How do you divide #(x^5 - 10x^2 - 8) ÷ (x - 2)#?

1 Answer
May 29, 2016

#(x^5-10x^2-8)/(x-2) = x^4+2x^3+4x^2-2x-4-16/(x-2)#

Explanation:

I like to long divide coefficients rather than including all of the terms in #x#. It's easier to typeset for one thing. The important thing to remember is to include #0#'s for any missing powers of #x#. In our example that means #x^4#, #x^3# and #x# in the dividend:

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The process is similar to long division of numbers.

Write the dividend under the bar and the divisor to the left.

Each term of the quotient is chosen so that when multiplied by the divisor, the leading term matches the leading term of the running remainder.

If you prefer to include the powers of #x# then it can be written like this:
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We find:

#(x^5-10x^2-8)/(x-2) = x^4+2x^3+4x^2-2x-4-16/(x-2)#