Question 187261
<pre><font size = 4 color = "indigo"><b>
Your 8 in the top should have been 40.

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

Factor the first two denominators:

{{{8/((x-2)(x+2))+ (2x)/((x+2)(x+2)) - 6/(x+2)}}}

Now let's create the LCD:

{{{(x-2)}}} appears 1 time in the first denominator,
0 time in the second denominator and 0 times in the
third denominator.  That's at most 1 time, so it
has to be included in the LCD 1 time

{{{(x+2)}}} appears 1 time in the first denominator,
2 times in the second denominator and 1 times in the
third denominator.  That's at most 2 times, so it
has to be included in the LCD 2 times

So the LCD contains {{{(x-2)}}} 1 time and {{{(x+2)}}}
2 times. So:

LCD = {{{(x-2)(x+2)(x+2)}}}

Each of the three terms must be rewritten so that each
fraction has the LCD as its denominator.

The first fraction is

{{{8/((x-2)(x+2))}}}

Its denominator needs another {{{(x+2)}}} factor to become
the LCD, so we multiply its numerator and denominator by
{{{(x+2)}}}

The second fraction is

{{{(2x)/((x+2)(x+2))}}} 

Its denominator needs an {{{(x-2)}}} factor to become
the LCD, so we multiply its numerator and denominator by 
{{{(x-2)}}}

The third fraction is

{{{-6/(x+2)}}}

Its denominator needs another {{{(x+2)}}} factor and
another {{{(x-2)}}} factor to become the LCD, so we 
multiply its numerator and denominator by 
{{{(x+2)(x-2)}}}

So

{{{8/((x-2)(x+2))+ (2x)/((x+2)(x+2)) - 6/(x+2)}}}

now becomes:

{{{(8(x+2))/((x-2)(x+2)(x+2))+ (2x(x-2))/((x+2)(x+2)(x-2)) - (6(x+2)(x-2))/((x+2)(x+2)(x-2))}}}

Multiply out the numerators only:

{{{(8x+16)/((x-2)(x+2)(x+2))+(2x^2-4x)/((x+2)(x+2)(x-2))-(6(x^2-4))/((x+2)(x+2)(x-2))}}}  

The last numerator still needs one more step to be multiplied
out:

{{{(8x+16)/((x-2)(x+2)(x+2))+ (2x^2-4x)/((x+2)(x+2)(x-2)) - (6x^2-24)/((x+2)(x+2)(x-2))}}}  

Now we combine all the numerators over the LCD:

{{{((8x+16)+ (2x^2-4x)- (6x^2-24))/((x-2)(x+2)(x+2))}}}


Remove all the parentheses on the top, but not on the bottom.

{{{(8x+16+ 2x^2-4x-6x^2+24)/((x-2)(x+2)(x+2))}}}

Combine all like terms on the top:
 
{{{(4x+40-4x^2)/((x-2)(x+2)(x+2))}}}

If you like you may arrange the top in descending powers of x:

{{{(-4x^2+4x+40)/((x-2)(x+2)(x+2))}}}

and you see that there is a 40 where you had 8.  If
you like you can factor -4 out of the top, and as you
did,  write {{{(x+2)^2}}} instead of {{{(x+2)(x+2)}}} in
the bottom:

{{{(-4(x^2-x-10))/((x-2)(x+2)^2)}}}

But we find that the quadratic in the parenthese does not
factor, so that's as far as we can go.

Edwin</pre>