Question

12*​x^​5-​x^​3+​6*​x/​(4*​x^​3

Answer 1

\[12​x^5−​x^3+​\frac{ 6x }{ 4x​^3} or \frac{ 12​x^5−​x^3 +6x }{ 4x​^3 }\]

Answer 2

and what do you want to know about it?

derivatives
zero points
graph?

Answer 3

12x^5-x^3+6x divided by 4x^3 is the problem and i have no i dea what the firsts step in solving it is

Answer 4

xo the second?
\[\frac{ 12​x^5−​x^3+6x }{ 4x​^3 }\]
what does it equal?
zero?

Answer 5

what do you want to solve it for, what does it equate?
this is just a function

Answer 6

yes, i want to know what it equals.

Answer 7

That's what you need to give,
Or is there some domain for x given?

Answer 8

that is some values of x?
Is that given?

Answer 9

what? no i just need to simplify the problem as much as possible.

Answer 10

@abster i apologize if i am confusing you, but the values of x are not given, you just have to simplify to get the answer

Answer 11

OK, just a sec

Answer 12

\[\frac{ 12​x^5-​x^33+​6​x }{ 4x^3 } =\frac{ 12​x^​4-​x^​2+​6x }{ 4x^2 }\]

Answer 13

is the first step (scratch one x

Answer 14

you wouldnt divide 12 and 4?

Answer 15

sorry I made a mistake
\[\frac{ 12​x^5-​x^33+​6​x }{ 4x^3 } =\frac{ 12​x^​4-​x^​2+​6 }{ 4x^2 }\\]]

Answer 16

\[\frac{ 12​x^5-​x^33+​6​x }{ 4x^3 } =\frac{ 12​x^​4-​x^​2+​6 }{ 4x^2 }\]

Answer 17

okay so how did you get x^2+6?

Answer 18

that stupid equation editer is not behaving very well
\[\frac{ 12​x^5-​x^3+​6​x }{ 4x^3 } =\frac{ 12​x^4-​x^2+​6 }{ 4x^2 } =3x^2 -\frac{ 1 }{4 } +\frac{ 1.5 }{x^2 }\]

Answer 19

It depends what you consider simple

Answer 20

Oh, and it is only if x is unequal to zero, (there the function does not exist

Answer 21

oh okay, i think i understand, thanks so much!

Answer 22

if you would want to know the zeropoints, you might be after something like
(ax² + b)(cx² + d)/4x²
=
(acx⁴ +(bc + ad)x² + bd)/4x²

had you have to find a, b, c, and d such that
ac = 12
bc + ad = -1
bd = 6

but I don't see an easy solution to that, so I don't think that is what they are after