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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
 one year ago
 one year ago
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
 one year ago
 one year ago

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phiBest ResponseYou've already chosen the best response.4
can you factor the quadratic ?
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
can you factor all of it? :S is it (z+2)(z+1)?
 one year ago

phiBest ResponseYou've already chosen the best response.4
you can check: multiply (z+2)(z+1) = z^2 +2z + z +2 = x^2 +3z+2 yes that works. how about the bottom ?
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
hm, I'm not sure..
 one year ago

phiBest ResponseYou've already chosen the best response.4
look for a common factor (hint: z)
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
i don't know how to factor the bottom half, i'm really bad at factoring
 one year ago

phiBest ResponseYou've already chosen the best response.4
example: x(x+2) = x^2 +2x or undoing it x^2 +2x =x(x+2)
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
x(x+3)? i really don't know :/
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
oh, so i got it right? :D
 one year ago

phiBest ResponseYou've already chosen the best response.4
yes, except they are using z z(z+3) can be written as z^2 +3z (distribute the z) so yes
 one year ago

phiBest ResponseYou've already chosen the best response.4
so what do we have so far?
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
well that's easy :) what's next?
 one year ago

phiBest ResponseYou've already chosen the best response.4
write down what we have after all the factoring , and see what we have
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
just what we wrote up there? :/
 one year ago

phiBest ResponseYou've already chosen the best response.4
like this \[ \frac{z\cdot z}{(z+1)} \cdot \frac{(z+2)(z+1) }{z(z+3)} \]
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
then cross out z+1 and z...which would be z+2 / z+3?
 one year ago

phiBest ResponseYou've already chosen the best response.4
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
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
restriction is that x cannot be 0 or 3?
 one year ago

phiBest ResponseYou've already chosen the best response.4
yes, but I think there is another z up top. Check your canceling.
 one year ago

phiBest ResponseYou've already chosen the best response.4
yes, up top we had z^2 or z*z and only 1 z in the bottom
 one year ago

phiBest ResponseYou've already chosen the best response.4
no 1. You almost have it. \[ \frac{z\cdot z \cdot (z+2)(z+1) }{z(z+3)(z+1)} \]
 one year ago

phiBest ResponseYou've already chosen the best response.4
and yes, the restricted values are 0, 3, and 1 what do you have for the final answer?
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
hm, z^2+2z/z+3? :S
 one year ago

phiBest ResponseYou've already chosen the best response.4
yes, with restrictions that z is not 0, 1 or 3
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
you are awesome! tytyty :D x
 one year ago

phiBest ResponseYou've already chosen the best response.4
though I would put parens around it just to be clear. (z^2+2z)/(z+3) and no :S (ha ha)
 one year ago

phiBest ResponseYou've already chosen the best response.4
Here are some videos http://www.khanacademy.org/math/algebra/rationalexpressions/simplifyingrationalalg/v/simplifyingrationalexpressionsintroduction they are very helpful
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
haha, you mean yes! thanks :)
 one year ago

heathernellyBest ResponseYou've already chosen the best response.1
i'll check them out! :D
 one year ago
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