Question

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

Answer 1

can you factor the quadratic ?

Answer 2

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

Answer 3

you can check: multiply (z+2)(z+1) = z^2 +2z + z +2 = x^2 +3z+2
yes that works.

how about the bottom ?

Answer 4

hm, I'm not sure..

Answer 5

look for a common factor (hint: z)

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

oh, so i got it right? :D

Answer 10

yes, except they are using z

z(z+3) can be written as z^2 +3z (distribute the z)
so yes

Answer 11

so what do we have so far?

Answer 12

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

Answer 13

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

Answer 14

just what we wrote up there? :/

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

restriction is that x cannot be 0 or -3?

Answer 19

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

Answer 20

oh

Answer 21

-1?

Answer 22

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

Answer 23

no -1. You almost have it.

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

Answer 24

and yes, the restricted values are 0, -3, and -1

what do you have for the final answer?

Answer 25

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

Answer 26

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

Answer 27

you are awesome! tytyty :D x

Answer 28

though I would put parens around it just to be clear.
(z^2+2z)/(z+3)

and no :S (ha ha)

Answer 29

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

they are very helpful

Answer 30

haha, you mean yes! thanks :)

Answer 31

i'll check them out! :D