First, we rearrange the inequality with a zero on the right:

x2 − 2x − 3 > 0

which can be factored to give:

`(x + 1)(x - 3) > 0`

Setting both factors to zero, we get:

`(x + 1) = 0 and (x - 3) = 0`

`x = -1 and x = 3`

Therefore the critical values are

`x = -1 and x = 3`.

These critical values divide the number line into 3 intervals:

`x < -1`,

`-1 < x < 3`, and

`x > 3`.

Next, we need to determine the sign (plus or minus) of the function in each of the 3 intervals.

For the first interval, `x < -1`,

The value of `(x + 1)` will be negative (substitute a few values of `x` less than `-1` to check),

The value of `(x − 3)` will also be negative

So in the interval `x < -1`, the value of the function x2 − 2x − 3 will be

negative × negative = positive

We continue doing this for the other 2 intervals and summarise the results in this table:

 Interval `(x + 1)` `(x - 3)` sign of f(x) `x < -1` − − + `-1 < x < 3` + − − `x > 3` + + +

We are solving for

`(x + 1)(x - 3) > 0`

The intervals that satisfy this inequality will be those where f(x) has a positive sign.

Hence, the solution is: `x < -1` or `x > 3`.

Here is the graph of our solution:

Open image in a new page