OK
Polynomial division....the idea is to multiply the "n" of the divisor by a term that will "match" the first term
For instance.....we want to multiply "n" by 9n^4 to match the first term , 9n^5
We just keep carrying this idea through each time
For the second term....we want to multiply "n" by -2n^3 to "match" -2n^4.....etc......
9n^4 - 2n^3 + 10n^2 + n + 6
(n + 9) [ 9n^5 + 79n^4 - 8n^3 + 91n^2 + 15n + 46 ]
-( 9n^5 + 81n^4 )
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-2n^4 - 8n^3
- (-2n^4 - 18n^3)
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10n^3 + 91n^2
-(10n^3 + 90n^2)
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n^2 + 15n
-( n^2 + 9n)
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6n + 46
-( - 6n + 54)
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-8
So....the result is 9n^4 - 2n^3 + 10n^2 + n + 6 R [ -8 / (n + 9) ]
Synthetic is usually much faster....we want to divide by the thing that makes (n + 9) = 0....this is - 9
So we have
-9 [ 9 79 - 8 91 15 46 ]
-81 18 -90 -9 -54
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9 -2 10 1 6 -8
The residual polynomial is 9n^4 - 2n^3 + 10n^2 + 1n + 6 R [ -8 / ( n + 9) ]...just as we found before !!!
Here's a good primer on polynomial diivsion :
http://www.mesacc.edu/~scotz47781/mat120/notes/divide_poly/long_division/long_division.html
Here's a good primer on synthetic division : http://mesacc.edu/~scotz47781/mat120/notes/divide_poly/synthetic/synthetic_division.html