Question 1048600
<pre><b>

{{{(x+4)/(x-5)-((x-6)/(x+1))/(x+1)}}}

{{{(x+4)/(x-5)-((x-6)/(x+1))*(1/(x+1))}}}

{{{(x^""+4)/(x^""-5)-(x^""-6)/(x+1)^2)}}}

LCD = (x-5)(x+1)<sup>2</sup>

{{{((x+4)(x+1)^2-(x-6)(x-5))/((x-5)(x+1)^2)}}}

{{{((x+4)(x^2+2x+1)-(x^2-11x+30))/((x-5)(x+1)^2)}}}

{{{((x^3+2x^2+x+4x^2+8x+4)-(x^2-11x+30))/((x-5)(x+1)^2)}}}

{{{(x^3+2x^2+x+4x^2+8x+4-x^2+11x-30)/((x-5)(x+1)^2)}}}

{{{(x^3+5x^2+20x-26)/((x-5)(x+1)^2)}}}

Now we analyze the polynomial {{{x^3+5x^2+20x-26}}}

candidates for zeros are ±1, ±2, ±13, ±26

We begin by trying +1 using synthetic division with
the factor theorem:

1|1  5  20 -26
 |<u>   1   6  26</u>
  1  6  26   0

That was fortunate.  The very first candidate for a
zero succeeded!

So {{{x^3+5x^2+20x-26}}} factors as {{{(x-1)(x^2+6x+26)}}}
The quadratic does not factor, so the final result is:

{{{((x-1)(x^2+6x+26))/((x-5)(x+1)^2)}}}

Edwin</pre></b>