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Step-by-step Solution

Sort $mn^2x^2-3mn^3x-2+\frac{1}{8}x^3$ ascending with respect to $x$

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Final Answer

$m\left(-3n^3x+n^2x^2\right)-2+\frac{1}{8}x^3$
Got a different answer? Try our new Answer Assistant!

Step-by-step Solution

Problem to solve:

$order\left(m n^2 x^2-3m n^3\cdot x-2+\frac{1}{8} x^3,x,2\right)$
1

Apply the formula: $\mathrm{order}\left(x,a,b\right)$, where $a=x$, $b=2$ and $x=mn^2x^2-3mn^3x-2+\frac{1}{8}x^3$

$-2-3mn^3x+mn^2x^2+\frac{1}{8}x^3$
2

Factoring by $m$

$m\left(-3n^3x+n^2x^2\right)-2+\frac{1}{8}x^3$

Final Answer

$m\left(-3n^3x+n^2x^2\right)-2+\frac{1}{8}x^3$
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Got another answer? Verify it!

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1
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5
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7
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9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

Tips on how to improve your answer:

$order\left(m n^2 x^2-3m n^3\cdot x-2+\frac{1}{8} x^3,x,2\right)$

Time to solve it:

~ 0.06 s