dividing polynomials that don't divide evenly.

Posted on 2014-08-19
Last Modified: 2014-08-19
Using the long division method how would you solve(x^2-9-x)/(x+1) ?  For example:
 example of a polynomial being divided by another polynomial.Unlike the above example x+1 won’t divide evenly into x^2-9-x, so what would be the result?  Am I stuck with a remainder value or is there some kind of decimal point system?  

Could someone please provide me with an example of how this would work (please include all steps).

Question by:NevSoFly
    LVL 17

    Accepted Solution

    No, that is the answer. You will have a remainder of 7 as there is no value to evenly divide those polynomials.
    x^2-9-x/x+1= x+2 r7
    LVL 26

    Expert Comment

    There is an example here:

    >>  x+1 won’t divide evenly into x^2-9-x
          You have to keep your terms in order:   x+1 won’t divide evenly into   x² - x - 9

    The final answer will be of the form:     x - 2  +  ( ????)/(x+1)    

    The ???? is the remainder
    LVL 17

    Expert Comment

    Correction: x-2 r -7

    Author Comment

    So in that case there is no way to divide x^2-9-x by let's say a-b.  Is this correct?

    It seems confusing to me because I know that x^2/a can be written as x^2a^-1.  I tried to do a similar problem like this but it would go on for eternity.
    LVL 26

    Assisted Solution

    You can divide a quadratic (or higher degree) polynomial by a any lesser degree polynomial.  Both polynomials must use the same variable (x in your case).

    Author Closing Comment

    thanks, I got my answer and a bit of explanation.

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