Question 1017815
<pre>
{{{4/(a^2-a-2) }}}{{{""+""}}}{{{ 3/(a^2-4)}}}

Factor the denominators:

{{{4/((a+1)(a-2)) }}}{{{""+""}}}{{{ 3/((a-2)(a+2))}}}

The LCD is {{{(a+1)(a-2)(a+2)}}}

The denominator in the first term needs the
factor (a+2) in order to equal the LCD, so
we multiply the first term by {{{red(((a+2))/((a+2)))}}}
which equals 1 and therefore will not change
the value when we multiply the first term by it.

The denominator in the second term needs the
factor (a+1) in order to equal the LCD, so
we multiply the second term by {{{red(((a+1))/((a+1)))}}}
which equals 1 and therefore will not change
the value when we multiply the second term by it.

{{{expr(4/((a+1)(a-2)))*red(expr(((a+2))/((a+2)))) }}}{{{""+""}}}{{{ expr(3/((a-2)(a+2)))*red(expr(((a+1))/((a+1))))}}}

Rewrite each multiplication of fractions as single fractions:

{{{(4red((a+2)))/((a+1)(a-2)red((a+2))) }}}{{{""+""}}}{{{ (3red((a+1)))/((a-2)(a+2)red((a+1)))}}}

Distribute to remove the parentheses in the numerators, but
do not multiply the denominators out.

{{{(4a+8)/((a+1)(a-2)(a+2)) }}}{{{""+""}}}{{{ (3a+3)/((a-2)(a+2)(a+1))}}}

Notice that the denominators of both terms are the same except
for the order of factors, so we will add the terms of the numerator
and have just one fraction term.  We will choose the second
denominator, since that's the order of factors that you have in
your answer:

{{{(4a+8+3a+3)/((a-2)(a+2)(a+1))}}}

Combine like terms in the numerator:

{{{(7a+11)/((a-2)(a+2)(a+1))}}}

Edwin</pre>