## heathernelly 2 years 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

1. heathernelly

2. phi

can you factor the quadratic ?

3. heathernelly

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

4. 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 ?

5. heathernelly

hm, I'm not sure..

6. phi

look for a common factor (hint: z)

7. heathernelly

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

8. phi

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

9. heathernelly

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

10. heathernelly

oh, so i got it right? :D

11. phi

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

12. phi

so what do we have so far?

13. heathernelly

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

14. phi

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

15. heathernelly

just what we wrote up there? :/

16. phi

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

17. heathernelly

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

18. 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

19. heathernelly

restriction is that x cannot be 0 or -3?

20. phi

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

21. heathernelly

oh

22. heathernelly

-1?

23. phi

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

24. phi

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

25. phi

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

26. heathernelly

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

27. phi

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

28. heathernelly

you are awesome! tytyty :D x

29. phi

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

30. phi

31. heathernelly

haha, you mean yes! thanks :)

32. heathernelly

i'll check them out! :D