I got up to simplifying it down to ...

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

$$\frac{ (k-2)(k-1) }{ (k-5)(k-1) }$$

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

http://s3.amazonaws.com/awesome_screenshot/3982475?AWSAccessKeyId=0R7FMW7AXRVCYMAPTPR2&Expires=1392732541&Signature=cTrX%2FKzbOAJR5ephkpj8n63I5lo%3D

Question

I got up to simplifying it down to ...

$$\frac{ k^2 - k - 2 }{ k^2 - 4k - 5 }$$

$$\frac{ (k-2)(k-1) }{ (k-5)(k-1) }$$

$$\frac{ k - 2 }{ k - 5 }$$

http://s3.amazonaws.com/awesome_screenshot/3982475?AWSAccessKeyId=0R7FMW7AXRVCYMAPTPR2&Expires=1392732541&Signature=cTrX%2FKzbOAJR5ephkpj8n63I5lo%3D

ganeshie8 · Answer

$\large \frac{ k^2 - k - 2 }{ k^2 - 4k - 5 }  $

$\large \frac{ (k-2)(k\color{red}{+}1) }{(k-5)(k\color{red}{+}1) }  $

ganeshie8 · Answer

looks u made a typo while writing out the factored form ?

ganeshie8 · Answer

before cancelling common factors, u need to find the restricted values

ganeshie8 · Answer

to find the restricted values, set the denominator to 0 :

$\large    (k-5)(k+1) = 0 $

anonymous · Answer

So option d should be the answer, since denominator can't be zero.

anonymous · Answer

Thx @ganeshie8. Could you please look up my question (probability)?

anonymous · Answer

Oops you're right... I swear watching my signs will be the death of me lol.

Um I think $$k \neq -1 $$ and $$k \neq -5$$ = 0

anonymous · Answer

and @d3Xter that makes sense...

ganeshie8 · Answer

restricted values :

$\large    (k-5)(k+1) = 0$

$\large    (k-5) = 0~~~~~OR~~~~~(k+1) = 0$

$\large    k =5~~~~~OR~~~~~k = -1$

anonymous · Answer

Thank you >.<