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# Help 20 ​

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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.

$$(z-2)(z+4)$$

$$z^2-2z-8$$

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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