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anonymous

  • 5 years ago

How do you use synthetic division to find P(2) and P(-1)? P(x) = x^5 - 4x^3 + 2x I have this so far, but I can't remember how to find P(-1) 2 | 1 0 -4 0 2 0 ........2 4 0 0 4 _____________ ....1 2 0 0 2 | 4 So, P(2) = 4

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  1. anonymous
    • 5 years ago
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    list the coefficients and put -2 in the box: 1 0 -4 0 2 0 -2

  2. anonymous
    • 5 years ago
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    the remainder will be P(2)

  3. anonymous
    • 5 years ago
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    do you know the procedure for synthetic division?

  4. anonymous
    • 5 years ago
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    1 0 -4 0 2 0 -2 ________________________

  5. anonymous
    • 5 years ago
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    oops

  6. anonymous
    • 5 years ago
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    1 0 -4 0 2 0 -2 -2 4 0 0 -2 _________________________________ 1 -2 0 0 2 -2

  7. anonymous
    • 5 years ago
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    last line is a mistake. it should be -2*2=-4 and 0 + (-4) = -4 sorry

  8. anonymous
    • 5 years ago
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    so P(2)=-4

  9. anonymous
    • 5 years ago
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    Oh wow..thank you so much...i just couldn't figure that out

  10. anonymous
    • 5 years ago
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    would you like me to write the next one?

  11. anonymous
    • 5 years ago
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    same coefficients but this time you put 1 in the box

  12. anonymous
    • 5 years ago
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    Yes please..that would help

  13. anonymous
    • 5 years ago
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    I'm actually studying for my final

  14. anonymous
    • 5 years ago
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    ok list the coefficients again 1 0 -4 0 2 0 1 ___________________________________

  15. anonymous
    • 5 years ago
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    bring down the one from the top line 1 0 -4 0 2 0 1 ___________________________________ 1

  16. anonymous
    • 5 years ago
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    1*1=1 goes on the second line 1 0 -4 0 2 0 1 1 ___________________________________ 1

  17. anonymous
    • 5 years ago
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    add 0+1=2 1 0 -4 0 2 0 1 1 ___________________________________ 1 1

  18. anonymous
    • 5 years ago
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    1*=1 1 0 -4 0 2 0 1 1 1 ___________________________________ 1 1

  19. anonymous
    • 5 years ago
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    -4+1=-.3 1 0 -4 0 2 0 1 1 1 ___________________________________ 1 1 -3

  20. anonymous
    • 5 years ago
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    1*-3=-3 1 0 -4 0 2 0 1 1 1 -3 ___________________________________ 1 1 -3

  21. anonymous
    • 5 years ago
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    0+-3=-3 1 0 -4 0 2 0 1 1 1 -3 ___________________________________ 1 1 -3 -3

  22. anonymous
    • 5 years ago
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    1*-3=-3 1 0 -4 0 2 0 1 1 1 -3 -3 ___________________________________ 1 1 -3 -3

  23. anonymous
    • 5 years ago
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    2+-3=-1 1 0 -4 0 2 0 1 1 1 -3 -3 ___________________________________ 1 1 -3 -3 -1

  24. anonymous
    • 5 years ago
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    1*-1=-1 1 0 -4 0 2 0 1 1 1 -3 -3 -1 ___________________________________ 1 1 -3 -3 -1

  25. anonymous
    • 5 years ago
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    0+-1=-1 1 0 -4 0 2 0 1 1 1 -3 -3 -1 ___________________________________ 1 1 -3 -3 -1 -1

  26. anonymous
    • 5 years ago
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    done

  27. anonymous
    • 5 years ago
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    on the other hand if i had to evaluate P(-1) i would just plug it in because it is easier: \[P(-1)=(-1)^5 -4(-1)^3 + 2(-1)=-1+4-2=-1\]

  28. anonymous
    • 5 years ago
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    okay, so P(-1) = -1

  29. anonymous
    • 5 years ago
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    yes this is correct by either calculation

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