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Factors of a polynomial

  1. Oct 16, 2004 #1
    [tex]3x^2 + 4x + C \equiv A(x + 1)^2 + B(x + 1) + 7[/tex]
    Find all values of A, B and C.
    Could someone teach how to do this?
     
  2. jcsd
  3. Oct 16, 2004 #2

    arildno

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    Rewrite
    [tex]x^{2}=((x+1)-1)^{2}[/tex]
    and do the necessary operations.
     
  4. Oct 16, 2004 #3
    Huh? Sorry I don't understand what u mean.
     
  5. Oct 16, 2004 #4

    arildno

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    [tex]((x+1)-1)^{2}=(x+1)^{2}-2(x+1)+1[/tex]
     
  6. Oct 16, 2004 #5
    How is [tex]((x+1)-1)^{2}=(x+1)^{2}-2(x+1)+1[/tex]
    related to [tex]3x^2 + 4x + C \equiv A(x + 1)^2 + B(x + 1) + 7 ?[/tex]
    Sorry but i don't get u.
     
  7. Oct 16, 2004 #6

    arildno

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    It's equal to [tex]x^{2}[/tex]!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
    Make a similar rewriting of x:
    [tex]x=((x+1)-1)[/tex]
    Now, substitute these expressions for [tex]x,x^{2}[/tex] into your LEFT-HAND SIDE.
    Reorganize the terms you get, and derive conditions so that your new expression equals your ORIGINAL RIGHT-HAND SIDE.
    This will determine A,B,C.
     
  8. Oct 16, 2004 #7

    Hurkyl

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    Alternatively, you could expand the right hand side, and equate coefficients to solve for A, B, and C.
    Or you could plug in a few values for x to generate equations.
     
  9. Oct 16, 2004 #8
    I probably wasn't clear. Firstly, how does [tex]x^{2}=((x+1)-1)^{2}.[/tex] Secondly, which ones do i substitute in for x? Is it [tex]((x+1)-1)^{2} = 3 or (x + 1)^2 ?[/tex]
     
  10. Oct 16, 2004 #9

    arildno

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    You know that 1-1=0, right?
    So (x+1)-1=x+1-1=x.
    To be nice, I'll do this for once:
    We have:
    [tex]3x^{2}=3((x+1)-1)^{2}=3(x+1)^{2}-6(x+1)+3[/tex]
    [tex]4x=4((x+1)-1)=4(x+1)-4[/tex]
    [tex]C=C=C[/tex]
    Now, add these equations together, downwards. The outermost terms then turn into:
    [tex]3x^{2}+4x+C=3(x+1)^{2}-2(x+1)+(C-1)[/tex]
    Do you understand this?
     
  11. Oct 16, 2004 #10

    Hurkyl

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    What does (x + 1) - 1 equal? Then what does ((x+1) - 1)^2 equal?
     
  12. Oct 16, 2004 #11
    I'm just curious, has this got something to do with modulus?
     
  13. Oct 16, 2004 #12

    arildno

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    Not that I know of..
    It is simply to substitute "equal for equal"
     
  14. Oct 16, 2004 #13
    <Sarcastic mode ON>
    arildno you have to be a little more diplomatic, I think
    <Sarcastic mode OFF>
     
  15. Oct 16, 2004 #14
    you guys are making the problem too complicated
    since
    [tex]3x^2 + 4x + C = A(x + 1)^2 + B(x + 1) + 7[/tex]
    expand [tex]A(x + 1)^2 + B(x + 1) + 7[/tex]
    then you have [tex]Ax^2 + 2Ax + A + Bx + B + 7[/tex]
    and [tex]3x^2 + 4x + C = Ax^2 + 2Ax + A + Bx + B + 7[/tex]
    equate the coeffients of the powers you have
    [tex]3 = A[/tex] for [tex]x^2[/tex]
    [tex]2A + B = 4[/tex] for [tex]x^1[/tex]
    [tex]A + B + 7 = C[/tex] for [tex]x^0[/tex]
    solve the system and you got the answer :smile:
     
    Last edited: Oct 16, 2004
  16. Oct 17, 2004 #15
    Ahh... Sorry but i was a bit slow on "(x+1)-1" part (sorry if i pissed u arildno :frown: ). Thanks for the help guys!
     
  17. Oct 17, 2004 #16

    arildno

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    I wasn't exactly pissed off; rather, I felt resignation sneak up on me..:wink:
     
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