# How do you solve (x+1)(x-3)>0?

Oct 26, 2016

#### Explanation:

Look at (x + 1)(x - 3) = 0

This means that the function is 0 at $x = - 1 \mathmr{and} x = 3$.

Also, it means that the sign of the corresponding factor changes sign at that value of x.

At values of x < -1:

Both $\left(x + 1\right)$ and $\left(x - 3\right)$ are negative. A negative multiplied by a negative is a positive, therefore, x < -1 is one of the regions where $\left(x + 1\right) \left(x - 3\right) > 0$. Let's make a note of that:

$x < - 1$

At values between -1 and 3:

$\left(x + 1\right)$ is positive but $\left(x - 3\right)$ is still negative. A positive multiplied by a negative is negative, therefore, this is NOT a region for $\left(x + 1\right) \left(x - 3\right) > 0$

At values $x > 3$:

Both $\left(x + 1\right)$ and $\left(x - 3\right)$ are positive. A positive multiplied by a positive is a positive, therefore, x > 3 is one of the regions where (x + 1)(x - 3) > 0. Let's make a note of that:

$x < - 1 \mathmr{and} x > 3$

We have no more regions to investigate, therefore, the above is our answer.