How do you divide $( x^5 - x^3 - x^2 - 17x - 15 )/(x^2 - 2 )$ ?

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Answer 1 · 2016-01-05T23:27:27

Long divide the coefficients to find:

$(x^5-x^3-x^2-17x-15)/(x^2-2) = (x^3+x-1) + (-15x+13)/(x^2-2)$

...not forgetting to include $0$ 's for any missing powers of $x$ .

I like to long divide the coefficients, not forgetting to include $0$ 's for any missing powers of $x$ ...

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This is similar to long division of numbers.

Reconstructing polynomials from the resulting sequences we find that the quotient is $x^3+x-1$ with remainder $-15x+13$

That is:

$(x^5-x^3-x^2-17x-15)/(x^2-2) = (x^3+x-1) + (-15x+13)/(x^2-2)$

Or if you prefer:

$(x^5-x^3-x^2-17x-15) = (x^2-2)(x^3+x-1) + (-15x+13)$

How do you divide ( x^5 - x^3 - x^2 - 17x - 15 )/(x^2 - 2 )( x^5 - x^3 - x^2 - 17x - 15 )/(x^2 - 2 )?