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Homework Help: Equations of Higher Degree

  1. Mar 7, 2007 #1
    1. The problem statement, all variables and given/known data

    Determine the remainder using the remainder theorem
    P(x) = 2x^3 + 3x^2 + 4x - 10; D(x) = x + 1

    2. Relevant equations

    Remainder Theorem

    3. The attempt at a solution
    x = -1
    P(x) = -2 + 3 - 4 - 10
    R -13


    Can you have a negative remainder?


    The next question says:
    Decide whether or not the number is a zero of the polynomial.

    Is that just another way of saying find the remainder?
     
    Last edited: Mar 7, 2007
  2. jcsd
  3. Mar 7, 2007 #2

    Dick

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    Sure you can have a negative remainder. Is the 'number' you are referring to -1? If so, IS it a root?
     
  4. Mar 7, 2007 #3
    Oh ok, so -13 is the right answer to that one?


    No, the next question is referring to a different polynomial, though the question was stated different. Can I use the remainder theorem again, or is it asking me to do something different?
     
  5. Mar 7, 2007 #4

    Dick

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    What's the remainder of x-r if r is a root? And if you don't believe -13 is the remainder - check it with a long division.
     
    Last edited: Mar 7, 2007
  6. Mar 7, 2007 #5

    HallsofIvy

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    The remainder theorem which you mention but don't cite (that would have been a good idea) says that if P(x) is divided by (x-a) then the remainder is P(a). That's true because P(x)/(x-a)= Q(x)+ remainder r is the same as saying P(x)= (x-a)Q(x)+ r. What happens when you set x= a?
     
  7. Mar 7, 2007 #6
    I'm lost =/

    My teacher didn't go over the theorem, he just told us how to use the f(x) = blah

    Sorry I didn't site it.

    I'll work it out in long division to see if I got the right answer, but I'm thinking I did since you can have a - remainder.

    Now I'm not sure what to do about the question that asks if a certain number is a zero of the polynomial... is that the same as saying find the remainder?
     
  8. Mar 7, 2007 #7

    Dick

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    Look at what Halls said. Dividing a polynomial P(x) by (x-a) and getting a remainder r and a quotient Q(x) means you can write the polynomial P(x)=Q(x)*(x-a)+r, right? Make sense? Just the same as division by numbers? Putting x=a in that gives:

    P(a)=r

    since a-a=0. That is the Remainder theorem. So to see if r=0 (which is the same as saying that a is a root, yes?), you can either put 'a' into the polynomial and see if you get zero or you can divide the polynomial by (x-a) and see if the remainder is zero. Same thing. Do actually try out the long division to prove it to yourself.
     
  9. Mar 7, 2007 #8
    What is a?

    When i divided it with long division I got -9 =/
    But when i used synthetic division i got -13 again...

    The question about finding a zero is

    P(x) = -x^4 + 9x^2 - 9x + 27; 3
    So I did used
    P(3) = -(3)^4 + 9(3)^2 - 9(3) + 27 = 0
    so yes it is a zero?

    There is yet another problem that says find the quotient and remainder of the following problems and whatnot... i can get the remainder but how do i get the quotient?
    Is that where I subtract a root from the equation since I just took one out?
    If so here is what I did:

    Problem: (3x^5 + 4x^4 + 2x^2 - 1) / (x + 2)

    Solution (using synth div.)

    3 -2 4 -6 R11
    so
    3x^3 - 2x^2 + 4x - 6 R11 ??
     
    Last edited: Mar 7, 2007
  10. Mar 8, 2007 #9

    HallsofIvy

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    In the statement of the problem you said the the polynomial was 2x3+ 3x2+ 4x- 10. Here you use "-2 + 3 - 4 -10" which corresponds to -2x3+ 3x2- 4x- 10. Which is it?

    ??? If you were asked to find a zero, where did you get "3" from? If you were asked to determine whether or not 3 is a zero then it's just a matter of understanding what a "zero" of a polynomial is.

    Are you saying you do not know the meaning of the word "quotient"? If you were to divide 1233 by 5 what would be the quotient and what would be the remainder?
     
  11. Mar 8, 2007 #10
    It's the one stated in the problem, but i did (-1)^3 * 2.

    Yes, I needed to determine if 3 was a zero.

    I actually typed the problem wrong... but it's ok I have it right now.

    thanks for the help.
     
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