+0  
 
0
149
1
avatar+3654 

Help 20

NotSoSmart  Mar 5, 2018
 #1
avatar+68 
0

\(\frac{z^2-4}{z-3}/\frac{z+2}{z^2+z-12}\)

Switch the division sign to multiplication and switch the top and bottom equations on the second fraction.

\(\frac{z^2-4}{z-3}*\frac{z^2+z-12}{z+2}\)

Simplify the polynomial

\(\frac{(z+2)(z-2)}{z-3}*\frac{(z+4)(z-3)}{z+2}\)

You can cross out from top to bottom, top left to bottom right, and bottom left to top right.

\((z-2)(z+4)\)

Multiply your answer.

\(z^2-2z-8\)

---------

To find out the restrictions, you need to take the bottom number of each fraction and make it equal to zero.

\(z-3=0\)

\(z=3\)

...

\(z^2+z-12=0\)

\((z+4)(z-3)=0\)

\(z+4=0, z-3=0\)

\(z=-4, z=3\)

You can't use -4 or 3 as an answer to this problem.

CoopTheDupe  Mar 5, 2018
edited by CoopTheDupe  Mar 5, 2018

14 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.