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anonymous

  • one year ago

Just wanted to check my answers if they are right. (See below)

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  1. anonymous
    • one year ago
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    1.)\[\frac{ x-4 }{ x^2-3x-4 }\]

  2. anonymous
    • one year ago
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    My answer: \[\frac{ 1 }{ (x+1) }\]

  3. anonymous
    • one year ago
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    1 is correct

  4. anonymous
    • one year ago
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    Thanks! 2.) \[\frac{ x^3-8 }{ x-2 }\]

  5. anonymous
    • one year ago
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    My answer:\[(x+2)^2\]

  6. anonymous
    • one year ago
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    I think u can simplify more than that

  7. anonymous
    • one year ago
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    How would I simplify that?

  8. anonymous
    • one year ago
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    hold on

  9. anonymous
    • one year ago
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    something is wrong

  10. anonymous
    • one year ago
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    I don't think I factored the numerator right.

  11. anonymous
    • one year ago
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    start by putting the 8 into 2^3

  12. anonymous
    • one year ago
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    Okay. Then what?

  13. anonymous
    • one year ago
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    and use the difference of cubes which is \[a^3−b^3=(a−b)(a^2+ab+b^2)\]

  14. anonymous
    • one year ago
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    it will get you \[(x−2)(x^2+(x)(2)+2^2)\]

  15. anonymous
    • one year ago
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    Simplify 2^2 to 4 and regroup terms

  16. anonymous
    • one year ago
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    \[\frac{ (x−2)(x^2+2x+4) }{ x-2 }\]

  17. anonymous
    • one year ago
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    now just cancel the x-2

  18. anonymous
    • one year ago
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    did you get it?

  19. anonymous
    • one year ago
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    Ya I got it. Wasn't that the same answer I got?

  20. anonymous
    • one year ago
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    \[(x+2)^2\]

  21. anonymous
    • one year ago
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    It's different

  22. anonymous
    • one year ago
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    I wrote the same answer.

  23. anonymous
    • one year ago
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    \[x^2+2x +4 \neq (x+2)^2\]

  24. anonymous
    • one year ago
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    Ya sorry about that. Just noticed.

  25. anonymous
    • one year ago
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    I would have to use the quadratic formula for this right?

  26. Zale101
    • one year ago
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    \(a^3-b^3=(x-a)(a^2+ab+b^2)\) \(x^3-2^3=(x-2)(x^2+2x+2^2)=x^2+2x+4\) \(\Large\frac{x^3-8}{x-2}=\frac{x^2-2^3}{x-2}=\frac{(x-2)(x^2+2x+4)}{x-2}\)

  27. Zale101
    • one year ago
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    \(\Large =x^2+2x+4\)

  28. anonymous
    • one year ago
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    Ya I got that so far. Thanks!

  29. anonymous
    • one year ago
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    \[\frac{ -2\pm \sqrt{2^2-4(1)(4)} }{ 2(1) }\]\[\frac{ -2\pm \sqrt{4-16} }{ 2 }\]\[\frac{ -2\pm \sqrt{-12} }{ 2 }\]

  30. anonymous
    • one year ago
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    The answer is going to have an i?

  31. Zale101
    • one year ago
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    Yes. That's why it cant be factored. If you get complex numbers when factoring, then the polynomial is irreducible. x^2+2x+4 is a prime polynomial.

  32. anonymous
    • one year ago
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    So I don't have to simplify that then? Do I just put x^2+2x+4?

  33. Zale101
    • one year ago
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    Yes :)

  34. anonymous
    • one year ago
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    Alright thanks! 3.)\[\frac{ 5-x }{ x^2-25 }\]

  35. anonymous
    • one year ago
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    My answer:\[\frac{ -1 }{ x+5 }\]

  36. anonymous
    • one year ago
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    4.) \[\frac{ x^2-4x-32 }{ x^2-16 }\]

  37. alekos
    • one year ago
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    3 is correct

  38. anonymous
    • one year ago
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    My answer:\[\frac{ x-8 }{ x-4 }\]

  39. anonymous
    • one year ago
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    Thanks! What about #4?

  40. alekos
    • one year ago
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    yes. well done

  41. anonymous
    • one year ago
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    Thanks you guys!

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