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Polynomial, Division, Remainder

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data
    When a polynomial is divided by (x+2), the remainder is -19. When the same polynomial is divided by (x-1), the remainder is 2. Determine the remainder when the polynomial is divided by (x+2)(x-1).

    EDIT: Took out my attempts lol, there were way off.

    This was a "Math Contest" question on our textbook so we didn't have to do it because of it's difficulty but I was curious how to get the answer. The answer at the back of the textbook is 7x-5.

    I'd really appreciate it if someone could give me a hand with this problem. I can't sleep 'til I figure out the answer. :D Thank you.
     
    Last edited: Oct 5, 2009
  2. jcsd
  3. Oct 4, 2009 #2

    lanedance

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    I'm not sure what you're doing in 1 & 2, but i don't think its correct... see below

    do you know or can you use the polynomial remainder theorem? could be useful

    basically say you have F(x), and remainder r when dividing by (x-a)

    Then
    F(x) = g(x).(x-a) + r

    so evaulating at x = a gives:
    F(a) = g(x).0 + r = r

    comparing this to 1&2 when, you divide by (x-a) you get
    F(x)/(x-a) = g(x) + r/(x-a)

    and this is the meaning of your remainder r
     
    Last edited: Oct 4, 2009
  4. Oct 5, 2009 #3
    Thanks for the reply. And yeah, I didn't think I was doing it right either.

    I got as far as P(-2) = -19 and P(1) = 2.

    Not really sure where I need to go after that.

    The whole mess in the first post.. I was trying to make both sides equal, like one would do when solving for two different variables but obviously I was wrong.
     
  5. Oct 5, 2009 #4
    If anyone was interested, here is how the answer is achieved.

    y = mx+b is the standard form for a linear equation

    So we divide the polynomial by a linear equation in the form (x-b) where b is equal to 1
    (mx +b) / (x-1) = m with remainder m + b

    (mx+b) / (x+2) = m with remainder -2m + b


    We solve for the variable b in each case.
    m + b = 2
    b = 2-m

    -2m + b = -19
    b = 2m - 19

    Let b = b and solve for m
    2-m = 2m-19
    3m = 21
    m = 7

    Substitute m = 7 into b = 2-m
    b=2-m
    b=2-7
    b=-5

    But these into the form P(x) = mx +b
    P(x) = mx +b
    P(x) = 7x -5

    All credit goes to Bobbym of the MathIsFun Forum.
     
  6. Oct 5, 2009 #5

    lanedance

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    hey Anakin_k, i think this makes a few assumptions you should be clear about though, just need to justify why you are equating certain things

    now say you have F(x) and divide by (x-1)(x+2), getting remainder R(x)= a.x + b for some constants a, b.

    Note that the degree of the remainder is always less than that of the divisor, justifying ax+b.

    Then F(x) can be written as the following:

    F(x) = g(x)(x-1)(x+2) + R(x)
    with g(x) the quotient (or the division result), R(x) the remainder

    Now evaluating at -2 & 1 gives
    F(-2) = g(-2).(-2-1).0 + R(-2) = R(-2) = -19
    F(1) = g(1).0.(1+2) + R(1) = R(1) = 2

    so solving
    R(-2) = a(-2)+b = -19
    R(1) = a(1)+b = 2

    gives the desired result
    R(x) = 7x-5
     
    Last edited: Oct 5, 2009
  7. Oct 5, 2009 #6

    lanedance

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    updated last post for clarity
     
  8. Oct 5, 2009 #7
    Thanks for posting that.

    Just asking. You put g(0) in there, would it not be g(-2) and g(1) respectively? Just asking.

    Edit: Also, why was a linear equation (y=ax +b) of all options? Not sure about that.
     
    Last edited: Oct 5, 2009
  9. Oct 5, 2009 #8

    lanedance

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    yeah updated g(0) as you suggest - cheers

    The answer is because the remainder R(x) is always of degree less than the divisor d(x).

    In this case d(x) = (x+2)(x-1) = x^2 + x -2, which d(x) has degree 2 (x^2 term)

    So in general, the solution for the remainder has degree <=1, which is why we use the arbitrary degree 1 form for our solution, R(x) = ax + b


    To see why the degree of the remainder is always less than that of d(x), say you have F(x) and want to divide by d(x), the process will give you:

    F(x)/d(x) = g(x) + R(x)/d(x)
    where g(x) is the quotient
    and R(x) is the remainder

    now say R(x) is the same degree or greater than d(x), then we could perform polynomial divsion on R(x), and find
    R(x)/d(x) = q(x) + E(x)/d(x)
    where q(x) is the quotient
    and E(x) is the remainder

    but then the original division becomes:
    F(x)/d(x) = g(x) + R(x)/d(x) = (g(x) + q(x)) + E(x)/d(x)

    This shows R(x) in this case, was not really the true remainder, nor was g(x) the true quotient.
    This shows that the degree of the remainder must be less than that of the divisor.

    Thus justifying the linear solution used for R(x)
     
  10. Oct 5, 2009 #9

    Hurkyl

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    By the way, are you familiar with the Chinese Remainder theorem? It applies here.... (The extended Euclidean algorithm is useful too)
     
  11. Oct 5, 2009 #10
    @lanedance: I asked my teacher the same question and he gave the same answer as you. The degree of the remainder has to be one less than that of the divisor. Cheers!

    @:Hurkyl: I came across is while surfing Wikipedia but I have no practice with it at all.
     
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