Factor Theorem

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  1. krisshP's Avatar
    1 12-07-2012  15:04
    • Peer Of The TSR Realm
    • Posts: 1,806
    Factor Theorem
    Below is the question which I'm stuck on:

    Find a and b given that
    (x+2) and (3x+1) are factors of f(x)=3x^3+ax^2+bx-2

    The following is my working out which is a bit simplified:

    3\times(-2)^3+a(-2)^2+b(-2)-2=0

    3\times-8+4a-2b=2

    2a-b=13
    --------------------------------------------------
    3\times(-\frac{1}{3^3})+a(-\frac{1}{3^2})+b(-\frac{1}{3})=2

    -\frac{3}{27}-\frac{a}{9}-\frac{b}{3}=2

    -\frac{1}{9}-\frac{a}{9}-\frac{b}{3}=2

    -\frac{a}{9}-\frac{3b}{9}=\frac{19}{9}
    --------------------------------------------------
    -a-3b=19

    2a-b=13
    I solved the above simultaneous equations to get the following:

    a=\frac{40}{14}

    b=-\frac{51}{7}

    The values I get for a and b are wrong. What have I done wrong in my working out?????:confused::confused:

    Thanks a lot.
  2. flown_muse's Avatar
    2 12-07-2012  15:11
    • Peer Of The TSR Realm
    • Location: Scotland
    • Posts: 1,753
    Re: Factor Theorem
    In the second equation, it is (-1/3) all squared, you are just squaring the bottom so you haven't got rid of the negative.

    I think!?
    Post rating:
    1
  3. Phredd's Avatar
    3 12-07-2012  15:15
    • Adored and Respected Member
    • Location: Surrey, United Kingdom
    • Posts: 408
    Re: Factor Theorem
    the above poster is right, and you also made a mistake when solving the simultaneous equations you got, i think.
  4. flown_muse's Avatar
    4 12-07-2012  15:17
    • Peer Of The TSR Realm
    • Location: Scotland
    • Posts: 1,753
    Re: Factor Theorem
    (Original post by krisshP)
    Below is the question which I'm stuck on:

    Find a and b given that
    (x+2) and (3x+1) are factors of f(x)=3x^3+ax^2+bx-2

    The following is my working out which is a bit simplified:

    3\times(-2)^3+a(-2)^2+b(-2)-2=0

    3\times-8+4a-2b=2

    2a-b=13
    --------------------------------------------------
    3\times(-\frac{1}{3^3})+a(-\frac{1}{3^2})+b(-\frac{1}{3})=2

    -\frac{3}{27}-\frac{a}{9}-\frac{b}{3}=2

    -\frac{1}{9}-\frac{a}{9}-\frac{b}{3}=2

    -\frac{a}{9}-\frac{3b}{9}=\frac{19}{9}
    --------------------------------------------------
    -a-3b=19

    2a-b=13
    I solved the above simultaneous equations to get the following:

    a=\frac{40}{14}

    b=-\frac{51}{7}

    The values I get for a and b are wrong. What have I done wrong in my working out?????:confused::confused:

    Thanks a lot.
    Forgot to quote you in original post to draw attention!
    Last edited by flown_muse; 12-07-2012 at 15:18.
  5. krisshP's Avatar
    5 12-07-2012  15:20
    • Peer Of The TSR Realm
    • Posts: 1,806
    Re: Factor Theorem
    (Original post by flown_muse)
    In the second equation, it is (-1/3) all squared, you are just squaring the bottom so you haven't got rid of the negative.

    I think!?
    Thanks a lot

    I give it a shot and see what happens
  6. krisshP's Avatar
    6 12-07-2012  15:25
    • Peer Of The TSR Realm
    • Posts: 1,806
    Re: Factor Theorem
    (Original post by Phredd)
    the above poster is right, and you also made a mistake when solving the simultaneous equations you got, i think.
    The mistake in the simulataneous equations is probably due to error carried forward.

    thanks
  7. krisshP's Avatar
    7 12-07-2012  15:28
    • Peer Of The TSR Realm
    • Posts: 1,806
    Re: Factor Theorem
    Phredd and Flown_muse thanks alot

    I end up getting the following as the answer due to your help:

    a=4
    b=-5

    This answer is correct, so thanks a lot for your help
  8. flown_muse's Avatar
    8 12-07-2012  15:45
    • Peer Of The TSR Realm
    • Location: Scotland
    • Posts: 1,753
    Re: Factor Theorem
    (Original post by krisshP)
    Phredd and Flown_muse thanks alot

    I end up getting the following as the answer due to your help:

    a=4
    b=-5

    This answer is correct, so thanks a lot for your help
    Don't worry about it

    The number of times an arithmetic error has ruined everything for me is ridiculous!
  9. steve2005's Avatar
    9 12-07-2012  16:42
    • TSR Demigod
    • Location: LONDON
    Re: Factor Theorem
    (Original post by krisshP)
    Below is the question which I'm stuck on:

    Find a and b given that
    (x+2) and (3x+1) are factors of f(x)=3x^3+ax^2+bx-2

    The following is my working out which is a bit simplified:

    3\times(-2)^3+a(-2)^2+b(-2)-2=0

    3\times-8+4a-2b=2

    2a-b=13
    --------------------------------------------------
    3\times(-\frac{1}{3^3})+a(-\frac{1}{3^2})+b(-\frac{1}{3})=2

    -\frac{3}{27}-\frac{a}{9}-\frac{b}{3}=2

    -\frac{1}{9}-\frac{a}{9}-\frac{b}{3}=2

    -\frac{a}{9}-\frac{3b}{9}=\frac{19}{9}
    --------------------------------------------------
    -a-3b=19

    2a-b=13
    I solved the above simultaneous equations to get the following:

    a=\frac{40}{14}

    b=-\frac{51}{7}

    The values I get for a and b are wrong. What have I done wrong in my working out?????:confused::confused:

    Thanks a lot.
    You have turned an easy sum into one which is more difficult.
    Attached Thumbnails
    Click image for larger version.  Name: Screen shot 2012-07-12 at 16.40.50.png  Views: 21  Size: 16.7 KB  ID: 162720  
  10. krisshP's Avatar
    10 13-07-2012  14:59
    • Peer Of The TSR Realm
    • Posts: 1,806
    Re: Factor Theorem
    (Original post by flown_muse)
    Don't worry about it

    The number of times an arithmetic error has ruined everything for me is ridiculous!
    lol

    annoying arithemetic errors
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