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X^4 - 3x^3 + 2x - 6 by x - 1
Desi
Raat Key Rani !
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You need to use the polynomial long division process... it's hard to type that here, probably you have the procedure explained in a textbook..
If not there are sites online that explain it, for example:
http://www.sosmath.com/algebra/factor/fac01/fac01.html
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I can make an exercise on it, if that helps.
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Ok let me check that site out
Desi
Raat Key Rani !
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that seems a bit hard...just a bit
Desi
Raat Key Rani !
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@ Devanté -- sure!
@ unique: Yeah it's not a great explanation -- I would really prefer to have any sort of long division explained on a blackboard or something, a forum post is inadequate.
This site has nice explanations of long division of integers and decimals, but I don't think it explains polynomial long division anywhere...
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ok let me look back at my book
Desi
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Here you go: http://www.mathsisfun.com/forum/viewtopic.php?pid=45574#p45574
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Thanks
DEV.
lemme check it out
Desi
Raat Key Rani !
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_______________
x-1)x^4-3x^3+2x-6
x go into x^4 is x^3
______x^3_____
x-1)x^4-3x^3+2x-6
multiply x^3 by x-1 to get x^4-x^3, then you subtract this from the x^4-3x^3
____ x^3________
x-1)x^4-3x^3+2x-6
- (x^4-x^3)
-2x^3
x go into -2x^3? Answer: -2x^2
____x^3-2x^2-2x__This the quotient
x-1)x^4-3x^3+2x-6
- (x^4-x^3) v
-2x^3+2x
-(-2x^3+2x^2)
-2x^2+2x
-(-2x^2+2x)
0 - 6 This is the remainder.
Desi
Raat Key Rani !
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That's what I got as well.
You can always check these like this, by making sure this equation holds:
quotient + (remainded/divisor) = (dividend/divisor)
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