PLEASE HELP!!!
What is the product in simplest form? State any restrictions on the variable.
z^2/z+1 times z^2+3z+2/z^2+3z

- anonymous

- katieb

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- anonymous

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- phi

can you factor the quadratic ?

- anonymous

can you factor all of it? :S is it (z+2)(z+1)?

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## More answers

- phi

you can check: multiply (z+2)(z+1) = z^2 +2z + z +2 = x^2 +3z+2
yes that works.
how about the bottom ?

- anonymous

hm, I'm not sure..

- phi

look for a common factor (hint: z)

- anonymous

i don't know how to factor the bottom half, i'm really bad at factoring

- phi

example: x(x+2) = x^2 +2x
or undoing it x^2 +2x =x(x+2)

- anonymous

x(x+3)? i really don't know :/

- anonymous

oh, so i got it right? :D

- phi

yes, except they are using z
z(z+3) can be written as z^2 +3z (distribute the z)
so yes

- phi

so what do we have so far?

- anonymous

well that's easy :) what's next?

- phi

write down what we have after all the factoring , and see what we have

- anonymous

just what we wrote up there? :/

- phi

like this
\[ \frac{z\cdot z}{(z+1)} \cdot \frac{(z+2)(z+1) }{z(z+3)} \]

- anonymous

then cross out z+1 and z...which would be z+2 / z+3?

- phi

now you (hopefully) know that when you divide something by itself you get 1, so we an cancel terms that are in both the top and bottom

- anonymous

restriction is that x cannot be 0 or -3?

- phi

yes, but I think there is another z up top. Check your canceling.

- anonymous

oh

- anonymous

-1?

- phi

yes, up top we had z^2 or z*z and only 1 z in the bottom

- phi

no -1. You almost have it.
\[ \frac{z\cdot z \cdot (z+2)(z+1) }{z(z+3)(z+1)} \]

- phi

and yes, the restricted values are 0, -3, and -1
what do you have for the final answer?

- anonymous

hm, z^2+2z/z+3? :S

- phi

yes, with restrictions that z is not 0, -1 or -3

- anonymous

you are awesome! tytyty :D x

- phi

though I would put parens around it just to be clear.
(z^2+2z)/(z+3)
and no :S (ha ha)

- phi

Here are some videos
http://www.khanacademy.org/math/algebra/rational-expressions/simplifying-rational-alg/v/simplifying-rational-expressions-introduction
they are very helpful

- anonymous

haha, you mean yes! thanks :)

- anonymous

i'll check them out! :D

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