09-16-2002, 02:41 PM | #21 |
Drow Warrior
Join Date: December 11, 2001
Location: Canada, AB
Age: 38
Posts: 264
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So now x^3+5x^2+5x+1=(x+1)(x^2+4x+1)
And therefore (x^4+4x^3-4x-1)/(X-1)=the above. And what about those border values?
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09-16-2002, 05:44 PM | #22 |
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Never mind the none-PD stuff, I figured that out.
Now, I need a thing shortened: (X^3+3X-7)/(X^2-1) It's too long to divide without making an awful mess of left-over numbers. |
09-16-2002, 08:10 PM | #23 | ||
Lord Ao
Join Date: June 24, 2002
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Quote:
Remember .... Quote:
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09-16-2002, 09:13 PM | #24 |
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I know that I need filler. But I don't think that I can reduce it with the filler there....
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09-17-2002, 07:23 AM | #25 |
Jack Burton
Join Date: November 10, 2001
Location: Bathurst & Orange, in constant flux
Age: 37
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Actually, you don`t need the filler. It can help a little, but can also confuse matters.
To reduce a fraction, divide the top by the bottom; in this case x^3+3x+7=(x^2-1)(x-1)+(3x+6) |
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