Question 144818
{{{((x^2+5x+4)/(x+3)^2)/((x+1)/(x^2-9))}}}.


Remember when you learned to divide one fraction by another?  Just invert the denominator fraction and multiply:  {{{(1/2)/(3/4)=(1/2)(4/3)=4/6=2/3}}}


Do the same thing here:


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


To simplify, you need to factor:


{{{x^2+5x+4=(x+4)(x+1)}}}
{{{(x+3)^2=(x+3)(x+3)}}}, and 
{{{x^2-9=(x+3)(x-3)}}} (Difference of two squares)


Now put the factored forms back in and see what you have:
{{{((x^2+5x+4)/(x+3)^2)((x^2-9)/(x+1))=((x+4)(x+1)/(x+3)(x+3))((x+3)(x-3)/(x+1))= (x+4)(x+1)(x+3)(x-3)/(x+3)(x+3)(x+1)}}}


Then you can eliminate factors that are common to both numerator and denominator because {{{a/a=1}}} for any real {{{a}}}.


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


Depending on what your instructor expects, you can either leave that as the answer, or multiply the remaining binomials in the numerator:


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