Is x^3 - 6x^2 + 11x - 6 < 0

We know how to deal with quadratics but we usually do not deal with third degree equations. If you do get a third degree equation, it will have one very easy root i.e. 0 or 1 or -1 or 2 or -2. Try a few of these values to get the first root. Here x = 1 works. It is easy to see since you have two 6s and an 11 as the co-efficients.

So you know that (x-1) is a factor. Now figure out the quadratic which when multiplied by (x-1) gives the third degree expression

\((x - 1)(x^2 - 5x + 6) = x^3 - 6x^2 + 11x - 6\)
How do you get this quadratic? Let's see.

\((x - 1)(ax^2 + bx + c) = x^3 - 6x^2 + 11x - 6\)

Coefficient of x^3 on right hand side is 1. So you know that all you need is x^2 so that it multiplies with x to give x^3 on left hand side too. So a must be 1.

\((x - 1)(x^2 + bx + c) = x^3 - 6x^2 + 11x - 6\)

The constant term is easy to figure out too. It should multiply with -1 to give -6 on right hand side. Hence c must be 6.

\((x - 1)(x^2 + bx + 6) = x^3 - 6x^2 + 11x - 6\)
The middle term is slightly more complex. bx multiplies with x to give x^2 term on right hand side (i.e. -6x^2) but you also get the x^2 term by multiplying -1 with x^2. You need -6x^2 and you have -x^2 so you need another -5x^2 from bx^2. So b must be -5.

\((x - 1)(x^2 - 5x + 6) = x^3 - 6x^2 + 11x - 6\)
Now you just factorize the quadratic in the usual way and use the wavy curve to solve the inequality.

\((x - 1)(x - 2)(x - 3) < 0\)

2 < x < 3 or x < 1

(1) 1 < x <= 2
Between 1 to 2 (inclusive), the expression is not negative. It is 0 or positive.
Sufficient.

(2) 2 <= x < 3
Between 2 and 3, the expression is negative but it is 0 when x = 2. Hence insufficient.

Answer (A)