Question 31461
{{{ (2u^4-5u^3-12u^2+2u-8)/(u-4) }}}
When dividing by a binomial you MUST make sure that the numerator is in decending order, and that there are no missing exponential values.
This numerator meets that criteria.
Now ... I have not found a way to show the long division in the rendering page so this may seem really long, but bear with me and we will take it all step by step.
FIRST: look at the denominator.  you are going to divide by u first.  Ask yourself: what number times u equals 2u^4?  The answer: 2u^3
NEXT: multiply that value times the denominator
{{{ highlight(2u^3) (u-4) = 2u^4-8u^3 }}} Subtract this value from the numerator ...
{{{ (2u^4-5u^3-12u^2+2u-8) - (2u^4-8u^3) }}} dont forget to distribute the negative
{{{ 2u^4-5u^3-12u^2+2u-8-2u^4+8u^3 }}}
When you combine like terms ...
{{{ 3u^3-12u^2+2u-8 }}}
notice that the first term cancels.  We did this on purpose.
continue on as we did in the the first step, this time asking: what number times u equals 3u^3?  The answer: 3u^2
Multiply that value by the denominator.
{{{ highlight(3u^2) (u-4) = 3u^3-12u^2 }}} subtract this answer once again.
{{{ 3u^3-12u^2+2u-8 - (3u^3-12u^2) }}} watch the negatives
{{{ 3u^3-12u^2+2u-8-3u^3+12u^2 }}} combine like terms.
{{{ -12u^2+2u-8+12u^2 }}} notice that TWO terms will cancel in this step
{{{ 2u-8 }}} back to step one:
What number times u equals 2u?  The answer is 2.
Multiply it by the denominator.
{{{ highlight(2)(u-4) = 2u-8 }}} 
Subtract this value from what remains.
{{{ 2u-8 - (2u-8) }}} watch the negatives ....
{{{ 2u -8 - 2u + 8 }}} EVERYTHING cancels.
This means there is NO remainder.
Lets look at the numbers we came up with in each "First step" (the highlighted numbers)
{{{ 2u^3 }}} and {{{ 3u^2 }}} and {{{ 2 }}}
Add these numbers together
{{{ 2u^3 + 3u^2 + 2 }}} this is your answer