Question

What is the quotient in simplified form? State any restrictions on the variable. Show Work.

Answer 1

    a+2           a +1
= -------  / ------------------
    a -5      (a - 3)(a - 5)

    a+2           (a - 3)(a - 5)
= -------  *  ------------------
    a -5              a +1

    (a+2)(a - 3)
= ----------------
        a +1

a NOT equal 5 and -1

Answer 2

Answer:

$\frac{(a+2)}{(a-5)} /\frac{(a+1)}{(a^2-8a+15)}=\frac{a^2-a-6}{a+1}$

Restriction:

$a\neq -1$

$a\neq 5$

Step-by-step explanation:

we are given

$\frac{(a+2)}{(a-5)} /\frac{(a+1)}{(a^2-8a+15)}$

Since, it is division

so, we can flip it to get in multiplication

$\frac{(a+2)}{(a-5)} /\frac{(a+1)}{(a^2-8a+15)}=\frac{(a+2)\times (a^2-8a+15)}{(a-5)\times (a+1)}$

now, we can factor it

and then we can simplify it

$a^2-8a+15=(a-5)(a-3)$

$\frac{(a+2)}{(a-5)} /\frac{(a+1)}{(a^2-8a+15)}=\frac{(a+2)\times (a-5)\times (a-3)}{(a-5)\times (a+1)}$

now, we can cancel it

$\frac{(a+2)}{(a-5)} /\frac{(a+1)}{(a^2-8a+15)}=\frac{(a+2)\times (a-3)}{(a+1)}$

$\frac{(a+2)}{(a-5)} /\frac{(a+1)}{(a^2-8a+15)}=\frac{a^2-a-6}{a+1}$

Restriction:

we know that denominator can not be zero

so,

$a+1\neq 0$

$a\neq -1$

and factored term can not be 0 as well

$a-5\neq 0$

$a\neq 5$