To multiply and divide fractions with variables:
Multiply, and write your answer in simplest form: [beautiful math coming... please be patient] $$ \frac{x^29}{5x^2+20x+15} \cdot \frac{x+1}{x+4} $$
$\displaystyle \frac{x^29}{5x^2+20x+15} \cdot \frac{x+1}{x+4} $  $=$  $\displaystyle \frac{(x3)(x+3)}{5(x^2+4x+3)}\cdot\frac{x+1}{x+4}$  factor: difference of squares (numerator), common factor (denominator) 
$=$  $\displaystyle \frac{(x3)(x+3)}{5(x+3)(x+1)}\cdot\frac{x+1}{x+4} $  factor the trinomial in the denominator  
$=$  $\displaystyle \frac{\hphantom{5}(x+1)(x+3)(x3)}{5(x+1)(x+3)(x+4)} $  multiply, reorder  
$=$  $\displaystyle \frac{(x3)}{5(x+4)} $  cancel the two extra factors of $\,1\,$ 
It is interesting to compare the original expression (before simplification),
and the simplified expression (after cancellation).
Although they are equal
for almost all values of $\,x\,$, they do differ a bit,
because of the cancellation:
VALUES OF $\,x\,$ 
ORIGINAL EXPRESSION:
$\displaystyle
\frac{x^29}{5x^2+20x+15}
\cdot
\frac{x+1}{x+4}
$
in factored form:
$\displaystyle
\frac{\hphantom{5}(x+1)(x+3)(x3)}{5(x+1)(x+3)(x+4)}$

SIMPLIFIED EXPRESSION:
$\displaystyle
\frac{(x3)}{5(x+4)}$

COMPARISON 
$x = 4$  not defined (division by zero) 
not defined (division by zero) 
behave the same: both are not defined 
$x = 1$  not defined (division by zero) 
[beautiful math coming... please be patient] $\displaystyle \frac{13}{5(1+4)} = \frac{4}{15}$ 
the presence of
$\,\frac{x+1}{x+1}\,$
causes a puncture point at $\,x = 1\,$; see the first graph below 
$x = 3$  not defined (division by zero) 
$\displaystyle \frac{33}{5(3+4)} = \frac{6}{5}$ 
the presence of
$\,\frac{x+3}{x+3}\,$
causes a puncture point at $\,x = 3\,$; see the first graph below 
all other values of $\,x\,$ 
both defined; values are equal 
behave the same: values are equal 
GRAPH OF:
$\displaystyle
\frac{\hphantom{5}(x+1)(x+3)(x3)}{5(x+1)(x+3)(x+4)}$

GRAPH OF:
$\displaystyle
\frac{(x3)}{5(x+4)}$



For more advanced students, a graph is displayed.
For example, the expression $\,\frac{x+1}{x+2}\cdot\frac{x+3}{x+1}\,$
is optionally accompanied by the
graph of $\,y = \frac{x+1}{x+2}\cdot\frac{x+3}{x+1}\,$.
A puncture point occurs at $\,x = 1\,$, due to the presence of $\,\frac{x+1}{x+1}\,$.
The graph of the simplified expression would not have this puncture point.
Horizontal/vertical asymptote(s) are shown in light grey.
Note: A puncture point may occasionally occur outside the viewing window;
use the arrows in the lowerright graph corner to navigate left/up/down/right.
Click the “show/hide graph” button if you prefer not to see the graph.
CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.However, you can check to see if your answer is correct. 
PROBLEM TYPES:
