Question 119164
You have done the first steps correctly, that is you put the terms in decending order of degree and put a zero coefficient placeholder in for the missing 4th degree term.  I presume you ended up with:


{{{(25x^5+0x^4-x^3-2x^2-8x)/(5x^2-4x)}}}


The next step is to factor the denominator.  {{{5x^2-4x=x(5x-4)}}}.  Now your problem looks like this:


{{{(25x^5+0x^4-x^3-2x^2-8x)/x(5x-4)}}}


But notice that there is an {{{x}}} in every term of the numerator which means you can divide through by the {{{x}}} in the denominator to get:


{{{(25x^4+0x^3-x^2-2x-8)/(5x-4)}}}


Now your polynomial long division shouldn't be so ugly.  I don't know how to render the process on this site, so I'll just talk you through it.  Write back if you have trouble understanding.


Step 1:  {{{5x}}} goes into {{{25x^4}}} {{{5x^3}}} times, so {{{5x^3}}} is the first term of your quotient polynomial.


Step 2:  {{{5x^3}}} times {{{-4}}} is {{{-20x^3}}} and {{{5x^3}}} times {{{5x}}} is {{{25x^4}}}, so {{{25x^4-20x^3}}} is your first partial product.


Step 3:  Subtract {{{25x^4-20x^3}}} from the first two terms of the dividend polynomial giving you {{{0x^4+20x^3}}}.


Step 4:  Bring down the next term, {{{-x^2}}} to form your next partial dividend {{{20x^3-x^2}}}


Step 6:  {{{5x}}} goes into {{{20x^3}}} {{{4x^2}}} times, so this is the next term of your quotient polynomial.  So far, your quotient should look like {{{5x^3+4x^2}}}.


Step 7:  {{{4x^2}}} times {{{-4}}} is {{{-16x^2}}} and {{{4x^2}}} times {{{5x}}} is {{{20x^3}}}, so {{{20x^3-16x^2}}} is your next partial product.


Step 8:  Subtract this from your previous partial dividend, {{{20x^3-x^2}}} (remembering to change the sign and add when you subtract), to get {{{0x^3+15x^2}}}.


Step 9:  Bring down the next term of the dividend, {{{-2x}}}, to form your next partial dividend, {{{15x^2-2x}}}.


Step 10:  {{{5x}}} goes into {{{15x^2}}} {{{3x}}} times giving you the next term of your quotient.  Your quotient should now look like {{{5x^3+4x^2+3x}}}.


Step 11:  {{{3x}}} times {{{-4}}} is {{{-12x}}} and {{{3x}}} times {{{5x}}} is {{{15x^2}}}, so {{{15x^2-12x}}} is your next partial product.


Step 12:  Subtract this from your previous partial dividend, {{{15x^2-2x}}}, resulting in {{{0x^2+10x}}}


Step 13:  Bring down the last term, {{{-8}}} to form your next partial dividend {{{10x-8}}}


Step 14:  {{{5x}}} goes into {{{10x}}} {{{2}}} times, so {{{2}}} is the last term of your quotient which should look like {{{5x^3+4x^2+3x+2}}}.


Step 15:  {{{2}}} times {{{-4}}} is {{{-8}}} and {{{2}}} times {{{5x}}} is {{{10x}}} so your next partial product is {{{10x-8}}}.


Step 16:  Subtract.  {{{(10x-8)-(10x-8)=0}}}.  There is no remainder, so you are done.


Hope this helps.
John