Question

Help?

Answer 1

What's the question

Answer 2

is the house on fire? call 911

Answer 3

flip

Answer 4

lol

Answer 5

You need to multiply by the reciprocal....so think you can set it up?

Answer 6

z^2-4 z+2
-------- / -------
z-3 z^2+z-12

Answer 7

That is the original question, right? lol

Answer 8

Yeah.

Answer 9

thats as far as I got. XD

Answer 10

Okay, so flip that second fraction and multiply it by the first one...

Answer 11

ahemm... can you say hmmm factor the 1st numerator?, that is \(z^2-4?\)

Answer 12

(z+4)(z-3)

Answer 13

(z+2)(z-2)/(z-3) * (z+4)(z-3)/(z+2)

Answer 14

They will be the same number, seeing as there is no _z part to it

Answer 15

hmmm? what hmm did you factor anyway?

Answer 16

That is correct

Answer 17

Okay next step?

Answer 18

That is also the same as
\[\frac{(z+2)(z-2)(z+4)(z-3)}{(z-3)(z+2)}\]
So do you see anything that can cancel out?

Answer 19

yeap, that is correct above

so \(\bf \cfrac{z^2-4}{z-3}\div\cfrac{z+2}{z^2+z-12}
\implies \cfrac{\frac{z^2-4}{z-3}}{\frac{z+2}{z^2+z-12}}
\\ \quad \\
\cfrac{z^2-4}{z-3}\cdot \cfrac{z+2}{z^2+z-12}\implies
\cfrac{(z-2)(\cancel{z+2})}{\cancel{z-3}}\cdot \cfrac{(z+4)(\cancel{z-3})}{\cancel{z+2}}\implies ?\)

Answer 20

Gahhhhleaaa you guys are no joke. okay so, the answer is: x *x = x^2 x * 4 = 4x -2 * x = -2x -2 x 4 = 8 Combine to get x^2 +2x -8

Answer 21

That looks right

Answer 22

And 0 is the restrictions.

Answer 23

But it's z, not x ;)

Answer 24

yes and no, is not "x" is "z", but you're correct, it's \(\bf z^2+2z-8\)

Answer 25

Okay thank you guys!

Answer 26

well... hold the mayo on the restrictions.. notice the original version

one sec

Answer 27

\(\bf \cfrac{(z-2)(z+2)}{z-3}\cdot \cfrac{(z+4)(z-3)}{z+2}\qquad \qquad {\color{brown}{ z}}=3\ or\ -2
\\ \quad \\
\cfrac{(z-2)(z+2)(z+4)(z-3)}{({\color{brown}{ 3}}-3)(z+2)}\implies
\cfrac{(z-2)(z+2)(z+4)(z-3)}{0}\impliedby undefined
\\ \quad \\
\cfrac{(z-2)(z+2)(z+4)(z-3)}{(z-3)({\color{brown}{ -2}}+2)}\implies
\cfrac{(z-2)(z+2)(z+4)(z-3)}{0}\impliedby undefined\)

Answer 28

hmmm shoot got a bit cut off.. lemme fix that

Answer 29

\(\bf \cfrac{(z-2)(z+2)}{z-3}\cdot \cfrac{(z+4)(z-3)}{z+2}\qquad \qquad {\color{brown}{ z}}=3\ or\ -2
\\ \quad \\
\cfrac{(z-2)(z+2)(z+4)(z-3)}{({\color{brown}{ 3}}-3)(z+2)}
\\ \quad \\
\implies
\cfrac{(z-2)(z+2)(z+4)(z-3)}{0}\impliedby undefined
\\ \quad \\
\cfrac{(z-2)(z+2)(z+4)(z-3)}{(z-3)({\color{brown}{ -2}}+2)}
\\ \quad \\
\implies
\cfrac{(z-2)(z+2)(z+4)(z-3)}{0}\impliedby undefined\)

so... notice, if "z" is either, 3 or -2, will make the fraction "undefined"
thus, those are the restrictions on the domain