Restrictions on Variables in Division

Date: 8/4/96 at 18:25:6
From: Shannon
Subject: Restrictions on Division

In simplifying rational expressions  I have to identify the 
"restrictions" on several expressions, but I'm having trouble. 
I know that:

1.  The restrictions are in the denominator, not the numerator
2.  It's not possible to have a term in the denominator containing a
    variable equal to zero.  If it does,  it becomes a restriction.
    
   For some reason I'm not doing it right.  I have provided 
   two sample questions.  I hope you can help.  Thanks.

a)  y^3 + 8           b)        x^3 + 64      
  ____________              ________________
  y^2 + 3y + 2              ax + 4b + 4a +bx

Date: 8/4/96 at 20:39:12
From: Doctor Paul
Subject: Re: Restrictions on Division

The two 'rules' you wrote down are correct.  Here's the idea:  The
'restrictions' to which you refer occur because the denominator of the
function equals zero.  At this point the function is 'sick' and 
many of the rules we use to analyze functions do not apply.

This is because it is 'illegal' to divide something by zero. There are 
explanations in the Dr. Math archive:

  http://mathforum.org/dr.math/problems/divide.zero.html   
  http://mathforum.org/dr.math/problems/harmon2.12.96.html   
  http://mathforum.org/dr.math/problems/sky.html   

In your first equation:

   y^3 + 8
-------------
y^2 + 3y + 2

you want to be able to identify the points where this function is 
'sick'.  The function will only be sick when you try to divide by 
zero.  In other words, it will only be sick when the denominator is 
zero. So to find the points where your function is sick we set the 
denominator equal to zero and solve:

y^2 + 3y + 2 = 0

This factors nicely:

(y + 1)(y + 2) = 0

So when y = -1 or when y = -2, your function is sick.  These would be 
the two points that have restrictions for this equation.  


Question two:

From the above we know that all we have to do is set the denominator 
equal to zero:

   ax + 4b + 4a +bx = 0
x(a + b) + 4(a + b) = 0
    (x + 4) (a + b) = 0

so when x = -4 or when a = -b, your function is sick.  Note that if 
x = -4, then the values of a and b don't matter... whatever they are, 
your function will have a zero in the denominator.

I hope this clears things up.

-Doctor Paul,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/