Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Polynomial problem

  1. Jan 26, 2008 #1
    Suppose a polynomial is divided by (x-1) and remainder=2 and when the same polynomial is divided by (x-2), remainder is 3. What is the remainder when the polynomial is divided by (x-1)(x-2)? Why?
     
  2. jcsd
  3. Jan 26, 2008 #2

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    This may seem vague, but a remainder may either be a constant, or the degree in the numerator is lower than the degree in the denominator.

    You can start with:
    P(x) = D(x)Q(x) + R(x),
    where P(x) is your polynomial and your dividend, D(x) is your divisor which might be a binomial or trinomial, Q(x) is the quotient, and R(x) is the remainder.

    I worked your problem and it should not be too complicated. You should understand something about using synthetic division to obtain points of the polynomial. If you divide using synthetic division and you obtain a nonzero remainder, then the divisor number and the remainder will give you a point on the polynomial. If the root used is a and the remainder is b, then a point on the polynomial is (a, b).
     
  4. Jan 27, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You know three things:

    Since the polynomial, P(x), divided by x-1, has a numeric remainder, it must be quadratic, of the form ax2+ bx+ c.

    Since the remainder, when divided by x-1, is 2, P(1)= a+ b+ c= 2.

    Since the remainder, when divided by x-2, is 3, P(2)= 4a+ 2b+ c= 3.

    Subtracting the first equation from the second gives 3a+ b= 1 so b= 1- 3a. Putting that into the first equation, a+ 1- 3a+ c= 2 or -2a+ c= 1 so c= 1+ 2a.

    Now, divide ax2+ (1-3a)x+ 1+ 2a by (x-1)(x-2)= x2+ 3x+ 2.
     
  5. Jan 28, 2008 #4
    Thanx! It didn't occur to me that p(x) couldn't be anything other than quadratic.
     
  6. Jan 28, 2008 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Actually, I wrote out a long response arguing that this was a bad problem because there were too many possibilities and the couldn't all give the same remainder.

    Until I did the critical division at the end and found out they did!
     
  7. Jan 28, 2008 #6

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    You could show us what you were writing anyway. The reason I might like to read it is that I did not make any assumptions about the initial unknown polynomial, and then I found a non-polynomial result (a binomial, in fact). I would try to be clear about what assumption I did make that I should not have made; or what assumption I should have made that I did not. My result was that P(x)= ( x+1). Not a bad result but it is not a polynomial. The division process lead only to a remainder, and no other coefficients. Should I have understood that I must expect at least a quadratic degree polynomial?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Polynomial problem
Loading...