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Homework Help: Remainder/factor theorem

  1. Sep 22, 2007 #1
    1. The problem statement, all variables and given/known data

    Once again I'm stuck....
    Find a and b if x^4+ax^3-2x^2+bx-8 is divisible by x^2-4.

    3. The attempt at a solution

    x can be + or - 2 so P(2)=0 and P(-2)=0
    I don't think I can solve these simultaneously since everything will cancel so how am I supposed to find a and b? Should I be trying to find another factor or something?

  2. jcsd
  3. Sep 22, 2007 #2
  4. Sep 22, 2007 #3


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    No, you are exactly right. There is no single answer. For example, suppose a= b= 0 so the equation is x^4- 2x^2- 8= (x^2-4)(x^2+ 2) which is divisible by x^2- 4. However, if a= 1 and b= -4, so that 8a+ 2b= 0, the equation is x^4+ x^3- 2x^2-4x- 8= (x^2- 4)(x^+ x+ 2). That is also divisible by x^2- 4!. In fact, it is divisble by x^2- 4 as long as 8a+ 2b= 0. That is, as long as b= -4a.
  5. Sep 22, 2007 #4
    So I can't really give just one value for a and one value for b? The back of the book says a=3,b=-12 but is this just one set of solutions?

    Thanks for the replies!
  6. Sep 22, 2007 #5
    Your two equations are correct.
    When you have a system of equations like that, and everything cancels out, you end up with an infinite number of solutions. Pick a value for a, let's say 1. If a=1 then b = -4. Or, if a=3, then b=-12. Try these two polynomials to see if they're both factorable.

    If you try a couple of polynomial (and maybe make up a couple more), you can do two things:
    1. Write down the relationship between a and b (you should be able to do that from the equations you already had.)
    2. See if you can develop a relationship between the choices of the polynomial and the quotient after long division.

    As far as (2.) goes, you should be able to show this same relationship by doing a long division.
  7. Sep 22, 2007 #6
    Ahhhh, I'm a little slower with the typing than Halls of Ivy.

    Yes, the answer in the back of the book is just one of many possible solutions.
  8. Sep 22, 2007 #7
    Woah that's so cool! Well not "cool" but you know... Thanks for the explanations!
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