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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

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  1. heathernelly
    • 2 years ago
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  2. phi
    • 2 years ago
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    can you factor the quadratic ?

  3. heathernelly
    • 2 years ago
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    can you factor all of it? :S is it (z+2)(z+1)?

  4. phi
    • 2 years ago
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    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
    • 2 years ago
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    hm, I'm not sure..

  6. phi
    • 2 years ago
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    look for a common factor (hint: z)

  7. heathernelly
    • 2 years ago
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    i don't know how to factor the bottom half, i'm really bad at factoring

  8. phi
    • 2 years ago
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    example: x(x+2) = x^2 +2x or undoing it x^2 +2x =x(x+2)

  9. heathernelly
    • 2 years ago
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    x(x+3)? i really don't know :/

  10. heathernelly
    • 2 years ago
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    oh, so i got it right? :D

  11. phi
    • 2 years ago
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    yes, except they are using z z(z+3) can be written as z^2 +3z (distribute the z) so yes

  12. phi
    • 2 years ago
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    so what do we have so far?

  13. heathernelly
    • 2 years ago
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    well that's easy :) what's next?

  14. phi
    • 2 years ago
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    write down what we have after all the factoring , and see what we have

  15. heathernelly
    • 2 years ago
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    just what we wrote up there? :/

  16. phi
    • 2 years ago
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    like this \[ \frac{z\cdot z}{(z+1)} \cdot \frac{(z+2)(z+1) }{z(z+3)} \]

  17. heathernelly
    • 2 years ago
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    then cross out z+1 and z...which would be z+2 / z+3?

  18. phi
    • 2 years ago
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    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
    • 2 years ago
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    restriction is that x cannot be 0 or -3?

  20. phi
    • 2 years ago
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    yes, but I think there is another z up top. Check your canceling.

  21. heathernelly
    • 2 years ago
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    oh

  22. heathernelly
    • 2 years ago
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    -1?

  23. phi
    • 2 years ago
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    yes, up top we had z^2 or z*z and only 1 z in the bottom

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

  25. phi
    • 2 years ago
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    and yes, the restricted values are 0, -3, and -1 what do you have for the final answer?

  26. heathernelly
    • 2 years ago
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    hm, z^2+2z/z+3? :S

  27. phi
    • 2 years ago
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    yes, with restrictions that z is not 0, -1 or -3

  28. heathernelly
    • 2 years ago
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    you are awesome! tytyty :D x

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

  30. phi
    • 2 years ago
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    Here are some videos http://www.khanacademy.org/math/algebra/rational-expressions/simplifying-rational-alg/v/simplifying-rational-expressions-introduction they are very helpful

  31. heathernelly
    • 2 years ago
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    haha, you mean yes! thanks :)

  32. heathernelly
    • 2 years ago
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    i'll check them out! :D

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