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# dividing polynomials that don't divide evenly.

Posted on 2014-08-19
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Using the long division method how would you solve(x^2-9-x)/(x+1) ?  For example:
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).

thx
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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
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LVL 26

Expert Comment

There is an example here:    http://www.sosmath.com/algebra/factor/fac01/fac01.html

>>  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
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LVL 17

Expert Comment

Correction: x-2 r -7
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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.
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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).
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Author Closing Comment

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