1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Polynomial Help

  1. Jan 6, 2005 #1
    Given that a polymial p(x) is

    [tex] p(x)= (x-1)(x-2) q(x) + 2x+3 [/tex]

    where q(x) is also a polynomial

    Find the remainder when p(x) is divided by (x-1)(x+2) where the remainder divided by (x-1) and (x+2) is both 5 and 7 respectively. I don't know even where to start ! so please help, thanks alot.
  2. jcsd
  3. Jan 6, 2005 #2
    Consider this,
    P(x) = (x-1)(x+2)Q(x) + ax + b
    Find P(1) and P(-2) (You know the remainders , since u know them , try to find a and b).

    -- AI
    Last edited: Jan 6, 2005
  4. Jan 6, 2005 #3


    User Avatar
    Science Advisor

    Check your problem again. I suspect that p(x)= (x-1)(x+2)q(x)+ 2x+ 3 (or, conversely, you want to divide by (x-1)(x-2)). That way, the remainders are 5 and 7 as you say, no matter what q(x) is. The "quotient" when divided by (x-1)(x+2) is q(x) and the remainder is just 2x+ 3.
  5. Jan 6, 2005 #4
    Halls, actually its really meant to be (x-2) but yes, the answer you gave is right too, but one thing still baffles me. If say,

    [tex] \frac {(x-1)(x-2) q(x) + 2x+3}{(x-1)(x-2)}[/tex]

    I cannot factor out (x-1)(x-2) so that they can cancel out in the fraction and then I get 2x+3.

    And I understand why when you divide (x-1) and (x-2) individually q(x) can be ignored because it is multiplied by zero. But why is q(x) ignored when it is divided by the product of (x-1) and (x-2)?
  6. Jan 6, 2005 #5
    One way to think abt it,
    Dividend = Divisor * Quotient + Remainder.

    Another way to think abt it,
    Let Q'(x) = (x-1)(x-2)Q(x)
    P(x) = Q('x) + 2x + 3
    Q'(x) will give remainder 0 when divided by (x-1)(x-2) thereby P(x) will give remainder 2x+3.

    -- AI
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook