I know how to set it up but still having a hard time actually solving these types of problems.
RP......Do you want synthetic division or just regular polynomial division ???
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