• Use the values of $x$ found in Step 1 to break up the number line and plug in one value from each interval into $f(x)=(x+3)(x-1)$.
  
  

• Since $f(-3)=0$ and $f(1)=0$, pick one value less $-3$, one value between $-3$ and $1$, and one value greater than $1$.
  
  For example, since $f(-4)=5>0$, $f(x)>0$ for all $x<-3$. And since $f(2)=5>0$, $f(x)>0$ for all $x>1$. However, $f(0)=-3<0$, and therefore $f(x)<0$ for all $-3<x<1$.

The calculations of Method 1 are summarized in the following diagram.

Therefore, $x^2+2x-3>0$ whenever $x<-3$ or $x>1$.

Sketch the graph of $y=x^2+2x-3$ by drawing a parabola that opens upward and goes through the points $(-3,0)$ and $(1,0)$.

Finding values of $x$ such that $x^2+2x-3>0$ is equivalent to identifying the portions of the graph of $x^2+2x-3$ that are above the $x$-axis.

Based on the graph shown above, $x^2+2x-3>0$ whenever $x<-3$ or $x>1$.

Break up the number line at $x=-2$, $x=0$, and $x=2$ and plug in one value from each interval to determine the sign of $f(x)=\frac{x^2(x^2+3)}{(4-x^2{)}^3}$ on that interval.

The calculations of Method 1 are summarized in the following diagram.

Therefore, $\frac{x^2(x^2+3)}{(4-x^2{)}^3}<0$ when $x<-2$ or $x>2$.

Notice that $x^2$ and $x^2+3$ are both positive for $x\ne 0$, therefore $\frac{x^2(x^2+3)}{(4-x^2{)}^3}<0$ whenever $4-x^2<0$.

The graph of $y=4-x^2$ as shown above is below the $x$-axis (i.e., $4-x^2<0$) if $x<-2$ or $x>2$. Therefore, $\frac{x^2(x^2+3)}{(4-x^2{)}^3}<0$ when $x<-2$ or $x>2$.