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clamin

  • one year ago

PLEASE HELP!!! MEDAL!!! I Use synthetic division to solve. is this right??

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  1. clamin
    • one year ago
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    |dw:1443661956244:dw| did i do this righ??

  2. zepdrix
    • one year ago
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    No that's not quite right. If you want to do \(\large\rm \frac{-32}{3}+6\) then you need to get a common denominator before adding. This seems like a really difficult way to do synthetic though... What's the problem look like? Can't we get rid of those fractions maybe? :o

  3. clamin
    • one year ago
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    i dont know how to find common denominator.@zepdrix

  4. kva1992
    • one year ago
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    clamin what is the original problem?

  5. clamin
    • one year ago
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    |dw:1443663192218:dw| @kva1992

  6. kva1992
    • one year ago
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    alright so your set up is right only thing is the denominator zep was referring to so if yoru adding -32/3 + 6 it would be converted to -32/3+18/3 then you would subtract and what not

  7. kva1992
    • one year ago
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    does that make sense and do you see what i did?

  8. kva1992
    • one year ago
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    in order to subtract that problem you must convert it to an improper fraction

  9. kva1992
    • one year ago
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    the method you chose to solve it is correct you just didnt subtract and add properly

  10. kva1992
    • one year ago
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    hopefully that helped @clamin

  11. clamin
    • one year ago
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    where did you get that 18/3?? @kva1992

  12. kva1992
    • one year ago
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    change the 6 into a improper fraction since your using 3 as the denominator it would become 18/3

  13. kva1992
    • one year ago
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    change the 6 into a improper fraction since your using 3 as the denominator it would become 18/3

  14. freckles
    • one year ago
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    I like @zepdrix 's idea of getting rid of the nasty fractions in the top polynomial and then dividing using synthetic route

  15. freckles
    • one year ago
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    \[\frac{-6p^4-32p^3-50p^2-8p+33}{3(2p+6)} =\frac{-6p^4-32p^3-50p^2-8p+33}{6p+18}\]

  16. freckles
    • one year ago
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    now divide

  17. kva1992
    • one year ago
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    that works as well

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