Answer:
[tex]\frac{(a+2)}{(a-5)} /\frac{(a+1)}{(a^2-8a+15)}=\frac{a^2-a-6}{a+1}[/tex]
Restriction:
[tex]a\neq -1[/tex]
[tex]a\neq 5[/tex]
Step-by-step explanation:
we are given
[tex]\frac{(a+2)}{(a-5)} /\frac{(a+1)}{(a^2-8a+15)}[/tex]
Since, it is division
so, we can flip it to get in multiplication
[tex]\frac{(a+2)}{(a-5)} /\frac{(a+1)}{(a^2-8a+15)}=\frac{(a+2)\times (a^2-8a+15)}{(a-5)\times (a+1)}[/tex]
now, we can factor it
and then we can simplify it
[tex]a^2-8a+15=(a-5)(a-3)[/tex]
[tex]\frac{(a+2)}{(a-5)} /\frac{(a+1)}{(a^2-8a+15)}=\frac{(a+2)\times (a-5)\times (a-3)}{(a-5)\times (a+1)}[/tex]
now, we can cancel it
[tex]\frac{(a+2)}{(a-5)} /\frac{(a+1)}{(a^2-8a+15)}=\frac{(a+2)\times (a-3)}{(a+1)}[/tex]
[tex]\frac{(a+2)}{(a-5)} /\frac{(a+1)}{(a^2-8a+15)}=\frac{a^2-a-6}{a+1}[/tex]
Restriction:
we know that denominator can not be zero
so,
[tex]a+1\neq 0[/tex]
[tex]a\neq -1[/tex]
and factored term can not be 0 as well
[tex]a-5\neq 0[/tex]
[tex]a\neq 5[/tex]
