Question 152480
{{{((x^2+5x+4)/(x+1))/((x+3)^2/(x^2-9))}}} or

{{{(x^2+5x+4)/(x+1)}}}÷{{{(x+3)^2/(x^2-9)}}}
<pre><font size = 4 color="indigo"><b>
Invert the second fraction and change the
division to multiplication:

{{{(x^2+5x+4)/(x+1)}}}×{{{(x^2-9)/(x+3)^2}}}

Factor the two numerators, and write {{{(x+3)^2}}} as {{{(x+3)(x+3)}}}

{{{((x+4)(x+1))/(x+1)}}}×{{{((x-3)(x+3))/(x+3)(x+3)}}}

Put parentheses around the left denominator so
everything will be "packaged" in parentheses:

{{{((x+4)(x+1))/((x+1))}}}×{{{((x-3)(x+3))/(x+3)(x+3)}}}

Indicate the multiplication of the numerators 
and denominators, so it will all be one fraction:

{{{((x+4)(x+1)(x-3)(x+3))/((x+1)(x+3)(x+3))}}} 

Cancel the {{{(x+1)}}}'s

{{{((x+4)cross((x+1))(x-3)(x+3))/(cross((x+1))(x+3)(x+3))}}} 

Cancel the {{{(x+3)}}} in the top with one of the
{{{(x+3)}}}'s in the bottom:

{{{((x+4)cross((x+1))(x-3)cross((x+3)))/(cross((x+1))cross((x+3))(x+3))}}}

All that's left is

{{{((x+4)(x-3))/(x+3)}}}

You can leave it like that, or if you like
you can FOIL out the top as {{{x^2+x-12}}}
and leave it this way:

{{{(x^2+x-12)/(x+3)}}}

Edwin</pre>