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Help 20

NotSoSmart  Mar 5, 2018
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\(\frac{z^2-4}{z-3}/\frac{z+2}{z^2+z-12}\)

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.

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

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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

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