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Homework Help: Yet another factoring problem

  1. Feb 18, 2006 #1
    I asked my friends about this and they were stumped too.

    (a^2-5a)^2 + 8(a^2-5a)+12

    Since it's a trinomial, I tried to do decomposition but it didn't work. I tried to factor out the (a^2-5a) but I end up with a different answer then on the sheet.
     
  2. jcsd
  3. Feb 18, 2006 #2

    arildno

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    Set y=a^2-5a.
    Factorize your expresion in terms of y first
     
  4. Feb 18, 2006 #3
    so would it be ax^2+bx+c=0?
     
  5. Feb 18, 2006 #4

    arildno

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    x presumably meaning y, yes, you'll see how to factorize the polynomial once you've found the roots of your cited equation, with the relevant values for "a,b,c"
    ("a" NOT meaning the same here as above).
     
  6. Feb 18, 2006 #5
    would it be 1,8 and 12?
     
  7. Feb 18, 2006 #6

    arildno

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    That is correct.
     
  8. Feb 18, 2006 #7
    But where do I go from there? My friendsa nd I are stumped. Our teacher never taught us this >_>
     
  9. Feb 18, 2006 #8

    arildno

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    What are the roots of the equation:
    [tex]y^{2}+8*y+12=0[/tex]?
    How can you therefore factorize the polynomial [itex]y^{2}+8y+12[/itex]?
     
  10. Feb 18, 2006 #9
    (y+4)(y+3)=0
     
  11. Feb 18, 2006 #10

    arildno

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    No. Try again.
     
  12. Feb 18, 2006 #11
    err whoops
    (y+2)(y+6)=0
     
  13. Feb 18, 2006 #12
    I got the decomposition mixed up (which term to add up and which to multiply)
     
  14. Feb 18, 2006 #13

    arildno

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    So, you've factorized the left hand side of your EQUATION correctly.
    How is thereby the POLYNOMIAL [itex]y^{2}+8y+12[/itex] factorized?
     
  15. Feb 18, 2006 #14
    I'm not sure what you are asking for, I thought I just factored the polynomial
     
  16. Feb 18, 2006 #15

    arildno

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    OK:
    For ANY number y, we have the identity [itex]y^{2}+8y+12=(y+2)(y+6)[/tex]
    This is to factorize the POLYNOMIAL.


    The two numbers -2 and -6 are the solutions Y for the equation [itex]Y^{2}+8Y+12=0[/itex]

    Do you agree with that?
     
  17. Feb 18, 2006 #16
    Right, I understand that part. (Sorry if I'm a bit slow)
     
  18. Feb 18, 2006 #17

    arildno

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    That's okay.
    Now that you've factorized your polynomial in y, substitute into its factorized form (y+2)(y+6) on the y-places y=a^2-5a.
    What do you get then?
    In particular, can you do some further factorizations?
     
  19. Feb 18, 2006 #18
    so would it be (a^2-5a)(a+2)(a+6)?
     
  20. Feb 18, 2006 #19

    arildno

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    Eeh, why??
    Have you placed the expression a^2-5a into the y-places correctly, do you think?
     
  21. Feb 18, 2006 #20
    OH....is it...(a^2-5a)(a-2)(a-3)?
     
  22. Feb 18, 2006 #21

    arildno

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    No.
    You have (y+2)(y+6)
    If you substitute the y's in that expression using the equality y=a^2-5a, what do you get?
     
  23. Feb 19, 2006 #22

    VietDao29

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    You should read all the posts here again, Hollysmoke:
    You need to factor [tex](a ^ 2 - 5a) ^ 2 + 8(a ^ 2 - 5a) + 12[/tex].
    So you notice that a2 - 5a appears twice, you may want to make your expression have a nicer look.
    So you'll let: [tex]y = a ^ 2 - 5a[/tex], your expression suddenly becomes:
    y2 + 8y + 12, which can be factored to (y + 2) (y + 6). It looks nicer, right?
    Now, having let: y = a2 - 5a, you should substitute that back to the expression (y + 2) (y + 6) to get some expression in terms of a, instead of y.
    After substituting, can the expression can be factored more?
     
  24. Feb 19, 2006 #23
    oops sorry...I forgot to post my Thanks >_<
     
  25. Feb 19, 2006 #24

    arildno

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    Okay, do you agree that with substituting a^2-5a for y, we get:
    [tex](a^{2}-5a+2)(a^{2}-5a+6)[/tex] ?
     
  26. Feb 19, 2006 #25
    Yup. Then u use decomposition for the 2nd term, resulting in (a-2)(a-3)
     
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