting k = 1 in equation (1), we get

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

**Hence, x3 + 2x2 - 5x - 6 = (x + 1)(x - 2)(x + 3)**