Step-by-step Solution

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

Go!
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Final Answer

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

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

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

Time to solve it:

~ 0.03 s