Question 1032750
<pre><b>
{{{(x^2 + 3x - 54)/(x^2 + 5x + 6)}}}{{{"÷"}}}{{{(x^2 - 12x + 36)/(x^3 - 3x - 18)}}}

Invert the second fraction and change the division
to multiplication:

{{{(x^2 + 3x - 54)/(x^2 + 5x + 6)}}}{{{""*""}}}{{{(x^3 - 3x - 18)/(x^2 - 12x + 36)}}}

Factor each trinomial:

{{{x^2+3x-54}}}{{{""=""}}}{{{(x+9)(x-6)}}}

{{{x^2+5x+6}}}{{{""=""}}}{{{(x+3)(x+2)}}}

{{{x^2-12x+36}}}{{{""=""}}}{{{(x-6)(x-6)}}}

To factor {{{x^3 - 3x - 18}}}, consider as

{{{x^3 +0x - 3x - 18}}}

3|1  0 -3 -18
 |<u>   3  9  18</u>
  1  3  6   0

{{{x^3 - 3x - 18}}}{{{""=""}}}{{{(x-3)(x^2+3x+6)}}}

So original expression becomes:

{{{((x+9)(x-6))/((x+3)(x+2))}}}{{{""*""}}}{{{((x-3)(x^2+3x+6))/((x-6)(x-6))}}}

After all that factoring the only thing that cancels
is one of the (x-6)'s!!!:

{{{((x+9)cross((x-6)))/((x+3)(x+2))}}}{{{""*""}}}{{{((x-3)(x^2+3x+6))/(cross((x-6))(x-6))}}}

{{{((x+9)(x-3)(x^2+3 x+6))/((x+3)(x+2)(x-6))}}}

Edwin</pre></b>