Question

Simply the rational expression.state any restrictions on the variable. k2-k-2/k2-4k-5

Answer 1

We are given

$\frac{k^2-k-2}{k^2-4k-5}$

Firstly, we will factor numerator and denominator

$\frac{k^2-k-2}{k^2-4k-5}=\frac{(k-2)(k+1)}{(k-5)(k+1)}$

Restrictions:

we know that denominator can not be zero

so, we can set denominator =0

and solve for k to get restrictions

$(k-5)(k+1)=0$

$k=-1,k=5$

Simplification:

$\frac{k^2-k-2}{k^2-4k-5}=\frac{(k-2)(k+1)}{(k-5)(k+1)}$

we can see that k+1 gets cancelled

so, we get

$\frac{k^2-k-2}{k^2-4k-5}=\frac{(k-2)}{(k-5)}$..........Answer