Question 29257
You first want to factor out;
1){{{(a^2+2a)/(a^2+3a+2)}}}
For the numerator factor out a;
a(a+2)
THen for the denominator, you can't factor out a GCF, so you factor;
(a+2)(a+1)
Now you have;
{{{a(a+2)/(a+2)(a+1)}}}
Now cancel out (a+2) in numerator and denominator, and you are left with;
{{{a/(a+1)}}}
2){{{(3a^2+15a)/(a^3-25a)}}}
Factor out GCF;
{{{3a(a+5)/a(a^2-25)}}}
you can factor the denominator;
a(a+5)(a-5)
{{{3a(a+5)/a(a-5)(a+5)}}}
Cancel out a and (a+5);
{{{3/(a+5)}}}
3){{{(x+2)^2/(x^2-4)}}}
You can not factor out a GCF, so factor whole problem;
{{{(x+2)(x+2)/(x-2)(x+2)}}}
cancel out the (x+2) in numerator and denominator;
{{{(x+2)/(x-2)}}}
4){{{(x^2-16)/(x^2-4x)}}}
factor out the GCF in the denominator, and factor numerator;
{{{(x-4)(x+4)/x(x-4)}}}
cancel (x-4) in numerator and denominator;
{{{(x+4)/x}}}
Hope you understand
=)