Solve (x+1)(x+5) > 0: Finding Positive Values of a Quadratic Function

The given function is $f(x) = (x+1)(x+5)$. We need to determine for which values of $x$ this function is greater than zero.

First, let's find the roots of the function. Set each factor to zero to find the roots:
$x+1 = 0 \rightarrow x = -1$
$x+5 = 0 \rightarrow x = -5$

These roots divide the number line into three intervals: $x < -5$, $-5 < x < -1$, and $x > -1$.

Next, we will determine the sign of $f(x)$ in each interval:

• For $x < -5$, choose a test value like $x = -6$. Both $x+1$ and $x+5$ are negative, so their product $(x+1)(x+5) > 0$.
• For $-5 < x < -1$, choose a test value like $x = -3$. Here $x+1$ is negative, and $x+5$ is positive, making their product $(x+1)(x+5) < 0$.
• For $x > -1$, choose a test value like $x = 0$. Both $x+1$ and $x+5$ are positive, so their product $(x+1)(x+5) > 0$.

Therefore, $f(x) > 0$ when $x < -5$ or $x > -1$.

Thus, the solution is that $f(x) > 0$ for $x > -1$ or $x < -5$.