Polynom division

  1. 1. The problem statement, all variables and given/known data

    When we divide the polynom P(x) with x-a, we receive reminder [tex]\alpha[/tex], and if we divide P(x) with x-a, we receive reminder [tex]\beta[/tex]. What will be the reminder, if we divide P(x) with (x-a)(x-b)?

    2. Relevant equations



    3. The attempt at a solution

    [tex]P(x)=(x-a)f(x) + \alpha [/tex]

    [tex]P(x)=(x-b)c(x) + \beta [/tex]

    [tex]P(x)= (x-b)(x-a)g(x)+ R(x) [/tex]

    We need to find R(x)
     
    Last edited: Mar 10, 2008
  2. jcsd
  3. tiny-tim

    tiny-tim 26,041
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    Hi Physicsissuef!

    Hint: write P(x) = f(x)(x-a)(x-b) + px +q. :smile:
     
  4. Yes, I already knew it. What's next? :D
     
  5. tiny-tim

    tiny-tim 26,041
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    Physicsissuef, you were supposed show us what you'd already done, in your original post. :frown:

    Also the next thing you tried, and how it didn't work.

    That's what 3. The attempt at a solution is for.​

    If you don't, other PF members don't know where to start, and it wastes our time.

    Start again …
     
  6. Actually I don't know why px+q
    I will edited the first post. Look now. Sorry.
     
  7. tiny-tim

    tiny-tim 26,041
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    ok, your R(x) = px + q, for some constants p and q (because R(x) can't contain x^2, since it's a remainder after dividing by x^2 + …).

    So you should be able to write two equations involving (px + q) and alpha and beta.

    What are they? … :smile:
     
  8. Why after dividing x^2?
     
  9. tiny-tim

    tiny-tim 26,041
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    Remainders must be smaller

    Actually, "after dividing by x^2 + …"

    If you divide by (x^2 plus anything smaller), then the remainder has to be smaller.

    If the remainder was, say, 3x^2 + 5, then you'd just subtract 3 more lots of (x^2 + …), to give you a new remainder with only x and 1.

    Generally, the remainder after dividing by any polynomial beginning with x^n will be a polynomial beginning with x^(n-1) or less. :smile:
     
  10. And how do you know that it is x^2 ??
     
  11. tiny-tim

    tiny-tim 26,041
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    It's (x-a)(x-b), which is x^2 - (a+b)x + ab;
    so when you divide any polynomial by it, the remainder will always be either linear or constant.
     
  12. And how do you know that the polynom P(x) is in this form:
    [tex]ax^3+bx^2+cx+d[/tex], so you can write [tex]R(x)=px+q[/tex]
    ??
     
  13. tiny-tim

    tiny-tim 26,041
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    Sorry, Physicsissuef, you've completely lost me.

    Where did [tex]ax^3+bx^2+cx+d[/tex] come from? :confused:

    Your own equation, [tex]P(x)= (x-b)(x-a)g(x)+ R(x) [/tex], where R(x) is a remainder, shows that R(x) can't have anything higher than x.
     
  14. You said that [tex]R(x)=px+q[/tex], right?
    So when I divide some polynom with x^3 with x^2, I will receive R(x)=px+q, right?

    I asked how did you know that it is x^3?

    Maybe it is x^4?
     
  15. tiny-tim

    tiny-tim 26,041
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    Physicsissuef, divide anything with x^2 + …, you will receive R(x)=px+q.

    It could be x^3 or x^4 or x^307 … it doesn't matter … the remainder will always be of the form px +q (possibly, of course, with p or q = 0).
     
  16. So what I will substitute for p and q? Sorry, if I am getting annoyed.
     
  17. tiny-tim

    tiny-tim 26,041
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    I've lost the plot now …

    ah … we have to go back to post #6:
    Hint: divide (px + q) by (x - a) … what is the remainder? :smile:
     
  18. pa+q. Why I divide them?
     
  19. tiny-tim

    tiny-tim 26,041
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    Because that equals alpha … can you see why? :smile:

    And then, what equals beta?
     
  20. but didn't f(x)*(x-a)+px+q=P(x)

    So px+q equals alpha. Because alpha is the remainder.
     
  21. tiny-tim

    tiny-tim 26,041
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    You've got the principle, but you've got rather confused writing it.

    From post #1, P(x) = f(x)(x-a) + alpha.

    And so alpha equals pa + q (not px + q), from your post #17, because, as you say, alpha is the remainder.

    Right! Nearly there! alpha = pa + q.

    Now what does beta equal? :smile:
     
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