_______|1/2x^3 - 2x^2 - 7/2
_______2x^3 | x^6 - 4x^5 - 7x^3
_______ x^6
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___________- 4^5 - 7x^3
___________- 4^5
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________________- 7x^3
________________- 7x^3
----------------------------------------
___________________0
2007-12-12 13:46:12
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answer #1
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answered by @YD@y 4
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All terms in the numerator can be placed over the denominator:
x^6 / 2x^3 - 4x^5/2x^3 - 7x^3/2x^3
When you divide like bases, you subtract the exponents:
1/2 x^(6-3) - 2 x^(5-3) - 7/2 x^(3-3)
1/2 x^3 - 2 x^3 - 7/2 x^0. Anything to 0 is 1:
1/2 x^3 - 2 x^3 - 7/2
2007-12-12 13:41:31
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answer #2
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answered by engineer 2
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(x^6 - 4x^5 - 7x^3) / (2x^3)
You must divide the denominator term into each term in the numerator.
(1/2)x³ - 2x² - 7/2
.
2007-12-12 13:37:15
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answer #3
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answered by Robert L 7
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To do this problem, just do it term by term. That is, break up the numerator and divide each one.
x^6/(2x^3) = (1/2)x^3
-4x^5/(2x^3) = -2 x^2
-7 x^3/(2x^3) = (-7/2)
Recombining we have (1/2) x^3 - 2 x^2 -7/2
2007-12-12 13:36:42
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answer #4
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answered by Anonymous
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(x^6 - 4x^5 - 7x^3)/(2x^3)
(x^6)/(2x^3) - (4x^5)/(2x^3) - (7x^3)/(2x^3)
x^3/2 - 2x^2 - 7/2
2007-12-12 13:42:22
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answer #5
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answered by Anonymous
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x^6 - 4x^5 - 7x^3
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2x^3
2x^3-2x^2-3x
just divide then keep the variable and then subtract exponents!!
good luck!!
2007-12-12 13:37:30
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answer #6
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answered by ¸.•*´`*♥To0DaMnFaBuL0uS♥*´`*•.¸ 4
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Difficult to show it here but try:
http://www.purplemath.com/modules/polydiv2.htm
Good Luck
2007-12-12 13:37:16
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answer #7
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answered by Anonymous
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