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(x^6 - 4x^5 - 7x^3) / (2x^3)

Please show your work. I really want to learn something.

2007-12-12 13:30:13 · 7 answers · asked by Anonymous in Science & Mathematics Mathematics

7 answers

_______|1/2x^3 - 2x^2 - 7/2
_______2x^3 | x^6 - 4x^5 - 7x^3
_______ x^6
----------------------------------------
___________- 4^5 - 7x^3
___________- 4^5
-----------------------------------------
________________- 7x^3
________________- 7x^3
----------------------------------------
___________________0

2007-12-12 13:46:12 · answer #1 · answered by @YD@y 4 · 1 0

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 · answer #2 · answered by engineer 2 · 0 0

(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 · answer #3 · answered by Robert L 7 · 0 0

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 · answer #4 · answered by Anonymous · 0 0

(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 · answer #5 · answered by Anonymous · 0 0

x^6 - 4x^5 - 7x^3
-----------------------
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 · answer #6 · answered by ¸.•*´`*♥To0DaMnFaBuL0uS♥*´`*•.¸ 4 · 0 2

Difficult to show it here but try:
http://www.purplemath.com/modules/polydiv2.htm

Good Luck

2007-12-12 13:37:16 · answer #7 · answered by Anonymous · 0 0

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