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Factorising problem

  1. Jan 17, 2014 #1
    Having trouble solving this problem S2+6S+13

    For example I solved this one before and stuck on the above

    8S+38/2S2+5S-3=8S=38/(2S-1)(S+3)=A/2S-1+B/s+3

    A=12 B=-2

    How do I factorise S2+6S+13 to get value for A and B
     
    Last edited: Jan 17, 2014
  2. jcsd
  3. Jan 17, 2014 #2

    adjacent

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    Gold Member

    Solve?Do you mean factorize as in the title? That's not an equation
     
  4. Jan 17, 2014 #3
    ya sorry factorise
     
  5. Jan 17, 2014 #4

    Mark44

    Staff: Mentor

    As already noted, there's nothing to solve here - this is not an equation.
    You're missing lots of parentheses here. Whenever the numerator or denominator contains more than one term, put parentheses around the entire numerator or denominator.
    This quadratic can't be factored into factors with integer or even rational coefficients. About the best you can do is complete the square to write it in the form (S + <something>)2 + <something else>.

    Note: Homework-type questions must be posted in the Homework & Coursework section, not in the technical math sections.
     
  6. Jan 17, 2014 #5

    adjacent

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    Gold Member

    It's ##(S+3)^2 + 4##
     
  7. Jan 17, 2014 #6
    How do you get that?
     
  8. Jan 17, 2014 #7

    adjacent

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    wolframalpha.com but you should not use it to do all your homeworks.Learn to use your brain
     
  9. Jan 17, 2014 #8

    Mark44

    Staff: Mentor

    See post #4.
     
  10. Jan 17, 2014 #9

    Ray Vickson

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    Homework Helper

    Your lack of parentheses makes your expressions unreadable (or, rather, readable and saying something you might not mean). Literally, your expression is
    [tex]8S+\frac{38}{2S^2}+5S-3 [/tex]
    when read using standard rules for mathematical expressions.
    Did you mean
    [tex] \frac{8S + 38}{2S^2 +5S -3} ?[/tex]
    Did you mean
    [tex] \frac{8S + 38}{2S^2 +5S}-3 ? [/tex]
    If you meant the first, use parentheses like this: (8S+38)/(2S^2 + 5S - 3)---simple, and completely unambiguous. If you meant the second, write it as (8S+38)(/(2S^2 + 5S) - 3, which is unambiguous when read by standard rules (although some might engage in "overkill" and write, instead: [(8S+38)(/(2S^2 + 5S)] - 3).
     
  11. Jan 17, 2014 #10

    [tex] \frac{8S + 38}{2S^2 +5S -3} [/tex]

    Sorry but I cant write these expressions. my time is taken up studying them.
     
  12. Jan 17, 2014 #11
    There are a couple of ways to do this. The most reliable is using the solution to the quadratic equation: S2+6S+13 = 0 (A=1, B=6, C=13)

    The quadratic is S = (-B +/- Sqrt(B2-4AC))/2A

    Once you have the solutions, the factors will be (S-solution1)(S-solution2).

    The reason this works is that you know that if (S-solution1)(S-solution2)=0, then at least one of terms must be zero and therefor S must be one of the solutions to the quadratic.

    BTW: There is no nice [strike]integer[/strike] real solution to this.
     
    Last edited: Jan 17, 2014
  13. Jan 17, 2014 #12

    Mark44

    Staff: Mentor

    If you want help from us, you need to clearly communicate what the problem is you're working on. No one is demanding that you become expert in using LaTeX to write these fractions - just use enough parentheses so that we're all on the same page as to what the problem is.

    This means writing (8S+38)/(2S^2 + 5S - 3) instead of 8S+38/2S2+5S-3, as you wrote.
     
  14. Jan 17, 2014 #13
    I should have said there is no nice real. Your solution will look like this:
    (S+u+vi)(S+u-vi)

    I'll check this thread in another 10 or 15 minutes. If you haven't solved for u and v by then, I'll just post the answer.
     
  15. Jan 17, 2014 #14

    Mark44

    Staff: Mentor

    I'm not sure, but I suspect that the posted problem is part of a larger problem involving an integral or a Laplace transform. If so, factorization as you show above won't be of much help, but writing the denominator as (x + number1)2 + number2 will be useful
    Please don't. Per forum rules (under Homework Help Guidelines at https://www.physicsforums.com/showthread.php?t=414380):
     
  16. Jan 17, 2014 #15
    Only Anthony2013 knows for sure.
    OK - but I pretty much out of hints.
    Ohh, both u and v are integers.
     
  17. Jan 17, 2014 #16
    Laplace guys, sorry about confusion on my part. Typo on my work (3) should be S(3)
     

    Attached Files:

  18. Jan 17, 2014 #17
    Actually, I should have caught it when you were asking for the A and B.
     
  19. Jan 17, 2014 #18

    Mark44

    Staff: Mentor

    Just as I thought. In this case you don't want to factor your denominator. Instead, complete the square as already mentioned in posts #4 and #14. Since you either don't know or don't remember this technique, here's a link: http://www.purplemath.com/modules/sqrquad.htm. See the 2nd example on this page.
     
  20. Jan 17, 2014 #19
    Don't remember 15 years out of the game, dropped back in lately. (old age as well)
    Thanks for help
     
  21. Jan 17, 2014 #20

    Mark44

    Staff: Mentor

    Once you get your expression into the form I recommended, you should be able to look up the inverse Laplace in a table to get the solution of your differential equation.
     
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